Does the start equal the end?

Question

The Task

In this challenge, your task is to write a program or function which takes in a String and outputs a truthy or falsey value based on whether the first character and the last character of the input String are equal.

Input

You may take input in any way reasonable way. However, assuming that the input is present in a predefined variable is not allowed. Reading from a file, console, command line, input field etc., or taking input as a function argument is allowed.

Output

You may output in any reasonable format, except for assigning the result to a variable. Writing to a file, console, command line, modal box, function return statements etc. is allowed.

Additional Rules

• The input can be empty String as well, for which you should return a falsey value.

• Single-Char Input Strings should have a truthy result.

• Your program should be case-sensitive. helloH should output a falsey value.

• You can only have a single Truthy value and a single Falsey value. For example, outputting false for an Input String and 0 for another input String as Falsey values is not allowed.

• Standard loopholes are not allowed.

Test Cases

Input    ->    Output

"10h01"        Truthy
"Nothing"      Falsey
"Acccca"       Falsey
"wow!"         Falsey
"wow"          Truthy
"H"            Truthy
""             Falsey

This is code-golf, so the shortest code in bytes wins!

Answer 1

Jelly, 3 bytes

=ṚḢ

Try it online!

Answer 2

Python 3, 23 bytes

s=input()
s[0]!=s[-1]<e

Output is via exit code, so 0 (success) is truthy and 1 (failure) is falsy. If this is acceptable, a byte can be saved.

Try it online!

How it works

First of all, if s is an empty string, s[0] will raise an IndexError, causing the program to fail.

For non-empty s, if the first and last characters are equal, s[0]!=s[-1] will evaluate to False, so the program exits cleanly and immediately.

Finally, if the characters are different, s[0]!=s[-1] will evaluate to True, causing the compairson s[-1]<e to be performed. Since e is undefined, that raises a NameError.

If backwards compatibility with Python 2 is not desired,

s[0]!=s[-1]<3

works as well, since comparing a string with an integer raises a TypeError.

Answer 3

JavaScript, 19 bytes

a=>a.endsWith(a[0])

Answer 4

Retina, 13 12 bytes

^(.)(.*\1)?$

Try it online! Includes test suite. Edit: Saved 1 byte thanks to @Kobi.

Answer 5

Mathematica, 15 bytes

#&@@#===Last@#&

Takes an array of chars. Throws errors when the input is empty but can be ignored.

Answer 6

05AB1E, 4 bytes

S¬Q¤

Try it online! or Try All Tests

S    # Split the input into individual characters
 ¬   # Get the first character
  Q  # Check all characters for equality to the first
   ¤ # Get the last value i.e. head == tail

Answer 7

C++, 39 bytes

[](auto s){return s[0]&&s[0]==s.back();}

Full programs:

#include <string>
#include <iostream>

using namespace std;

int main()
{
    string t = "";
    auto f = [](auto s){return s[0]&&s[0]==s.back();};
    cout << f(t);
}

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Answer 8

Brachylog, 4 bytes

h~t?

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Explanation

h       The head of the Input...
 ~t?    ...is the tail of the Input

Answer 9

Python 2, 26 25 24 bytes

Thanks to @Dennis for saving a byte!

lambda x:""<x[:1]==x[-1]

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Answer 10

Java, 81 77 bytes

• -4 bytes, thanks @KevinCruijssen

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boolean f(String s){int l=s.length();return l>0&&s.charAt(l-1)==s.charAt(0);}

• Returns true if they're equal, otherwise false, false for empty string

Array Version, 60 bytes

boolean f(char[]s){int l=s.length;return l>0&&s[0]==s[l-1];}

Answer 11

MATL, 5 bytes

&=PO)

Try it at MATL Online!

Explanation

       % Implicitly grab input as a string (of length N)
&=     % Perform an element-wise equality check yielding an N x N matrix
P      % Flip this matrix up-down
O)     % Get the last value in the matrix (column-major ordering)
       % Implicitly display the result

In the case, that an empty input string must be handled, then something like the following (8 bytes) would work

&=POwhO)

This solution simply prepends a 0 to the front of the N x N matrix such that for an empty input, when the matrix is 0 x 0, there's still a 0 value that is then grabbed by 0)

Try it at MATL Online

Answer 12

brainfuck, 43 bytes

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

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Explanation

The main loop is [>[->+<<->],]. After each iteration, the cell to the right of the current position is the first byte of the string, and the cell to the left is the difference between the most recently handled character and the first. <[[-]-<] converts the final result to -1 if nonzero, and the rest converts -1 and 0 to 48 and 49 ("0" and "1") respectively.

Answer 13

Haskell, 21 bytes

c takes a String and returns a Bool.

c s=take 1s==[last s]

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• If not for empty strings, this could have been 16 bytes with c s=s!!0==last s.
• take 1s gives a list that is just the first element of s unless s is empty, in which case it's empty too.
• last s would error out on an empty string, but Haskell's laziness saves it: A string with a single element is always different from the empty string, without evaluating its element.

Answer 14

Japt, 6 bytes

tJ ¥Ug

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Answer 15

APL (Dyalog), 4 bytes

⊃⌽=⊃

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Explanation

  =                     Compare
   ⊃                    The first element of the right argument with
 ⌽                      The right argument reversed
                        This will return an array of the length of the reversed argument. Each element in the resulting array will be either 0 or 1 depending on whether the element at that position of the reversed argument equals the first element of the original right argument
                        So with argument 'abcda', we compare 'a' with each character in 'adcba' which results in the array 1 0 0 0 1
⊃                       From this result, pick the first element.

Here is the reason this works on empty strings. Applying ⊃ to an empty string returns a space . But reversing an empty string still returns an empty string, so comparing an empty string with a non-empty string (in this case ) gives an empty numerical vector. And applying ⊃ to an empty numerical vector returns 0. Hence passing an empty string returns 0.

Answer 16

Java, 52 43 bytes

s->!s.isEmpty()&&s.endsWith(""+s.charAt(0))

To make it work, feed this into a function such as the following that makes a lambda "go":

private static boolean f(Function<String, Boolean> func, String value) {
  return func.apply(value);
}

Answer 17

Ruby, 26 24 bytes

Saved two bytes thanks to @philomory!

->e{!!e[0]>0&&e[0]==e[-1]}

First post on codegolf -))

Answer 18

PHP>=7.1, 23 Bytes

prints 1 for equal and nothing if the character is different

<?=$argn[0]==$argn[-1];

Answer 19

C#, 38 30 bytes

s=>s!=""&&s[0]==s[s.Length-1];

Saved 8 bytes thanks to @raznagul.

Answer 20

Swift, 57 bytes

var s=readLine()!,a=Array(s.characters);a[0]==a.last ?1:0

Answer 21

R, 40 bytes

function(x)x>""&&rev(y<-charToRaw(x))==y

Thanks to Nitrodon for -2 bytes.

Thanks to MickyT for -8 bytes.

Test:

f=function(x)x>""&&rev(y<-charToRaw(x))==y
test <- c("10h01", "Nothing", "Acccca", "wow!", "wow", "H", "")
sapply(test, f)
all(sapply(test, f) == c(T, F, F, F, T, T, F))

Output:

> f=function(x)x>""&&rev(y<-charToRaw(x))==y
> test <- c("10h01", "Nothing", "Acccca", "wow!", "wow", "H", "")
> sapply(test, f)
  10h01 Nothing  Acccca    wow!     wow       H         
   TRUE   FALSE   FALSE   FALSE    TRUE    TRUE   FALSE 
> all(sapply(test, f) == c(T, F, F, F, T, T, F))
[1] TRUE

Answer 22

><>, 39 33 bytes

 2i&01. >~&-?v1v
  i:1+?!^01. >0>n;

This is my first time both using ><> and playing code golf, so helpful suggestions would be appreciated.

The code is in three basic sections.

2i&01. Pushes an arbitrary number (2 in this case, this causes an empty string to print 0) onto the stack and puts the input's first character in the register.

>i:1+?!^01. Main loop. Pushes the next character onto the stack. If the string has been read completely, then go to the last section

>~&-?v1v
     >0>n;  Compare the first and last characters. Print 1 if they're the same, 0 if not

Answer 23

Google Sheets, 33 Bytes

Takes input from cell [A1] and outputs 1 for truthy input and 0 for falsey input.

=(A1<>"")*Exact(Left(A1),Right(A1

It is noted that the parentheticals in Exact( and Right( are left unclosed as Google Sheets automatically corrects this as soon as the user has input the formula text and pressed enter to leave that cell.

Output

Answer 24

R, 50 43 41 40 64

Second solution with 41 bytes for a callable function - thanks to @niczky12 & @Giuseppe - amended for x=""

r=function(x,y=utf8ToInt(x))ifelse(x=="","FALSE",(y==rev(y))[1])

First with 50 bytes but not for the challenge

function(x){charToRaw(x)[1]==rev(charToRaw(x))[1]}

Answer 25

JavaScript, 20 bytes

Add f= at the beginning and invoke like f(arg).

_=>_[0]==_.slice(-1)

f=_=>_[0]==_.slice(-1)

i.oninput = e => o.innerHTML = f(i.value);

<input id=i><pre id=o></pre>

Explanation

This function takes in an argument _. In the function body, _[0]==_.slice(-1) checks whether the first element of _ (at 0th index) equals the last element of it, and returns the appropriate true or false boolean.

Answer 26

Fireball, 4 bytes

d1╡├

Explanation:

d      Duplicate implicit input
 1╡    Get the first character
   ├   Check whether the input ends with the first character

Alternative program:

d↔♥├

Answer 27

JavaScript (ES6), 25 bytes

s=>/^(.)(.*\1)?$/.test(s)

21 bytes if we can return true for the empty string.

s=>s[0]==[...s].pop()

---

Try it

f=
s=>/^(.)(.*\1)?$/.test(s)
o.innerText=f(i.value="10h01")
i.oninput=_=>o.innerText=f(i.value)

<input id=i><pre id=o>

Answer 28

Octave, 16 bytes

@(s)s(1)==s(end)

It takes a string s as input, and compares the first s(1) element with the last s(end).

This could be @(s)s(1)-s(end) if it was OK to swap true/false to false/true.

Answer 29

GNU grep, 12 bytes

^(.)(.*\1)?$

Run in extended or PCRE mode.

I don't know if this is considered cheating or not.

Answer 30

Common Lisp, 83 74 61 58 bytes

Original: 83 bytes

I've just started learning Common Lisp, so I feel like I'm bringing a putter to a driving range. There must be some kind of recursive macro wizardry or array manipulation possible here that I'm not seeing.

This is an anonymous function that accepts a string as its input:

(lambda (s) (let ((n (- (length s) 1))) (when (> n 0) (eq (char s 0) (char s n)))))

Prettified:

(lambda (s)
  (let ((n (- (length s) 1)))
    (when (> n 0)
      (eq (char s 0)
          (char s n)))))

Would love to see a slicker solution!

Revision 1: 74 bytes

Gotta love those standard library functions!

Ugly:

(lambda (s) (when (> (length s) 0) (eq (elt s 0) (elt (reverse s) 0))))

Pretty:

(lambda (s)
  (when (> (length s) 0)
    (eq (elt s 0)
        (elt (reverse s) 0))))

Revision 1.5: 61 bytes

Whitespace!

(lambda(s)(when(>(length s)0)(eq(elt s 0)(elt(reverse s)0))))

Revision 2: 58 bytes

Ugly:

(lambda(s)(and(>(length s)0)(not(mismatch s(reverse s)))))

Pretty:

(lambda (s)
  (and (> (length s) 0)
       (not (mismatch s (reverse s)))))

That's all for now! I think I'm smarter already.

Answer 31

S.I.L.O.S, 81 bytes

loadLine
a=256
c=get a
lblb
t=s
s=get a
a+1
if s b
t-c
if t d
i+1
lbld
printInt i

Try it online!

Answer 32

AWK, 29 34 bytes

This one might be cheating slightly, because it requires invoking AWK with the option:

`-F ''`

In GNU Awk you can use the long-form synonyms:

`--field-separator=''`

So I added 5 bytes to the total to account for this.

Ugly:

NR==1{a=$1}END{print(a==$NF)}

Pretty:

NR == 1
{
    a = $1
}

END
{
    print(a == $NF)
}

Answer 33

C, 50 40 36 bytes

Saved 10 bytes thanks to Dennis.

#define f(s)(*s&&*s==s[strlen(s)-1])

Equates to 0 if the first and last characters are different, or if the string is empty.

You could call f with something like:

int main(void)
{
    char s[100] = {0};
    gets(s);
    printf("%d\n",f(s));
}

Or, try it online!

Answer 34

shortC, 27 26 25 24 bytes

f(C*s){T*s&&*s==s[Ss)-1]

How it works:

f(C*s){                     declare int-returning function that takes a string
       T                    return
        *s                  provided string has any length
          &&                and
            *s==s[Ss)-1]    the first character equals the last one

Try it online!

Answer 35

Python, 51 bytes

s=input()
if s:print(s[0]==s[-1])
else:print(False)

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Answer 36

Bash, 37 bytes

[ -n "$1" ]&&[ ${1:0:1} == ${1: -1} ]

Takes command line input and returns with exit status 0 (truthy) or 1 (falsy).

Test with:

bash test.sh "helloH" ; echo $?

Answer 37

Pyth, 9 bytes

.xqhzez!1

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Could make it 5 bytes if I didn't have to deal with "", probably even less if I was good at Pyth.

Answer 38

Casio Basic, 46 bytes

StrLeft s,1,a
StrRight s,1,b
Print judge(a=b)

Strings aren't very nice to work with in this language. We need to take the first character from the left and right of the string, assign them to a and b, then print whether a is equal to b.

45 bytes for the code, 1 byte to enter s as parameter.

Answer 39

Python 3, 31 24 Bytes

lambda a:a[0]==a[-1]!=''

Old code:

def f(a):return a[0]==a[-1]!=''

This is pretty self explanatory; It takes a string a and checks if the first and last chars are equal, but I then had to add the !='' in order to satisfy the requirement "The input can be empty String as well, for which you should return a falsey value" because Python returns True for an empty string.

EDIT:

-7 Bytes thanks to @numbermaniac

Answer 40

Ruby, 16 17 bytes

Outputs true or false. Now with fixed syntax error.

p !! ~/./&~/#$&$/

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Answer 41

Python, 31 bytes

lambda x:bool(x)and x[0]==x[-1]

Answer 42

J, 8 bytes

1 byte thanks to Kritixi Lithos.

1{.{.=|.

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Answer 43

Python 2, 36 35 bytes

-1 byte thanks to Erik the Outgolfer

lambda s:s[0]==s[-1]if s else False

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Answer 44

Standard ML - 52 54 bytes

open String
fn s=>s<>""andalso sub(s,0)=sub(s,size s-1)

Answer 45

QBIC, 12 11 bytes

?_s;|=_s_fA

Explanation:

?              PRINT
     =         -1 if equal, 0 if not, between
 _s;|          QBIC's Substring function takes a variable amount of parameters.
                 With only a string as argument it takes the leftmost char of it.
                 The ; takes a string from the cmd line and names it A$
                 | closes the call to Substring
      _s_fA    _f flips string A, _s without arguments takes the left 1 char again.

---

Original 12 byter, which takes a substring of 1 from the left and 1 from the right:

?_s;|=_sA,-1

Explanation of the second substring:

      _sA,-1   Another call to Substring
                 A$ is implicitly defined by the ; in the other substring
                 -1 sets the starting index at the last position
                 No argument for length = 1 char by default.
               No closing |, auto-added at EOF

Answer 46

CJam, 6

q_W>#!

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Explanation:

q_     read and duplicate the input
W>     get the substring starting from the last position (W=-1)
        the result is empty if the input is empty
#      find the position of that substring within the initial string
        a bit surprising, position of empty string within itself is -1 (not found)
!      negate (0->1, non-zero->0)

Note: I consider the behavior of # with an empty substring to be a bug in CJam⩽0.6.5 and I will probably fix it in the next version. It's useful for this challenge though.

Answer 47

Ruby, 22 20 bytes

->x{x[0]===x[-1]||p}

Answer 48

Excel, 38 bytes

=1*IFERROR(CODE(A1)=CODE(RIGHT(A1)),0)

Surprisingly longer than I expected, in order to get the same truthy/falsy for all cases. Text conditionals ignore case by default in excel, so "A"="a" is TRUE. Empty cells yield an error for CODE(). Multiplied by 1 to force everything to a number, rather than having TRUE, FALSE, or 0 cases.

Answer 49

Batch, 52 bytes

@set s=!=
@set/ps=
@if "%s:~,1%"=="%s:~-1%" echo 1

Outputs 1 if equal, nothing if not. set/p doesn't change the variable if nothing is entered, so I can initialise it to a failure case, and != seemed appropriate.

Answer 50

Perl 6,  18  17 bytes

{?m/^(.)[.*$0]?$/}

Test it

{?/^(.)[.*$0]?$/}

Test it

Expanded:

{  # bare block lambda with implicit parameter ｢$_｣

  ?        # Boolify the following

  /        # match implicitly against ｢$_｣


    ^      # beginning of string

    (.)    # character ｢$0｣

    [

      .*   # followed by any number of characters
      $0   # and itself

    ]?     # optionally (Str with a single char)

    $  # end of string
  /
}

Answer 51

C, 34 bytes

f(char*s){s=*s&&*s==s[puts(s)-2];}

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C, 35 bytes

f(char*s){s=*s&&!strrchr(s,*s)[1];}

Try it online

Answer 52

Java 8, 29 bytes

s->s.matches("^(.)(.*\\1)?$")

Port of @Neil's Retina answer.

Try it here.

Answer 53

Brain-Flak, 51 bytes

Includes +1 for -a

{(<([({})<{({}<>)<>}<>>]{})((){[()](<{}>)}{})>)}{}

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Outputs 1 on top of the stack for truthy, and either nothing or 0 on top of the stack for falsy. These are consistent with Brain-Flak's "if" statement: {...}.

{(<                                          >)}{} # If there is input...
     ({})<            >                            #   Evaluate to the first char after...
          {({}<>)<>}<>                             #     reversing the entire stack
   ([                  ]{})((){[()](<{}>)}{})      # Check if the top 2 are equal
                                                   # I.e. first == last

Answer 54

Perl 5: 28 bytes

exit$ARGV[0]=~/^(.)(.*\1)?$/

Similar to perl 6 but it seems to be shorter as a program.

Answer 55

PowerShell, 42 bytes

$a=(read-host);$a-ne""-and$a[0]-ceq$a[-1]

Answer 56

Brain-Flak, 55 bytes

(({})<(())>)({}[<{({}<>)<>}><>{}]<>)((){[()](<{}>)}{})

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Answer 57

Actually, 10 bytes

;;lb)F@N=*

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Explanation:

;;lb)F@N=*
;;          copy input twice
  lb        length of input, cast to boolean
    )F      first character
      @N    last character
        =   compare equality
         *  multiply by boolean casted length (to make empty strings falsey)

Answer 58

Excel VBA, _{52 48 40 38 37 36} 34 Bytes

Anonymous VBE immediate window function that takes input of type variant and expected type variant\String from cell [A1] on the ActiveSheet object and outputs boolean response to the VBE immediate window.

?[A1]<>""=([Left(A1)]=[Right(A1)])

Answer 59

Perl 5, 22+1 (-p flag)=23 bytes

/^(.).*(.)/;$_=$1 eq$2

Outputs 1 for truthy and an empty string for falsey.

Answer 60

REXX 32 Bytes

a=arg(1)
say abbrev(a,right(a,1))

Tests if the last character is a valid abbreviation of the whole string.

Answer 61

Clojure: 21, 27 or 34 bytes

depending on the handling of "" test case

true for "aba" and "", false for "abc":

#(=(first %)(last %))

true for "aba", nil for "", false for "abc":

#(first(map =(reverse %)%))

true for "aba", nil for "" and "abc":

#(or(first(map =(reverse %)%))nil)

Answer 62

Chip, 91+3 = 94 bytes

*Z~.
,-{mA
>{xmB
|BA|
|CD|AvB
>{xmC+G
>-{mD+H
>{-mE+~t
>x{mF^S
|EF|
|HG|
>x{mG
>{-mH
Z--)~a

+3 for -z

Outputs the byte 0x0 for falsey, and 0x1 for truthy. This is a lot bigger than I was hoping, sadly.

Try it online! (Note about the TIO: it includes an extra line e*f to map the output to ASCII digits. TIO also includes the verbose flag -v, which gives extra debug output via stderr.)

The first line produces a signal on only the first byte, allowing us to store that byte's bits, and to detect the empty string. (If we could give a truthy value for the empty string, -3 bytes.)

The last line deals with output, producing truthy only if the first and most recent bytes match, and if it isn't the first byte. Output is given one byte after the end of the input, with the help of -z. If not, we would be unable to detect the end of the string. (If we swapped truthy and falsey, -2 bytes, or if combined with empty string savings above, -4 for the both.)

The blob to the right, surrounding the +'s, is what triggers the end of input behavior. This actually looks for a zero byte, meaning that incorrect results may occur if one is given as input.

The remainder of the elements perform the actual comparison. This comparison performed is equivalent to, in C-ish: !(input[0] xor input[n]). In Chip, however, this must be performed for each bit individually, hence the eight sets of memory cells m, xor-gates {, and so on.

There is an interesting caveat to this implementation, in that it can handle 8 bits, but is unaware of unicode. So, effectively, this compares the first and last bytes, rather than chars.

Answer 63

J, 12 bytes

'({.={:)*.*@#

{. means start, = means equals, and {: means end. *.*@# means "logical and with the length of the string", i.e., if the length is 0, it returns 0.

Answer 64

Axiom, 30 bytes

f(a)==(#a=0=>false;a.1=a.(#a))

this below seems not to be ok

f(a)==(#a=0=>false;a.#a=a.1)

Answer 65

TI-Basic (TI-84 Plus CE), 17 bytes

sub(Ans,1,1)=sub(Ans,length(Ans),1

Run with "string":prgmNAME. Returns 1 for true and 0 for false.

Answer 66

Powershell, 30 bytes

"$args"|%{!($_[0]-$_[-1]+!$_)}

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Answer 67

PowerShell, 25 Bytes

($a="$args")[0]-ceq$a[-1]

gets the first char and last char, performed a case-sensitive comparison of them.

Answer 68

q/kdb+, 21 20 18 bytes

Solution:

{#:[x]&(1#x)~-1#x}

Example:

q){#:[x]&(1#x)~-1#x}"abc"
0
q){#:[x]&(1#x)~-1#x}"abca"
1
q){#:[x]&(1#x)~-1#x}""
0

Explanation:

Take the first and last elements of the list, check for equality (return boolean 1 or 0), then check length of string, return the minimum of these two results.

{                } / anonymous lambda function
             -1#x  / take (#) 1 item from end of list
       (1#x)       / take (#) 1 item from start of list
            ~      / are they equal
 #:[x]             / count (#:) length of list x
      &            / minimum

Answer 69

Tcl, 34 bytes

proc C s {regexp ^(.)(.*\\1)?$ $s}

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Answer 70

Vim, 30 bytes

:s/\v^(.)(.*\1)?$/1
:s/^..\+/

Leaves you with 1, if the string starts and ends with the same character, or nothing, if it doesn't.

Answer 71

Julia 0.6, 29 bytes

f(x)=x==""?false:x[1]==x[end]

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Answer 72

Husk, 4 bytes

Γ·=→

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Takes a string as an argument to the program, returns 0 for falsy cases and 1 for truthy cases.

Explanation:

Γ is list deconstruction. The version I'm using (listN) takes a binary function and returns a function that takes a list. This new function returns the default type (0 in this case) for an empty string, and otherwise applies the given function to the head and tail of the string. The binary function I'm giving it is = (equality), composed on the second argument (·) with → (last element of a list).

Answer 73

SmileBASIC 3, 36 bytes

In SmileBASIC, 0 is false and 1 (or nonzero) is true.

LINPUT S$L=LEN(S$)?L&&S$[0]==S$[L-1]

Explainer:

 LINPUT S$                'read line from console to string S$
 L=LEN(S$)                'store length of line in L
 ? L && S$[0] == S$[L-1]
'| |          |
'| |          \-----------first char equals last char
'| \----------------------length is false if zero, && shortcuts
'\------------------------print

Answer 74

Brachylog, 4 bytes

h.&t

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and

Brachylog, 4 bytes

h~t?

Try it online!

are both predicates which try to unify the first and last elements of the input, outputting through success or failure (which prints true. or false. when it's run as a program).

Answer 75

PowerShell, 24 bytes

!("$args"[0,-1]|gu).Rank

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Answer 76

Golang, 27 bytes

if n[0:1] == n[(len(n))-1:]

Full Program

package main
import (
    "fmt"
)
func f(x string) string {
    if x[0:1] == x[(len(x))-1:] {
        return "Truthy"
    } else {
        return "Falsey"
    }
}
func main() {
    var n string
    fmt.Scanf("%q", &n)
    fmt.Printf("%q", f(n))
}

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Answer 77

Cubix, 20 bytes

@O0UA?n\.\1.1\2ntc?U

Returns 0 for false and 1 for true.

Cubified:

    @ O
    0 U
A ? n \ . \ 1 .
1 \ 2 n t c ? U
    . .
    . .

explanation:

A - read in input as chars; so input of abc has the stack -1,99,98,97
? - turn left if negative (empty string), right if positive
if negative:
   0 - push 0
   U - left u-turn
   O - output as int
   @ - end program
if positive:
\ - reflect so we are moving to the right
2 - push 2
n - negate top of stack 
t - pop top of stack, move that item from the bottom to the top; moves the last character to top of stack
c - xor
? - turn right if positive, go straight if negative
if 0:
   U - left U-turn
   1 - push 1
   \ - reflect so we point up
   O - output 1
   @ - end program
if positive
   n - negate
   \ - reflect
   n - negate
   ? - righthand turn
   0 - push 0
   U - lefthand U-turn
   O - output
   @ - end program

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Answer 78

Tcl, 59 bytes

proc C s {expr {[string in $s 0]==[string in $s e]&$s!=""}}

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Answer 79

C (gcc), 45 bytes

f(s,j)char*s;{for(j=*s;*s;s++);j=j&&j==*--s;}

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Answer 80

SmileBASIC, 29 bytes

DEF E S?S>""&&POP(S)==S[0]END

Pretty simple, but it's important that the POP is done after the first character is checked, so it will work on 1 character long strings.

Answer 81

ActionScript 2.0, 49 57bytes

function a(b){trace(b!=""?b.substr(0,1)==b.slice(-1):0);}

Old version (v) doesn't output false for "", but the new, longer version does.

function a(b){trace(b.substr(0,1)==b.slice(-1));}

I'm fairly sure this is the shortest way to do it, but...

The semicolon can be removed and it'll still compile in JPEXS.

This is a pretty simple function - if the argument is not "", output whether the (0, 1) substring of the argument (basically a way of getting the first character) is equal to the last character, obtained by slicing the string at its last character, otherwise 0. Tracing is shorter than returning.

Answer 82

><>, 11 bytes

l?!<{:}=n;!

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Takes input through the -s flag. You'll need to remove the flag for an empty input.

How it works

l?!<  Go right if there is input, left is there is none
    {:  If there is input, rotate the stack and dupe the first element
        This creates two copies of a one character input
      }=n;  Print if the last character is equal to the first
l  If there is no input, push the length of the stack (0)
        n;!  Wrap around and print the 0

Answer 83

jq, 23 characters

(20 characters code + 3 characters command line option)

.>""and.[:1]==.[-1:]

Sample run:

bash-4.4$ jq -R '.>""and.[:1]==.[-1:]' <<< $'10h01\nNothing\nAcccca\nwow!\nwow\nH\n'
true
false
false
false
true
true
false

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jq, 25 characters

(22 characters code + 3 characters command line option)

endswith(.[:1])and.>""

Just to show that jq also has endswith() function. But unfortunately all strings are considered to end with empty string, so is not shorter. ☹

Answer 84

GolfScript, 7 bytes

)\(@=\;

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Explanation

)       # The last character
 \(     # And the first character
   @=\; # Are they equal

Answer 85

Burlesque, 6 bytes

l_-]==

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l_ # Non-destructive tail, leaving the rest of the string on top of the stack
-] # Destructive head
== # Equal