What? No error?

Question

Your task is simple. Write a program that should obviously produce an error on first glance either when compiled or run, but either doesn't or produces some other unrelated error. This is a popularity contest, so be creative.

Answer 1

C++

Make sure you compile the following code in standard conforming mode (for example, for g++ use the -ansi flag):

int main()
{
  // why doesn't the following line give a type mismatch error??/
  return "success!";
}

How it works:

The ??/ is a trigraph sequence that is translated into a backslash which escapes the following newline, so the next line is still part of the comment and therefore won't generate a syntax error. Note that in C++, omitting the return in main is well defined and equivalent to returning 0, indicating a successful run.

Answer 2

Ruby

Always a fan of this one.

x = x

No NameError. x is now nil.

This is just a "feature" of Ruby :-)

Here's a more mundane one that's gotten me before:

x = 42

if x < 0
  raise Exception, "no negatives please"
elseif x == 42
  raise Exception, "ah! the meaning of life"
else  
  p 'nothing to see here...'
end 

Prints "nothing to see here."

It's elsif, not elseif. (and it's certainly not elif - woe to the wayward python programmer (me)!) So to the interpreter elseif looks like a normal method call, and since we don't enter the x<0 block, we go straight on to else and don't raise an exception. This bug is incredibly obvious in any syntax-highlighting environment, thankfully (?) code golf is not such an environment.

Answer 3

C?

Pretty normal code here...

void main() = main--;

It's Haskell, not C. It defines a function named "void" that takes two arguments. The first is named "main" and if the second (unnamed) is an empty tuple, it returns the "main" variable. "--" starts a comment in Haskell, so the ";" is commented out.

Answer 4

JavaScript

var а = 100;
if (typeof a !== 'undefined') throw 'This should always throw, right?';
console.log('How am I still alive?');

Here's how it works:

The first a is actually an а (that is, Cryllic Unicode "a").

Answer 5

JavaScript

When I was providing the following code I was told many times "It must be a typo! How can it work?".

console.log( 42..toString(2) );

The description below was copied exactly from one the recent cases.

As you probably know, in JavaScript everything except literals is an object. Numbers are objects as well. So theoretically (and practically) you may get properties or call methods of any non-literal via dot notation, as you do 'string'.length or [1,2,3].pop(). In case of numbers you may do the same but you should keep in mind that after a single dot the parser will look for a fractional part of the number expecting a float value (as in 123.45). If you use an integer you should "tell" the parser that a fractional part is empty, setting an extra dot before addressing a property: 123..method().

Answer 6

bash

#!/bin/bash

[ 1 < 2 ] && exit

for i in `seq 1 $[2 ** 64]`
    do "$0" | "$0"
done

while [[ false ]]
    do :
done

if maybe
    do [: [: [: [: [; [; [; [; ;] ;] ;] ;] :] :] :] :]
fi

Results

• You might expect the script not to produce any errors at all, since it exits after the first command. It doesn't.

• You might expect the typical error messages caused by an ongoing fork bomb due to the for loop. There's no fork bomb.

• You might expect bash to complain about the missing maybe command or the whole bunch of syntax error inside the if block. It won't.

• The only error message the script might produce ends in 2: No such file or directory.

Explanation

• [ isn't special to bash, so < 2 performs, as usual, redirection. Unless there is a file with name 2 in the current directory, this will cause an error.

• Due to that error above, the command before && will have a non-zero exit status and exit will not be executed.

• The for loop isn't infinite. In fact, there's no loop at all. Since bash cannot compute the 64th power of 2, the arithmetic expression's result is 0.

• [[ false ]] tests if false is a null string. It isn't, so this while loop is infinite.

• Because of the above, the if statement never gets executed, so no errors get detected.

Answer 7

Perl

use strict;
use warnings;
Syntax error!
exit 0;

Source and explanation: https://stackoverflow.com/questions/11695110

Answer 8

Java

class Gotcha {
    public static void main(String... args) {
        try  {
            main();
        } finally {
            main();
        }
    }
}

No stack overflows here; move along.

At first glance, this should produce a StackOverflowError, but it doesn't! It actually just runs forever (for all practical purposes at least; technically it would terminate after a time many orders of magnitude longer than the age of the universe). If you want to know how/why this works, see this. Also, if you happen to be wondering why we can call main() without arguments when the main method generally would need a String[] argument: it's because we've declared it to be variable-argument here, which is perfectly valid.

Answer 9

CoffeeScript

What? No error? Yep, this code does not have any bugs, why would it?

? followed by a space is operator that calls a function, but only if it exists. JavaScript doesn't have a function called What, therefore the function isn't called, and its arguments are simply ignored. The other words in the code are function calls that actually aren't called, because What function doesn't exist. At end, ? is existence operator, as it is not used in call function. Other sentence enders, such as . or ! would not work, as . is for methods, and ! is not operator (which cannot be used after an identifier). To read how CoffeeScript converted this to JavaScript, visit http://coffeescript.org/#try:What%3F%20No%20error%3F%20Yep%2C%20this%20code%20does%20not%20have%20any%20bugs%2C%20why%20it%20would%3F.

Answer 10

VBScript

The & operator in VBScript is string concatenation but what on earth are the && and &&& operators? (Recall that the "and" operator in VBScript is And, not &&.)

x = 10&987&&654&&&321

That program fragment is legal VBScript. Why? And what is the value of x?

The lexer breaks this down as x = 10 & 987 & &654& & &321. An integer literal which begins with & is, bizarrely enough, an octal literal. An octal literal which ends with & is, even more bizarrely, a long integer. So the value of x is the concatenation of the decimal values of those four integers: 10987428209.

Answer 11

Objective-C

Not a big deal, but it has surprised me while trying to put a link inside a comment:

http://www.google.com
        return 42;

http is a code label here, such labels are used in goto instructions

Answer 12

C#

class Foo
{
    static void Main(string[] args)
    {
        Bar();
    }

    static IEnumerable<object> Bar()
    {
        throw new Exception("I am invincible!");
        yield break;
    }
}

Because the Bar method does a yield, the method doesn't actually run when called, it returns an enumerator which, when iterated,s runs the method.

Answer 13

C

main=195;

Works on x86 platforms, where 195 is the opcode for ret. Does nothing,

Answer 14

Java

Probably too obvious.

public static void main(String[] varargs) throws Exception{
    char a, b = (char)Integer.parseInt("000d",16);
    // Chars have \u000d as value, so they're equal
    if(a == b){
        throw new Exception("This should be thrown");
    }
}

What?

Throws a syntax error after \u000d. \u000d is the unicode for a new line. Even though it is commented out, the Java compiler treats what is after this as code since it isn't commented out anymore.

Answer 15

C++

#include <iostream>

int succ(int x)
{
  return x + 1;
}

int succ(double x)
{
  return int(x + 1.0);
}

int succ(int *p)
{
  return *p + 1;
}

int main()
{
  std::cout << succ(NULL) << '\n';
}

Why?

NULL is an intergal constant, so it matches the int overload strictly better than the int* one. Still, most programmers have NULL associated with pointers, so a null pointer dereference can be expected.

Answer 16

Python

print """""quintuple-quoted strings!"""""

Perfectly valid, but the output is hard to guess. The first 3 " characters start a multiline string and the next two are part of the string. At the end, the first three "s terminate the string and the last two are an empty string literal that gets concatenated by the parser to the multiline string.

Answer 17

JavaScript

if (1/0 === -1/0) {
  throw "Surely there's an error in here somewhere...";
}

How it works:

There's positive and negative infinity in JS, and no error for dividing by zero.

Answer 18

C++

Mixing trigraphs and space-less lambdas can be quite confusing and definitely look erroneous to people who are not aware of trigraphs:

int main()
{
    return??-??(??)()??<return"??/x00FF";??>()??(0??);
}

How it works:

Some sequences consisting of 3 symbols, beginning with ??, are called trigraphs and will be substituted by a fully-compliant preprocessor. Preprocessed, the line in question looks as follows: return ~[] (){ return "\x00FF"; }()[0]; As one can see, this is nothing but a superfluous lambda function returning a string consisting of the 0xFFth character. The [0] just extracts that character and ~ NOTs it, so 0 is returned.

Answer 19

VBA/VB6

Private Sub DivByZero()

    Dim x() As String
    x = Split(vbNullString, ",")

    Debug.Print 1 / UBound(x)

End Sub

Splitting an empty comma delimited string should give an empty array. Should be an obvious division by zero error, right?

Nope. Surprisingly, when any zero length string is split the runtime gives you an array with a lower bound of 0 and an upper bound of -1. The code above will output -1.

Answer 20

Javascript

5..toString();
5 .toString();

Gives: 5

Whereas:

5.toString();

Gives SyntaxError

How it works:

JavaScript tries to parse dot on a number as a floating point literal

Answer 21

HTML

First post here, I'm not sure I get this or not, but here goes.

<html>
    <head></head>
    <body>
        <?php $_POST['non-existant'] = $idontexisteither ?>
    </body>
</html>

---

It's a .html file...

Answer 22

PHP (40 bytes)

<?for(;;$e.=$e++)foreach($e::$e()as&$e);

This was the answer I gave in this question: Insanity Check Program

The idea was to make a code that produced errors.

The 1st error that we will think of, is a syntax error.

There are no syntax errors...

Other would be that the class/function doesn't exist.

It doesn't run that far...

Other would be a time-out or a memory overflow, but, again, it doesn't reach that far...

Test the code here: http://writecodeonline.com/php/ (remove the <? on the beginning to test).

Answer 23

VBScript

Visual Basic 6 users will know that

If Blah Then Foo Bar

is legal, as is

If Blah Then 
    Foo Bar
End If

But what about

If Blah Then Foo Bar End If

? Turns out that is legal in VBScript but not in VB6. Why?

It's a bug in the parser; the intention was to reject this. The code which detects the End If was supposed to also check whether it was a multi-line If statement, and it did not. When I tried to fix it and sent out a beta with the fix, a certain influential industry news organization discovered that they had this line of code in one of their VBScript programs and said they would give the new version a low rating unless we un-fixed the bug, because they didn't want to change their source code.

Answer 24

C

This reminded me of an error I ran into when I learned C. Sadly the original variant doesn't seem to work with a current GCC, but this one still does:

#define ARR_SIZE 1234
int main() {
    int i = ARR_SIZE;
    int arr[ARR_SIZE];
    while(i >= 0) {
        (--i)[arr] = 0;
    }
    i = *(int*)0;
}

This obviously segfaults because we dereference a null pointer, right?

Wrong - actually, it's an infinite loop as our loop condition is off by one. Due to the prefix decrement, i runs from 1023 to -1. This means the assignment overwrites not only all elements in arr, but also the memory location directly before it - which happens to be the place where i is stored. On reaching -1, i overwrites itself with 0 and thus the loop condition is fulfilled again...

This was the original variant I which I can't reproduce anymore:

The same thing worked with i going upwards from 0 and being off by one. The latest GCC always stores i before arr in memory; this must have been different in older versions (maybe depending on declaration order). It was an actual error I produced in one of my first toy programs dealing with arrays.

Also, this one's obvious if you know how pointers work in C, but can be surprising if you don't:

You might think that the assignment to (--i)[arr] throws an error, but it's valid and equivalent to arr[--i]. An expression a[x] is just syntactic sugar for *(a + x) which computes and dereferences the pointer to the indexed element; the addition is of course commutative and thus equivalent to *(x + a).

Answer 25

Java

public class WhatTheHeckException extends RuntimeException {
    private static double d;        // Uninitialized variable
    public static void main(String... args) {
        if (d/d==d/d) throw new WhatTheHeckException();
        // Well that should always be true right? == is reflexive!

        System.out.println("Nothing to see here...");
    }
}

Why this works:

Unitialized fields have default values. In this case d is just 0. 0/0 = NaN in double division, and NaN never equals itself, so the if returns false. Note this would not work if you had 0/0==0/0, as at would be integer 0/0 division would WOULD throw an ArithmeticException.

Answer 26

VBScript

function[:(](["):"]):[:(]=["):"]:
end function
msgbox getref(":(")(":)")

'Output: :)

What it does:

Function, Sub and Variable Names in VBScript can be anything if you use square brackets. This script makes a function called :( and one argument "):" but because they do not follow normal naming convention they are surrounded by square brackets. The return value is set to the parameter value. An additional colon is used to get everything on one line. The Msgbox statement gets a reference to the function (but does not need the brackets) and calls it with a smiley :) as parameter.

Answer 27

C++11

struct comp {
    comp operator compl () { return comp { }; }
    operator comp () { return comp { }; }
    compl comp () { return; comp { }; }
};

int main() {
    comp com;
    compl com;
}

Compiles and runs without any warnings with g++ -pedantic-errors -std=c++11.

compl is a standard alternative spelling for ~, just like not is an alternative for !. compl is used here to first override operator~ and then define a destructor. Another trick is that operator comp is a conversion function from the type comp to itself. Surprisingly the standard does not forbid such a conversion function - but it does say that such a function is never used.

Answer 28

C#

Actually I caught myself on mistakenly doing just that :)

public static object Crash(int i)
{
    if (i > 0)
        return i + 1;
    else
        return new ArgumentOutOfRangeException("i");
}

public static void Main()
{
    Crash(-1);
}

throw, not return.

Answer 29

Java

enum derp
{

    public static void main(String[] a)
    {
        System.out.println(new org.yaml.snakeyaml.Yaml().dump(new java.awt.Point()));
    }
}

And how that one works:

Firs you think the Enum is not valid but its valid; then you think it will print a standard Point objects attributes but Gotcha! due to how Snakeyaml serializes you get a smooth StackOverFLow error

And another one:

enum derp
{

    ;public static void main(String[] a)
    {
        main(a);
    }
    static int x = 1;

    static
    {
        System.exit(x);
    }
}

you think a Stackoverflow will happen due to the obvious recursion but the program abuses the fact that when you run it the static{} block will be executed first and due to that it exits before the main() loads

enum derp
{

    ;
        public static void main(
            String[] a)
    {
        int aa=1;
        int ab=0x000d;
        //setting integer ab to \u000d  /*)
        ab=0;

        /*Error!*/
        aa/=ab;
    }
    static int x = 1;
}

this one relies on that /*Error*/-commented out code as closing point for the comment opened before the ab=0; the explain about the integer ab to 0x000d hides the newline to activate the commentout of the next line

Answer 30

C

Strings and arrays in c can be pretty confusing

main(){
  int i=0;
  char string[64]="Hello world;H%s";
  while(strlen(&i++[string])){
    i[' '+string]=string[i]-' ';
  }
  5[string]=44;
  return printf(string,'!'+string);
}

Answer 31

Java

  public static void main(String[] args) throws Exception {
        Integer a = 1, b = 1;
        Integer c = 200, d = 200;
        if ( a != b || c == d) {
            throw new Exception("This should be thrown");
        }
    }

The Integers from the range of -127 .. 127 are cached so Integer.valueOf(127) == Integer.valueOf(127) is true but Integer.valueOf(128) == Integer.valueOf(128) is false.

Answer 32

JavaScript

/21 + 21     // SyntaxError in this

SyntaxError could be expected, but this expression works and even returns true.

Answer 33

python

 = 3

You expect

SyntaxError: unexpected indent

Instead you get

SyntaxError: invalid character in identifier

Also this:

" " == " "
False

Unicode spaces

Answer 34

APL

table ← {
  ⍵⍴⍳×/⍵
  [{(]}) ⍝ obviously a syntax error right?
}

Result:

      table 6 8
 1  2  3  4  5  6  7  8
 9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32
33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48

This does fail:

table ← {
  [{(]})
  ⍵⍴⍳×/⍵
}

But this runs fine again:

table ← {
  0:[{(]})
  ⍵⍴⍳×/⍵
}

APL parsing is lazy. At least in Dyalog APL. (Really!) Also, a d-fn returns the value of the first line that gets a result. ⍵⍴⍳×/⍵ always returns a value (it's not an assignment and it doesn't have a guard), so, ⍵⍴⍳×/⍵ is run and returns a value, and the second line is never even parsed, thus, no error. In the second example, the line [{(]}) is run first, so it fails. But the third example works, because it has a guard (0:), and 0 is false so it's skipped.

Answer 35

Casio Calculator

It's quite hard to count bytes/tokens in this "language" - I've given the number of keypresses required, excluding Shift, Alpha (the second shift key) and = at the end - this certainly fits into 1 byte per keypress.

Tested on the fx-85GT PLUS model, which is a standard, non-graphing, "non-programmable" scientific calculator. Other models will work.

These calculators have two possible input/output formats: MathIO, which is default, and displays expressions as they are written on paper (e.g. fractions display properly), and LineIO, which uses the more conventional method of representing expressions as a single string of characters without vertical dimension - e.g. square root is √(...), fractions are 1⎦2 (that's meant to be like an underscore with a vertical bar of equal length attached to the right, like _| but as a single character).

Code

You'd think that this monstrosity would give many syntax errors. How to enter it:

((.!%PowerNegative-.DegreeRight (close-paren if in linear mode)%Degree

Press = to evaluate it.

What it looks like in linear mode:

((.!%^(--.°)%°

Decimal points without numbers, multiple percent signs, multiple degree signs, too many minuses, unclosed parentheses...

Output

0°0'36"

. evaluates to 0, just as .5 is 0.5 and 5. is 5.
Parentheses are automatically closed at the end of the expression.
((.!%^(--.°) on its own evaluates to 1 - the degree symbol disappears.
Although the calculator uses different symbols for negation and subtraction, the latter can be used for both, while the first only negates. --. is just 0.
Multiple percent symbols can be used in a single number - 1%% evaluates to 0.0001.
Without the final °, the expression evaluates to 0.01.

Answer 36

C

I don't belive that nobody posted this yet (sorry for archaeology):

const int main[] = {};

C use the same namespace for function names and variable names. So this will compile and link as valid executable but will return SIGSEGV at runtime.

Answer 37

Java (java.util.regex.Pattern)

This post's effectiveness depends on how much you know about Pattern class documentation. The behavior shown below also depends on the quirks in OpenJDK's Java Class Library (JCL) implementation of Pattern class.

1. Inline flags (valid ones as shown in documentation (?idmsuxU-idmsuxU))

   Pattern.compile("(?t)"); // Throws PatternSyntaxException. Nothing surprising here
   
   Pattern.compile("(?c)"); // Compiles normally (!)
   

   c flag was intended to be used as inline flag for CANON_EQ, but it currently has absolutely no effect, due to the way CANON_EQ is handled.

   There is absolutely no reason to use this in your code.

2. Character class intersection (shown in documentation as [\p{L}&&[^\p{Lu}]] or [a-z&&[def]], i.e. nesting seems to be required from the example)

   Pattern.compile("[\\p{IsAlphabetic}&&\\pM]"); // Compiled normally (!)
   

   And the compiled Pattern also works when matching against "\u0345".

   However, it is still recommended that you follow the documentation's way of writing regex.

3. Character class in \p notation (shown in documentation to specify a POSIX character class, a java.lang.Character class, or for Unicode script/block/category/binary property)

   Pattern.compile("\\p{Letter}"); // Throws PatternSyntaxException. Nothing surprising here
   
   Pattern.compile("\\p{all}"); // Both compiles normally (!)
   Pattern.compile("\\p{L1}");
   

   all and L1 are hidden names usable by \p:

   • \p{all} is equivalent to (?s:.).
   • \p{L1} is equivalent to [\x00-\xFF].

   However, since this is not specified in the documentation, I strongly advise you against using them even if you know you can use them.

Answer 38

C++

How many times have you been told to be careful to avoid index out of bounds errors with your for loops?

#include<iostream>
#include<string>
using namespace std;

int main() {
    string s="text here";
    for(int i=0;i<=s.size();i++)
        cout<<s[i];
    return 0;
}

In C++ (and most other languages), strings are just special char-arrays. In C++, a string always has a buffer of a space at the end, which isn't usually noticed because all the functions, etc. account for it. Here I am just accessing the space directly, so it just prints a space instead of crashing with and index out of bounds error.

Answer 39

PHP

Some slightly non-standard syntax here:

<html>
<body>
<p>Error?</p>
<?php 

echo beach, far away in time;

?>
<p>What error??</p>

What's going on?

There's a three-per-em space after the opening <?php tag. The PHP parser treats this as part of the tag name, and therefore fails to recognise this code block as PHP. Instead it passes it straight to your web browser. Your browser doesn't understand it either, so all you see is the HTML text that surrounds it. (An ordinary non-breaking space would have worked too, but Markdown converts those to regular spaces so we need to use something different.)

Answer 40

Python 2

isSorted = lambda a: all(x>y for x,y in zip(a[1:],a[:-1]))
print isSorted( [(1+3j),(1+2j,),(2+1j)] )

The surplus comma after 1+2j does not raise a syntax error, because this is Python’s way of creating a tuple with one element. But most importantly, this avoids that in isSorted complex numbers are compared and thus a TypeError is raised. This is because apart from two complex numbers you can almost compare everything in Python, including complex numbers and tuples.

Answer 41

Perl

if (false) {
    # Shouldn't happen
}
else {
    die "Huge error";
}

In Perl, false is not a keyword, and it's parsed a bareword ("false"). As "false" is true, the program exits without any exception (die keyword). Both strict and warnings would notice this problem, which shows why using use strict; use warnings; in Perl is a good idea.

Answer 42

Java

It is more convoluted than I would like, but here it goes.

package misc;

import java.io.FileWriter;

public class HelloWorld
{
    public static String writeFile(String fileName)
    {
        try {
            FileWriter fw = null;
            try {
                if( fileName==null ) return "bad parameter!";

                fw = new FileWriter(fileName);
                fw.write("Hello, World!\n");
            } finally {
                fw.close();
            }
        } catch( Exception ex ){
        }

        return "OK";
    }

    public static void main(String[] args)
    {
        String result = writeFile("hello.txt"); // returns OK
        System.out.println(result);

        result = writeFile(null); // should return "bad parameter!", right?
        System.out.println(result);
    }
}

The second call to writeFile should complain, right? It doesn't. Both calls return "OK".

If fileName is null, the very first thing the method does is to return an error message. So how can it return "OK"?

The return "bad parameter!"; is executed. But before returning, it has to execute the final section. At that time, fw is null. It throws a NullPointerException and, well, forgets it was supposed to return. The exception is caught and ignored. The procedure returns "OK".

Answer 43

XML

And you thought you knew XML...

Lo§”“@¥…™¢‰–•~}ñKð}@on%L”…¢¢‡…nÈ…““–k@æ–™“„ZLa”…¢¢‡…n

copy the above in hello.xml and open with IE.

It is encoded in EBCDIC. The XML processor is supposed to detect it. It does in IE. But when you view it, it is usually displayed in UTF or latin, so it is all garbled.

Answer 44

Ruby

1 / 0.0  

This does not result in a ZeroDivisionError, but in

# => Infinity

Answer 45

C#

using System;
class DivideByZero
{
   static int \u004F = Convert.ToInt16('\u0031');

   static void Main(string[] args)
   {
      Console.WriteLine(1 / O);  // Divide by Zero
   }
}

This code is actually dividing by O(the letter) and not 0(the number). If you know your hexadecimal ASCII characters well, then you might notice that I am setting the value of O(the letter) to '1' converted to an int16. Perhaps even weirder to some people, the output of this program is (the number): 0

Answer 46

REBOL

do not run "this"

---

How it works:

do: execute the following code.
not: negate.
run: runs the following thing as a bash command.
"this": an illegal bash command, does nothing.

Answer 47

Ruby

class Foo

  def self.get_bar
    self.bar
  end

  private

  def self.bar
    1 + 1
  end

end

p Foo.bar

Prints 2. Yeah, Ruby just doesn't apply the private modifier to class methods if you declare them this way. If you instead do this in a class << self block it works fine.

Answer 48

Clojure

  (defn check-nil [v] {:pre [nil? v]} v)

  (check-nil nil) 

  (defn check-keyword [v] {:pre [keyword? v]} v)

  (check-keyword “sdf”) 

First one throwing exception but second one does not throw any exception. First one v itself is null that way it throw exception. The form should be {:pre [(nil? v)]} and {:pre [(keyword? v)]}

Answer 49

Haskell

main=print (0/0, 1/0)

It is funny because Haskell is such a mathematical language, but this won't produce a zero-division error. Instead it outputs (NaN,Infinity). This is because division isn't defined on integer anyways, so it just assumes they are floats, which can be NaN and Infinity.

Answer 50

C

#include <stdio.h>
#include <string.h>

int main(void)
{
    if(1 != 0)
        perror("Error?");

    return 0;
}

perror() produces a message describing the last error encountered during a call to a system or library function - In the absence of an error it will print "Success". As such the output of the program is: "Error?: Success"

Answer 51

This isn't anything new, but still, the syntax of C/C++ in gcc has always amazed me ;-) Of course, it compiles without any errors by gcc -Wall -pedantic-errors.

#include <ctype.h>
#include <stdio.h>

int main(int argc, char* argv[]) {
    char p = '!';

    switch (argc) {
        while (!isdigit(p))
    case 1:
        p += 1;
        do { break;
    case 2:
        p = '1';
        continue;
    default:
        p = *1[argv];
        } while (false);
    }

    printf("%c\n", p);

    return 0;
}

a[b] is just *(a+b) and cases in switch are just labels. As a bonus, continue works like break here, because the loop checks the condition before jumping. Also, it's nice trick to spell default as defau1t while using a font where 1 and l are hard to distinguish.

:-)

Answer 52

#include <stdio.h>

int main(void) {
    if ("hello"=="hello") {
        printf("Same!\n");
    } else {
        printf("Different!\n");
    }
    return 0;
}

---

Hint:

Oh ho! This is classic CS 101 stuff. Strings should not be compared directly, because they're pointers!

---

Answer:

. . . except that in this case, since they're the same literal, the compiler has optimized the literals to point to the same location in memory. Hence the output: "Same!\n".

Answer 53

Python

Looks like this should produce an indentation error:

def a():
 print "start"
 if True:
                    for x in range(2):
                     print x 
 if False:
            print "do not print this"

 a()

but it doesn't. And it prints "start" and then 0 and 1.

In fact, Python interpreter doesn't care about the size of indentation, and it is relative to the current block.

Answer 54

Haskell

{-# LANGUAGE FlexibleInstances, FlexibleContexts #-}

import Control.Monad

main = 0 -- just bogus value so we can compile and see if the below definitions make sense...
         -- will be an error if you actually try to run this as a program.

data Complex a = Complex { realPart :: a, imagPart :: a }

instance Monad Complex where  -- to get convenience functions `liftM` etc..
  return x = Complex x x
  Complex r i >>= f = f r

instance (Monad complex, Num a) => Num (complex a) where
                            -- allow `Complex` to be used with numerical operators
  fromInteger = return . fromInteger
  (+) = liftM2(+)
  (*) = liftM2(*)
  abs = liftM abs
  signum = liftM signum

instance (Monad Complex, Fractional a) => Fractional (Complex a) where
  fromRational = return . fromRational
  (/) = liftM2(/)

Indeed, it's normally not possible to define an effectful action such as main by simply writing = number. If you try to run main = 0 alone, it'll sure enough give an error (indeed at compile time already).
However, I made a mistake further down in the code: the Num (complex a) instance should have Complex in uppercase like everywhere else. But it's lowercase, so the compiler will interpret this as a type variable and define a Num instance not only for Complex, but for all Monad types. This includes particularly the IO monad, so all of the sudden the definition of main on top makes sense: 0 is now an acceptable value, and will simply be wrapped up and returned.

Answer 55

JavaScript

var bSum = function (result, current, index, source) {
    return  result + (current << ( soucre.length - index - 1))
}
[1,0,1,0,1,0].reduce (bSum,0) //42

This obviously fails with a ReferenceError: soucre is not defined right?

Nope,
because of the function expression, there is no ASI, and the ArrayLiteral becomes a Property Accesor. Since the function has no property 0. The above fails even before, throwing a
TypeError: Cannot call method 'reduce' of undefined

Answer 56

Bit late on this one, but whatever.

PHP

<?='this ';</script>is some<%echo('code');

outputs "this is some code" (at least on my system)

PHP allows <script>...</script> and <%...%> alongside <?php...?>, <?...?>, and <?=...?> for enclosing tags for reasons nobody quite understands, allows mixing of these tags in a single script (in this example, <?=...</script> for reasons nobody quite understands, and doesn't actually require a closing tag for reasons nobody quite understands. As such, this is actually two separate code segments (<?='this ';</script> and <%echo('code');), with is some just being garbage.

Answer 57

C++

Two plus two is minus four.

#include <cassert>

int main() {
  struct {
    int two : 2;
  } _;
  _.two = 2;

  assert(_.two + _.two == -4);

  return 0;
}

Returns successfully; there is no error. http://ideone.com/eHQcJQ

This is actually quite a useful (but dangerous) feature - controlled overflow of a limited size bitfield struct member.

Answer 58

JavaScript

As you've seen, using 99999.toString() gives an error, where 99999..toString() works.

Now, have a look at this code:

0322323022302343534534535043.toString()

The number here is an octal (base-8) number. The octal notation does not have decimals in JavaScript.

Answer 59

Javascript

a = 1
a+++1

Javascript now has a triple plus operator

Answer 60

Go

package main insofaras
import "fmt" notwithstanding
func whereas main() {
    fmt despiteallobjections.Println(
        thetruthofthematter "Hello, world!")
}

http://golang.org/src/cmd/gc/lex.c defines some strange keywords like despiteallobjections, which are ignored by the language.

Answer 61

Python

I made this mistake when I started coding in python. So, I guess its worth telling

a=b=[1,2,3]
b.append(1)
print(a[3])

It seems like it will produce IndexError: list index out of range right?

But, It won't! actually a,b both points to the same list so whatever we do with b will change the list. So we will see 1 as output

Having said that, lets see another example:

a=b=[1,2,3]  
b=[1,2,3,1]  
print(a[3])

You might expect to see 1 as output. But, you won't! Here things are different. Now, after b=[1,2,3,1] , b points to a completely different list. So, the first list remain unchanged and thus we get the IndexError: list index out of range

Answer 62

JavaScript:

function isInputValid(value) {
    return +value === 0;
}

if (!isInputValid(' \n')) throw 'Invalid input';

What happens is that the + unary operator tries to convert the operand to a number - and for some bizarre reason, any string containing only white-space is converted to 0.

---

if ('Test' === new String('Test')) throw 'The strings are equal';

The === operator compares the references, and the primitive string 'Test' is not equal to an instance of a String object; therefore, it doesn't throw any error.

Answer 63

JavaScript

• Code 1:

  (function() {
      return
      {
          error: undeclaredVariable
      };
  })();
  

• Code 2:

  (function() {
      return
      {
          error: undeclaredVariable,
          foo: 'bar'
      };
  })();
  

It seems both should throw ReferenceError: undeclaredVariable is not defined, but:

• The first one doesn't throw any error
• The second one throws SyntaxError: missing ; before statement

That's because JavaScript doesn't require ; at the end of lines, so a return followed by line break exits function without returning following object.

Answer 64

Python:

@type
@type
def f(x):
    return 0/0
f(0)(0)

f(anything)(object) returns type(object).

People tend to forget that decoraters don't always do what they seem to do ;)

Answer 65

Java

No error for any input, and the curly brackets don't even match

public class ShouldFail {

    public static void main(String[] args) {
        String secret = "v#19!e/\u0022;}/*sd@x";
        if (!secret.equals(args[0]))
            throw new RuntimeException("Invalid argument");
        }
        System.out.println("***/// argument is valid ///***"); 

}

\u0022 ends the string, the rest of the program is enclosed in /* comments */

Answer 66

HTML/PHP

 <?php

   echo '<pre>';
   print_r($_POST);
   echo '</pre>';

    ?>
        <form action="dummy.php" method="post">
        <dl> 
            <dd>select months</dd>
            <dt><select id="month" name="month" size="6" multiple="multiple">
                <option>January</option>
                <option>February</option>
                <option>March</option>
                <option>April</option>
                <option>May</option>
                <option>June</option>
                <option>July</option>
                <option>August</option>
                <option>September</option>
                <option>October</option>
                <option>November</option>
                <option>December</option>
            </select>
            </dt>                

            <dd></dd>
            <dt><input name="submit" id="submit" type="submit" value="submit"/>
        </dl>
        </form> 

You can select multiple months by holding down the shift key or ctrl key. You'd expect all selected options to arrive in $_POST after hitting the submit button.
Well...
Only one, no matter how many you pick.
NO error message, neither PHP nor HTML.

select id="month" name="month" size="6" multiple="multiple"
is the problem – PHP assigns them all into the same variable, so only the last of them survives.

name="month[]" instead of name="month" does the trick (now they become all entries in an array).

Answer 67

Emacs Lisp

It looks like there should be a divide by 0 error but there is not! (evals to 0)

(when (= 0 (- ? ? ))
  (print "(- ? ? ) does equal zero!")
  ;; look! no divide by 0 error!
  (/ 1 (- ? ? ))) 

chars in emacs lisp are written: '?{character}' so ?a in emacs lisp == 'a' in C. The bad part about this is that a ? followed by a space character is a valid way of writing ' '. This works with anywhite space character though. In the first (- ? ? ) I am doing space minus space which is 32 - 32 == 0. In the second (- ? ? ), the second whitespace character is actually the unicode char #x2001. so it is not 0

Answer 68

    public static void main(String[] args) {
    try{
        int a[] = new int[2];
        System.out.println("Accessing out of bounds :" + a[3]);
    }catch(NullPointerException e){
        System.out.println("But we dont catch out of bounds exceptions  :" + e);
    } finally {
        return;
    }
}

finally is always guaranteed to run

Answer 69

C

Here's another one:

#include <stdio.h>

int main()
{
  int a[2] = { 2, 6 };
  int typo = 2;

  /* Calculate a[1]/a[0] + typo
     to save a character (code golf!), write *a instead of a[0] */
  int r = a[1]/*a + tipo;
  /* the above should trigger an error because I wrote tipo instead
     of typo; why does it compile correctly? */

  -1; /* statement with no effect, but no warning about this? */

  printf("%d\n", r); /* and this even prints the correct value! */
  return 0;
}

Explanation:

The / from the intended division and the * from the intended pointer dereference together form the comment starter /*. Note that inside the comment, further /* are not parsed, so the comment continues until the end of the intended statement. Of course, due to thwe unintentionally long comment, the -0; is no longer a separate statement, but gets part of the previous definition, to form the complete definition int r = a[1] - 1. Due to the carefully chosen constants, this gives the same result as a[1]/a[0] + typo. However, when compiling using gcc with warnings on, you do get a warning of the nested /* (and another one about the unused typo variable).

Answer 70

Brainfuck

Of course, this varies by interpreter, but it fails to run on mine:

Author: Darkgamma (contact: darkgamma@email(dot)com)    

>++++++++[<++++++++>-]<++++++++.>+++++[<+++++>-]
<++++.>+++[<+++>-]<--..+++.>+++++++++[<---------
>-]<++.>+++++++[<+++++++>-]<++++++.>+++++[<+++++
>-]<-.+++.>++[<-->-]<--.>+++[<--->-]<+.>++++++++
[<-------->-]<---.

In Brainfuck, all text (save for the eight commands) is taken as a comment. So what's at play here? The trick's in the fact that some interpreters take "@" as end-of-source and stop interpreting after that point. Doesn't work on all interpreters but stumped me on mine until I figured out what was going on.

Answer 71

static void Main(string[] args)
    {
        object foo = 10;
        object bar = 10;

        if (foo == bar)
            throw new Exception("They're equal!");

        Console.WriteLine("Why am I here? Obviously {0} == {1}, right?", foo, bar);
        Console.ReadLine();
    }

Output:

Why am I here? Obviously 10 == 10, right?

The == operator compares the references of the two objects, not the unboxed value. The code above would throw an exception if it was "if (foo.Equals(bar))" instead

Answer 72

T-SQL

select CustomerID 
from NorthWind.dbo.Customers 
where not exists (select DivideByZero=1/0
                    ,BadConversion=convert(int,'xxx')
                    ,InvalidParameter=left('abc',-ShipVia)
                    ,DateOverflow=dateadd(year,9999,getdate())
                    ,ArithmeticOverflow=cast(1e308 as tinyint)
                    ,InvalidCursorRef=cursor_status('junk','junk')
                    ,BadSubquery=(select top (-OrderID) EmployeeID 
                                  from NorthWind.dbo.Employees)
              from NorthWind.dbo.Orders 
              where CustomerID=Customers.CustomerID)

This should have a lot of errors (named in the query), but because it's a subquery in an exists clause, it runs without error. Taken from http://bradsruminations.blogspot.com/2009/09/age-old-select-vs-select-1-debate.html.

Answer 73

JavaScript

Disturbing string to boolean conversion

booleanValue = Boolean("false");
if(!booleanValue){
    throw "I threw an error!"
}

Answer 74

Whoops! I'm such a n00b, I thought I could just paste a url right into the source code ..

int main()
{
  http://codegolf.stackexchange.com/
  printf("My first C program\n");
  return 0;
}

Answer 75

Java

public static void main(String[] args) {
     http://codegolf.stackexchange.com/questions/23250/what-no-error
}

http: is a label, the rest of the line is commented out

Answer 76

Python

#i don't think theres a module called antigravity..../
import antigravity

There's an easter egg in python.. im not going to tell u what it does

Answer 77

C

#include <stdio.h>

int g(int a) {
    return a + 1;
}

int f(int a) {
    g(a);
    return;
}

int main() {
    printf("%d\n", f(13));
}

f has to return an int, but returns void. This actually is an error and undefined behaviour, but with GCC on linux, this will work, since the return value of g will still be in place when control returns from f. This may - however - not be the case on any system.

Answer 78

In AppleScript

item -1 of {1,2}

  →  Expected error : AppleScript error : Impossible to get item -1 of {1, 2}.

  →  Actual result : 2

The negative number -1 has a special handling. This trick works even further :

item -2 of {1,2}

  →  Actual result : 1

But :

item 0 of {1,2}

  →  Actual result : AppleScript error : Impossible to get item 0 of {1, 2}.

Answer 79

AWK

This program works like cat utility in UNIX.

AWK! Run a program that works exactly like cat

Running (in shell):

awk 'AWK! Run a program that works exactly like cat' list of files (or nothing for STDIN)

Nothing in AWK is a concatentation operator, accessing unknown variables returns empty string, ! operator when seeing nothing returns 1. Contatenation of "AWK" variable and boolean inverse of concatenation of lots of strings gives 1 (think, $AWK + !($Run + $a + $program + ...)). AWK is line based, so 1 is taken as a condition in condition { code } block. Code block is optional, and when not specified, it prints the input line.

Answer 80

C or C++

Based on celtschk's answer:-

int main (int argc, char *argv [])
{
  return argc ["success!"];
}

(yes, I have done that in production code, mainly to ensure people had read the code)