Reverse Polish notation

Question

You must evaluate a string written in Reverse Polish notation and output the result.

The program must accept an input and return the output. For programming languages that do not have functions to receive input/output, you can assume functions like readLine/print.

You are not allowed to use any kind of "eval" in the program.

Numbers and operators are separated by one or more spaces.

You must support at least the +, -, * and / operators.

You need to add support to negative numbers (for example, -4 is not the same thing as 0 4 -) and floating point numbers.

You can assume the input is valid and follows the rules above

Test Cases

Input:

-4 5 +

Output:

1

---

Input:

5 2 /

Output:

2.5

---

Input:

5 2.5 /

Output:

2

---

Input:

5 1 2 + 4 * 3 - +

Output:

14

---

Input:

4 2 5 * + 1 3 2 * + /

Output:

2

Answer 1

Ruby - 95 77 characters

a=[]
gets.split.each{|b|a<<(b=~/\d/?b.to_f: (j,k=a.pop 2;j.send b,k))}
p a[0]

Takes input on stdin.

Testing code

[
  "-4 5 +",
  "5 2 /",
  "5 2.5 /",
  "5 1 2 + 4 * 3 - +",
  "4 2 5 * + 1 3 2 * + /",
  "12 8 3 * 6 / - 2 + -20.5 "
].each do |test|
  puts "[#{test}] gives #{`echo '#{test}' | ruby golf-polish.rb`}"
end

gives

[-4 5 +] gives 1.0
[5 2 /] gives 2.5
[5 2.5 /] gives 2.0
[5 1 2 + 4 * 3 - +] gives 14.0
[4 2 5 * + 1 3 2 * + /] gives 2.0
[12 8 3 * 6 / - 2 + -20.5 ] gives 10.0

Unlike the C version this returns the last valid result if there are extra numbers appended to the input it seems.

Answer 2

Scheme, 162 chars

(Line breaks added for clarity—all are optional.)

(let l((s'()))(let((t(read)))(cond((number? t)(l`(,t,@s)))((assq t
`((+,+)(-,-)(*,*)(/,/)))=>(lambda(a)(l`(,((cadr a)(cadr s)(car s))
,@(cddr s)))))(else(car s)))))

Fully-formatted (ungolfed) version:

(let loop ((stack '()))
  (let ((token (read)))
    (cond ((number? token) (loop `(,token ,@stack)))
          ((assq token `((+ ,+) (- ,-) (* ,*) (/ ,/)))
           => (lambda (ass) (loop `(,((cadr ass) (cadr stack) (car stack))
                                    ,@(cddr stack)))))
          (else (car stack)))))

---

Selected commentary

`(,foo ,@bar) is the same as (cons foo bar) (i.e., it (effectively^†) returns a new list with foo prepended to bar), except it's one character shorter if you compress all the spaces out.

Thus, you can read the iteration clauses as (loop (cons token stack)) and (loop (cons ((cadr ass) (cadr stack) (car stack)) (cddr stack))) if that's easier on your eyes.

`((+ ,+) (- ,-) (* ,*) (/ ,/)) creates an association list with the symbol + paired with the procedure +, and likewise with the other operators. Thus it's a simple symbol lookup table (bare words are (read) in as symbols, which is why no further processing on token is necessary). Association lists have O(n) lookup, and thus are only suitable for short lists, as is the case here. :-P

† This is not technically accurate, but, for non-Lisp programmers, it gets a right-enough idea across.

Answer 3

Python - 124 chars

s=[1,1]
for i in raw_input().split():b,a=map(float,s[:2]);s[:2]=[[a+b],[a-b],[a*b],[a/b],[i,b,a]]["+-*/".find(i)]
print s[0]

Python - 133 chars

s=[1,1]
for i in raw_input().split():b,a=map(float,s[:2]);s={'+':[a+b],'-':[a-b],'*':[a*b],'/':[a/b]}.get(i,[i,b,a])+s[2:]
print s[0]

Answer 4

c -- 424 necessary character

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define O(X) g=o();g=o() X g;u(g);break;
char*p=NULL,*b;size_t a,n=0;float g,s[99];float o(){return s[--n];};
void u(float v){s[n++]=v;};int main(){getdelim(&p,&a,EOF,stdin);for(;;){
b=strsep(&p," \n\t");if(3>p-b){if(*b>='0'&&*b<='9')goto n;switch(*b){case 0:
case EOF:printf("%f\n",o());return 0;case'+':O(+)case'-':O(-)case'*':O(*)
case'/':O(/)}}else n:u(atof(b));}}

Assumes that you have a new enough libc to include getdelim in stdio.h. The approach is straight ahead, the whole input is read into a buffer, then we tokenize with strsep and use length and initial character to determine the class of each. There is no protection against bad input. Feed it "+ - * / + - ...", and it will happily pop stuff off the memory "below" the stack until it seg faults. All non-operators are interpreted as floats by atof which means zero value if they don't look like numbers.

Readable and commented:

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

char *p=NULL,*b;
size_t a,n=0;
float g,s[99];
float o(){        /* pOp */
  //printf("\tpoping '%f'\n",s[n-1]);
  return s[--n];
};
void u(float v){  /* pUsh */
  //printf("\tpushing '%f'\n",v);
  s[n++]=v;
};
int main(){
  getdelim(&p,&a,EOF,stdin); /* get all the input */
  for(;;){
    b=strsep(&p," \n\t"); /* now *b though *(p-1) is a token and p
                 points at the rest of the input */
    if(3>p-b){
      if (*b>='0'&&*b<='9') goto n;
      //printf("Got 1 char token '%c'\n",*b);
      switch (*b) {
      case 0:
      case EOF: printf("%f\n",o()); return 0;
      case '+': g=o(); g=o()+g; u(g); break;
      case '-': g=o(); g=o()-g; u(g); break;
      case '*': g=o(); g=o()*g; u(g); break;
      case '/': g=o(); g=o()/g; u(g); break;
    /* all other cases viciously ignored */
      } 
    } else { n:
      //printf("Got token '%s' (%f)\n",b,atof(b));
      u(atof(b));
    }
  }
}

Validation:

 $ gcc -c99 rpn_golf.c 
 $ wc rpn_golf.c
  9  34 433 rpn_golf.c
 $ echo -4 5 + | ./a.out
1.000000
 $ echo 5 2 / | ./a.out
2.500000
 $ echo 5 2.5 / | ./a.out
2.000000

Heh! Gotta quote anything with * in it...

 $ echo "5 1 2 + 4 * 3 - +" | ./a.out
14.000000
 $ echo "4 2 5 * + 1 3 2 * + /" | ./a.out
2.000000

and my own test case

 $ echo "12 8 3 * 6 / - 2 + -20.5 " | ./a.out
-20.500000

Answer 5

Haskell (155)

f#(a:b:c)=b`f`a:c
(s:_)![]=print s
s!("+":v)=(+)#s!v
s!("-":v)=(-)#s!v
s!("*":v)=(*)#s!v
s!("/":v)=(/)#s!v
s!(n:v)=(read n:s)!v
main=getLine>>=([]!).words

Answer 6

MATLAB – 158, 147

C=strsplit(input('','s'));D=str2double(C);q=[];for i=1:numel(D),if isnan(D(i)),f=str2func(C{i});q=[f(q(2),q(1)) q(3:end)];else q=[D(i) q];end,end,q

(input is read from user input, output printed out).

---

Below is the code prettified and commented, it pretty much implements the postfix algorithm described (with the assumption that expressions are valid):

C = strsplit(input('','s'));         % prompt user for input and split string by spaces
D = str2double(C);                   % convert to numbers, non-numeric are set to NaN
q = [];                              % initialize stack (array)
for i=1:numel(D)                     % for each value
    if isnan(D(i))                   % if it is an operator
        f = str2func(C{i});          % convert op to a function
        q = [f(q(2),q(1)) q(3:end)]; % pop top two values, apply op and push result
    else
        q = [D(i) q];                % else push value on stack
    end
end
q                                    % show result

---

Bonus:

In the code above, we assume operators are always binary (+, -, *, /). We can generalize it by using nargin(f) to determine the number of arguments the operand/function requires, and pop the right amount of values from the stack accordingly, as in:

f = str2func(C{i});
n = nargin(f);
args = num2cell(q(n:-1:1));
q = [f(args{:}) q(n+1:end)];

That way we can evaluate expressions like:

str = '6 5 1 2 mean_of_three 1 + 4 * +'

where mean_of_three is a user-defined function with three inputs:

function d = mean_of_three(a,b,c)
    d = (a+b+c)/3;
end

Answer 7

Perl (134)

@a=split/\s/,<>;/\d/?push@s,$_:($x=pop@s,$y=pop@s,push@s,('+'eq$_?$x+$y:'-'eq$_?$y-$x:'*'eq$_?$x*$y:'/'eq$_?$y/$x:0))for@a;print pop@s

Next time, I'm going to use the recursive regexp thing.

Ungolfed:

@a = split /\s/, <>;
for (@a) {
    /\d/
  ? (push @s, $_)
  : ( $x = pop @s,
      $y = pop @s,
      push @s , ( '+' eq $_ ? $x + $y
                : '-' eq $_ ? $y - $x
                : '*' eq $_ ? $x * $y
                : '/' eq $_ ? $y / $x
                : 0 )
      )
}
print(pop @s);

I though F# is my only dream programming language...

Answer 8

Windows PowerShell, 152 181 192

In readable form, because by now it's only two lines with no chance of breaking them up:

$s=@()
switch -r(-split$input){
  '\+'        {$s[1]+=$s[0]}
  '-'         {$s[1]-=$s[0]}
  '\*'        {$s[1]*=$s[0]}
  '/'         {$s[1]/=$s[0]}
  '-?[\d.]+'  {$s=0,+$_+$s}
  '.'         {$s=$s[1..($s.count)]}}
$s

2010-01-30 11:07 (192) – First attempt.

2010-01-30 11:09 (170) – Turning the function into a scriptblock solves the scope issues. Just makes each invocation two bytes longer.

2010-01-30 11:19 (188) – Didn't solve the scope issue, the test case just masked it. Removed the index from the final output and removed a superfluous line break, though. And changed double to float.

2010-01-30 11:19 (181) – Can't even remember my own advice. Casting to a numeric type can be done in a single char.

2010-01-30 11:39 (152) – Greatly reduced by using regex matching in the switch. Completely solves the previous scope issues with accessing the stack to pop it.

Answer 9

Racket 131:

(let l((s 0))(define t(read))(cond[(real? t)
(l`(,t,@s))][(memq t'(+ - * /))(l`(,((eval t)(cadr s)
(car s)),@(cddr s)))][0(car s)]))

Line breaks optional.

Based on Chris Jester-Young's solution for Scheme.

Answer 10

Python 3, 119 bytes

s=[]
for x in input().split():
 try:s+=float(x),
 except:o='-*+'.find(x);*s,a,b=s;s+=(a+b*~-o,a*b**o)[o%2],
print(s[0])

Input: 5 1 1 - -7 0 * + - 2 /

Output: 2.5

(You can find a 128-character Python 2 version in the edit history.)

Answer 11

Python, 166 characters

import os,operator as o
S=[]
for i in os.read(0,99).split():
 try:S=[float(i)]+S
 except:S=[{'+':o.add,'-':o.sub,'/':o.div,'*':o.mul}[i](S[1],S[0])]+S[2:]
print S[0]

Answer 12

JavaScript (157)

This code assumes there are these two functions: readLine and print

a=readLine().split(/ +/g);s=[];for(i in a){v=a[i];if(isNaN(+v)){f=s.pop();p=s.pop();s.push([p+f,p-f,p*f,p/f]['+-*/'.indexOf(v)])}else{s.push(+v)}}print(s[0])

Answer 13

Perl, 128

This isn't really competitive next to the other Perl answer, but explores a different (suboptimal) path.

perl -plE '@_=split" ";$_=$_[$i],/\d||
do{($a,$b)=splice@_,$i-=2,2;$_[$i--]=
"+"eq$_?$a+$b:"-"eq$_?$a-$b:"*"eq$_?
$a*$b:$a/$b;}while++$i<@_'

Characters counted as diff to a simple perl -e '' invocation.

Answer 14

flex - 157

%{
float b[100],*s=b;
#define O(o) s--;*(s-1)=*(s-1)o*s;
%}
%%
-?[0-9.]+ *s++=strtof(yytext,0);
\+ O(+)
- O(-)
\* O(*)
\/ O(/)
\n printf("%g\n",*--s);
.
%%

If you aren't familiar, compile with flex rpn.l && gcc -lfl lex.yy.c

Answer 15

Python, 130 characters

Would be 124 characters if we dropped b and (which some of the Python answers are missing). And it incorporates 42!

s=[]
for x in raw_input().split():
 try:s=[float(x)]+s
 except:b,a=s[:2];s[:2]=[[a*b,a+b,0,a-b,0,b and a/b][ord(x)-42]]
print s[0]

Answer 16

PHP, 439 265 263 262 244 240 characters

<? $c=fgets(STDIN);$a=array_values(array_filter(explode(" ",$c)));$s[]=0;foreach($a as$b){if(floatval($b)){$s[]=$b;continue;}$d=array_pop($s);$e=array_pop($s);$s[]=$b=="+"?$e+$d:($b=="-"?$e-$d:($b=="*"?$e*$d:($b=="/"?$e/$d:"")));}echo$s[1];

This code should work with stdin, though it is not tested with stdin.

It has been tested on all of the cases, the output (and code) for the last one is here:
http://codepad.viper-7.com/fGbnv6

Ungolfed, 314 330 326 characters

<?php
$c = fgets(STDIN);
$a = array_values(array_filter(explode(" ", $c)));
$s[] = 0;
foreach($a as $b){
    if(floatval($b)){
        $s[] = $b;
        continue;
    }
    $d = array_pop($s);
    $e = array_pop($s);
    $s[] = $b == "+" ? $e + $d : ($b == "-" ? $e - $d : ($b == "*" ? $e * $d : ($b == "/" ? $e / $d :"")));
}
echo $s[1];

Answer 17

Python, 161 characters:

from operator import*;s=[];i=raw_input().split(' ')
q="*+-/";o=[mul,add,0,sub,0,div]
for c in i:
 if c in q:s=[o[ord(c)-42](*s[1::-1])]+s 
 else:s=[float(c)]+s
print(s[0])

Answer 18

Python 3, 126 132 chars

s=[2,2]
for c in input().split():
    a,b=s[:2]
    try:s[:2]=[[a+b,b-a,a*b,a and b/a]["+-*/".index(c)]]
    except:s=[float(c)]+s
print(s[0])

There have been better solutions already, but now that I had written it (without having read the prior submissions, of course - even though I have to admit that my code looks as if I had copypasted them together), I wanted to share it, too.

Answer 19

JavaScript ES7, 119 bytes

I'm getting a bug with array comprehensions so I've used .map

(s,t=[])=>(s.split` `.map(i=>+i?t.unshift(+i):t.unshift((r=t.pop(),o=t.pop(),[r+o,r-o,r*o,r/o]['+-*/'.indexOf(i)]))),t)

Try it online at ESFiddle

Answer 20

C 178

Sometime a program can be made a bit shorter with more golfing and sometimes you just take completely the wrong route and the much better version is found by someone else.

Thanks to @ceilingcat for finding a much better (and shorter) version

#import<stdlib.h>
float s[99],*p=s;t,z;main(c,v)char**v;{for(;--c;t=z?z-2?~z?z-4?p+=2,atof(*v):*p/p[1]:*p*p[1]:*p-p[1]:*p+p[1],*p=t)z=1[*++v]?-11:**v-43,p--;printf("%f\n",s[1]);}

Try it online!

My original version:

#include <stdlib.h>
#define O(x):--d;s[d]=s[d]x s[d+1];break;
float s[99];main(c,v)char**v;{for(int i=1,d=0;i<c;i++)switch(!v[i][1]?*v[i]:' '){case'+'O(+)case'-'O(-)case'*'O(*)case'/'O(/)default:s[++d]=atof(v[i]);}printf("%f\n",s[1]);}

If you are compiling it with mingw32 you need to turn off globbing (see https://www.cygwin.com/ml/cygwin/1999-11/msg00052.html) by compiling like this:

gcc -std=c99 x.c C:\Applications\mingw32\i686-w64-mingw32\lib\CRT_noglob.o

If you don't * is automatically expanded by the mingw32 CRT into filenames.

Answer 21

C, 232 229 bytes

Fun with recursion.

#include <stdlib.h>
#define b *p>47|*(p+1)>47
char*p;float a(float m){float n=strtof(p,&p);b?n=a(n):0;for(;*++p==32;);m=*p%43?*p%45?*p%42?m/n:m*n:m-n:m+n;return*++p&&b?a(m):m;}main(c,v)char**v;{printf("%f\n",a(strtof(v[1],&p)));}

Ungolfed:

#include <stdlib.h>

/* Detect if next char in buffer is a number */
#define b *p > 47 | *(p+1) > 47

char*p; /* the buffer */

float a(float m)
{
    float n = strtof(p, &p); /* parse the next number */

    /* if the next thing is another number, recursively evaluate */
    b ? n = a(n) : 0;

    for(;*++p==32;); /* skip spaces */

    /* Perform the arithmetic operation */
    m = *p%'+' ? *p%'-' ? *p%'*' ? m/n : m*n : m-n : m+n;

    /* If there's more stuff, recursively parse that, otherwise return the current computed value */
    return *++p && b ? a(m) : m;
}

int main(int c, char **v)
{
    printf("%f\n", a(strtof(v[1], &p)));
}

Test Cases:

$ ./a.out "-4 5 +"
1.000000
$ ./a.out "5 2 /"
2.500000
$ ./a.out "5 2.5 /"
2.000000
$ ./a.out "5 1 2 + 4 * 3 - +"
14.000000
$ ./a.out "4 2 5 * + 1 3 2 * + /"
2.000000

Answer 22

Python 2

I've tried out some different approaches to the ones published so far. None of these is quite as short as the best Python solutions, but they might still be interesting to some of you.

Using recursion, 146

def f(s):
 try:x=s.pop();r=float(x)
 except:b,s=f(s);a,s=f(s);r=[a+b,a-b,a*b,b and a/b]['+-*'.find(x)]
 return r,s
print f(raw_input().split())[0]

Using list manipulation, 149

s=raw_input().split()
i=0
while s[1:]:
 o='+-*/'.find(s[i])
 if~o:i-=2;a,b=map(float,s[i:i+2]);s[i:i+3]=[[a+b,a-b,a*b,b and a/b][o]]
 i+=1
print s[0]

Using reduce(), 145

print reduce(lambda s,x:x in'+-*/'and[(lambda b,a:[a+b,a-b,a*b,b and a/b])(*s[:2])['+-*'.find(x)]]+s[2:]or[float(x)]+s,raw_input().split(),[])[0]

Answer 23

Python - 206

import sys;i=sys.argv[1].split();s=[];a=s.append;b=s.pop
for t in i:
 if t=="+":a(b()+b())
 elif t=="-":m=b();a(b()-m)
 elif t=="*":a(b()*b())
 elif t=="/":m=b();a(b()/m)
 else:a(float(t))
print(b())

Ungolfed version:

# RPN

import sys

input = sys.argv[1].split()
stack = []

# Eval postfix notation
for tkn in input:
    if tkn == "+":
        stack.append(stack.pop() + stack.pop())
    elif tkn == "-":
        tmp = stack.pop()
        stack.append(stack.pop() - tmp)
    elif tkn == "*":
        stack.append(stack.pop() * stack.pop())
    elif tkn == "/":
        tmp = stack.pop()
        stack.append(stack.pop()/tmp)
    else:
        stack.append(float(tkn))

print(stack.pop())

Input from command-line argument; output on standard output.

Answer 24

R, 132 105 96 bytes

for(i in scan(,""))F="if"(grepl("[0-9]",i),c(as.double(i),F),c(get(i)(F[2],F[1]),F[-1:-2]));F[1]

Try it online!

Port of Amro's Matlab answer. I never realized the two languages had so much in common. Saved 27 bytes thanks to Giuseppe ! See edit history for test cases.

Answer 25

Runic Enchantments, 118 bytes

?!/rRlril1-{=
R:0q}"+0"{:}≠?\"-0"{:}≠?\"*0"{:}≠?\"/0"{≠?\R}l1=?!@
\r\ /<?6+{{~{~/?7-S{{~{~/?8 *{{~{~/?7,S{{~/\

Try it online!

Runic already operates in Reverse Polish Notation, so its just a matter of reading the input and converting from char-based symbols to actual operators.

Due to the near-perfect space usage on the lower line, ≠?\ works out to the same number of bytes as =?!\, but is easier to read.

Explanation

    /<                                                Entry
?!\rRlril1-{=                                         Read all inputs
/r/                                                   Reverse the stack (we need to parse from front to back, not back to front; mirrors flipped to preserve visual flow direction within the explanation)
R                                                     Initial parsing loop entry
 :                                                    Duplicate
  0q}                                                 Concat TOS with a 0 on the end
     "+0"                                             Construct string +0 (checking for addition)
         {:}                                          RoL, Dup, RoR (this allows the same i0 string concatenation to be reused)
            ≠?\                                       Compare, if not equal, continue
               "-0"{:}≠?\"*0"{:}≠?\"/0"{≠?\           Do the same for -, *, and /
                                           R          Bottom-of-loop entry
                                            }         Rotate numerical value to bottom of stack
                                             l1=?!@   If stack is only 1 object, print and terminate
           ~{~/                                       If equal, pop no-longer-needed i0 concatenations and the original input char
        +{{                                           Add most recent 2 values
      ?6                                              Skip to bottom-of-loop
               ?7-S{{~{~/?8 *{{~{~/?7,S{{~/           Similar chunks for -, *, and /
                                           \          Mirror up to bottom of main loop (after-math skips wind up here)
                                     ,S{{~            Note that division does not need to pop no-longer-needed concatenated string, as we used up the last one in the comparison

Invalid input characters will be treated as their decimal equivalents because chars implicitly convert to ints. Strings will crash.

Operators must be compared with their string equivalents, again, because chars implicitly convert to ints. A numerical 43 is identical to the character +. 0q is the shortest method of generating a unique string representation of the stack object that can be compared with known expected strings. Additionally a string may not be the ToS when constructing a new string or the two strings are automatically concatenated.

Answer 26

Scala 412 376 349 335 312:

object P extends App{
def p(t:List[String],u:List[Double]):Double={
def a=u drop 2
t match{
case Nil=>u.head
case x::y=>x match{
case"+"=>p(y,u(1)+u(0)::a)
case"-"=>p(y,u(1)-u(0)::a)
case"*"=>p(y,u(1)*u(0)::a)
case"/"=>p(y,u(1)/u(0)::a)
case d=>p(y,d.toDouble::u)}}}
println(p((readLine()split " ").toList,Nil))}

Answer 27

PHP - 259 characters

$n=explode(" ",$_POST["i"]);$s=array();for($i=0;$i<count($n);$s=$d-->0?array_merge($s,!$p?array($b,$a,$c):array($p)):$s){if($c=$n[$i++]){$d=1;$a=array_pop($s);$b=array_pop($s);$p=$c=="+"?$b+$a:($c=="-"?$b-$a:($c=="*"?$b*$a:($c=="/"?$b/$a:false)));}}echo$s[2];

Assuming input in POST variable i.

Answer 28

ECMAScript 6 (131)

Just typed together in a few seconds, so it can probably be golfed further or maybe even approached better. I might revisit it tomorrow:

f=s=>(p=[],s.split(/\s+/).forEach(t=>+t==t?p.push(t):(b=+p.pop(),a=+p.pop(),p.push(t=='+'?a+b:t=='-'?a-b:t=='*'?a*b:a/b))),p.pop())

Answer 29

C# - 323 284 241

class P{static void Main(string[] i){int x=0;var a=new float[i.Length];foreach(var s in i){var o="+-*/".IndexOf(s);if(o>-1){float y=a[--x],z=a[--x];a[x++]=o>3?z/y:o>2?z*y:o>1?z-y:y+z;}else a[x++]=float.Parse(s);}System.Console.Write(a[0]);}}

Edit: Replacing the Stack with an Array is way shorter

Edit2: Replaced the ifs with a ternary expression

Answer 30

Perl, 159

sub c{pop@s}sub z{push@s,$_[0]}while(<>){($_,$a)=/(\S*) ?(.*)/;z$_ if/\d/;z(('-'eq$_?- c:c)+c)if/[+-]/;z(('/'eq$_?1/c:c)*c)if/[*\/]/;redo if$_=$a;print c."\n"}

And here's the ungolfed version:

#!/usr/bin/perl

sub c{pop @stack}
sub z{push @stack,$_[0]}
while (<>) {
    ($_, $a) = /(\S*) ?(.*)/;
    z$_ if /\d/;
    z(('-'eq$_?- c:c)+c) if /[+-]/;
    z(('/'eq$_?1/c:c)*c) if /[*\/]/;
    redo if $_=$a;
    print c . "\n"
}

Not really much to look at, but I'm pretty happy with it. I managed to at least beat the Scheme version :)

Answer 31

Matlab, 228

F='+-/*';f={@plus,@minus,@rdivide,@times};t=strsplit(input('','s'),' ');i=str2double(t);j=~isnan(i);t(j)=num2cell(i(j));while numel(t)>1
n=find(cellfun(@(x)isstr(x),t),1);t{n}=bsxfun(f{t{n}==F},t{n-2:n-1});t(n-2:n-1)=[];end
t{1}

Ungolfed:

F = '+-/*'; %// possible operators
f = {@plus,@minus,@rdivide,@times}; %// to be used with bsxfun
t = strsplit(input('','s'),' '); %// input string and split by one or multiple spaces
i = str2double(t); %// convert each split string to number
j =~ isnan(i); %// these were operators, not numbers ...
t(j) = num2cell(i(j)); %// ... so restore them
while numel(t)>1
    n = find(cellfun(@(x)isstr(x),t),1); %// find left-most operator
    t{n} = bsxfun(f{t{n}==F}, t{n-2:n-1}); %// apply it to preceding numbers and replace
    t(n-2:n-1)=[]; %// remove used numbers
end
t{1} %// display result

Answer 32

K5, 70 bytes

`0:*{$[-9=@*x;((*(+;-;*;%)@"+-*/"?y).-2#x;x,.y)@47<y;(.x;.y)]}/" "\0:`

I'm not sure when K5 was released, so this might not count. Still awesome!

Answer 33

C# - 392 characters

namespace System.Collections.Generic{class P{static void Main(){var i=Console.ReadLine().Split(' ');var k=new Stack<float>();float o;foreach(var s in i)switch (s){case "+":k.Push(k.Pop()+k.Pop());break;case "-":o=k.Pop();k.Push(k.Pop()-o);break;case "*":k.Push(k.Pop()*k.Pop());break;case "/":o=k.Pop();k.Push(k.Pop()/o);break;default:k.Push(float.Parse(s));break;}Console.Write(k.Pop());}}}

However, if arguments can be used instead of standard input, we can bring it down to

C# - 366 characters

namespace System.Collections.Generic{class P{static void Main(string[] i){var k=new Stack<float>();float o;foreach(var s in i)switch (s){case "+":k.Push(k.Pop()+k.Pop());break;case "-":o=k.Pop();k.Push(k.Pop()-o);break;case "*":k.Push(k.Pop()*k.Pop());break;case "/":o=k.Pop();k.Push(k.Pop()/o);break;default:k.Push(float.Parse(s));break;}Console.Write(k.Pop());}}}

Answer 34

C++, 265 bytes

Worse than first thought, looking for improvements.

#include<iostream>
#include<string>
#include<stack>
#define G s.top();s.pop()
using namespace std;int main(){string t; stack<float>s;while(cin>>t)if(47<t.back()&t.back()<58)s.push(stof(t));else{float a,b=G;a=G;s.push(t=="+"?a+b:t=="-"?a-b:t=="*"?a*b:a/b);}cout<<G;}

Usage

Run -> input -> enter -> Ctrl+Z -> enter  (windows)
                      -> Ctrl+D           (unix)

Ungolfed

#include <iostream>
#include <string>
#include <stack>
#define G s.top(); s.pop()
using namespace std;

int main()
{
    string t;
    stack<float> s;
    while (cin >> t)
    {
        if (47 < t.back() & t.back() < 58)
            s.push(stof(t));
        else
        {
            float a, b = G;
            a = G;
            s.push(t=="+" ? a+b : t=="-" ? a-b : t=="*" ? a*b : a/b);
        }
    }
    cout << G;
}

Answer 35

Mathetmatica, 143 bytes

Last@Flatten[#~ImportString~"CSV"&/@StringSplit@#]//.{{a___,b_,c_,o_String,d___}:>{a,o[b,c],d},"+"->Plus,"*"->Times,"/"->Divide,"-"->Subtract}&

Anonymous function, takes a string as input (StringSplit[#]), splits at whitespace, ImportString[#,"CSV"] acts on each element separately (it's like a soft form of eval), converting number-like strings to numbers, however operators remain as strings.

//. - repeatedly apply replacement rule: go through the input list, until you hit an operator character, e.g. {1, 4, 2, / *} becomes {1 /[4,2] *}, then {*[1, /[4,2]]}.

String-forms of operators are replaced with the names of the actual Mathematica functions which do the required thing. Output is a number.

Answer 36

05AB1E, 32 30 bytes

#vyDd_iD'+Qi\+ëD'-Qi\-ë'*Qi*ë/

Try it online or verify all test cases.

Explanation:

#                  # Split the (implicit) input on spaces
 vy                # Loop `y` over each item:
   D               #  Duplicate `y` so its now twice on the stack
    d_i            #  If it's not a digit:
       D'+Qi       #   If it's a '+':
            \      #    Remove the duplicated '+' that's still on the stack
             +     #    Add the top two numbers together
       ëD'-Qi      #   Else-if it's a '-':
             \     #    Remove the duplicated '-' that's still on the stack
              -    #    Subtract the top two numbers from each other
       ë'*Qi       #   Else-if it's a '*':
            *      #    Multiply the top two numbers together
       ë           #   Else (it's a '/'):
        /          #    Divide the top two numbers
                   # (Implicitly output the result)

---

If eval would have been allowed, it's 14 bytes instead:

TF'-N+N'(+:}.V

Try it online or verify all test cases.

Explanation:

                # Take the string input (implicit)
TF         }    # Loop `N` in the range [0, 10):
  '-N+    :     #  Replace "-N" (where N is the current number of the loop) with:
      N'(+      #   "N(" (where N is again the current number of the loop)
.V              # Eval the string as 05AB1E code (and implicitly output the result)

Negative numbers are indicated with a trailing ( instead of leading - in 05AB1E. Apart from that the calculations are also done in Reverse Polish Notation style.

Answer 37

Kotlin, 175 bytes

{var s=List(0){0f}
for(t in it.split(" ")){try{s=s+t.toFloat()}catch(e:Exception){val(a,b)=s.takeLast(2)
s=s.dropLast(2)+listOf(a+b,a-b,a*b,a/b)["+-*/".indexOf(t)]}}
s.last()}

Try it online!

Answer 38

Dart, 196 bytes

F(l,n)=>l.removeAt(l.length-1-n);G(a,b,c)=>c=='+'?a+b:c=='-'?a-b:c=='*'?a*b:a/b;f(s){var l=[];s.split(' ').forEach((a){l.add('+-*/'.contains(a)?G(F(l,1),F(l,0),a):double.parse(a));});return l[0];}

Try it online!

Answer 39

Perl 5.10+: 95 bytes

perl '-anEmy@a;push@a,/\d/?$_:do{$j=pop@a;$i=pop@a;($i*$j,$i+$j,0,$i-$j,0,$i/$j)[-42+ord]}for@F;say$a[0]'

Size 95 bytes (diff with perl -e '' which is 10 bytes).

Ungolfed, expanding command line flags (thanks to perl -MO=Deparse '-anE...) and slightly edited:

use feature 'say';
while (defined($_ = readline)) {
    our @F = split(' ', $_, 0);
    my @a;
    push @a, /\d/ ? $_ : do {
        $j = pop @a;
        $i = pop @a;
        ($i * $j, $i + $j, 0, $i - $j, 0, $i / $j)[-42 + ord($_)]
    } foreach (@F);
    say $a[0];
}

Notes:

• See documentation of perl command line switches about -a, -n and -E
• ASCII values of the following chars are useful to understand the expression involving ord:

  * 42
  + 43
  - 45
  / 47
  

An alternate implementation using regexp matching and replacement: 100 bytes

perl '-pEwhile(s!\b([-0-9.]+)\s+([-0-9.]+)\s+([-+*/])!" ".($1*$2,$1+$2,0,$1-$2,0,$1/$2)[-42+ord$3]!e){}y/ //d'

Ungolfed:

use feature 'say';

while (defined($_ = readline ARGV)) {
    while (s[\b([-0-9.]+)\s+([-0-9.]+)\s+([-+*/])][' ' . ($1 * $2, $1 + $2, 0, $1 - $2, 0, $1 / $2)[-42 + ord($3)];]e) {
        ();
    }
    y/ //d;
} continue {
    die "-p destination: $!\n" unless print $_;
}

Answer 40

Javascript, 112 bytes

r=c=>(s=[],f=(a,b,o)=>eval(`${b}${o}${a}`),c.split(' ').forEach(e=>s.push(!+e?f(s.pop(),s.pop(),e):+e)),s.pop())

Zero error checking, uses the eval function, but it works with all valid RPN expressions.

r is a function (eg. r('2 3 +') returns 4)

Answer 41

PHP, 172 bytes (Demo)

<?foreach(preg_split('~ +~',fgets(STDIN))as$v)if(floatval($v))$s[]=$v;else{[$e,$d]=array_splice($s,-2);$s[]=$v=="+"?$e+$d:($v=="-"?$e-$d:($v=="*"?$e*$d:$e/$d));}echo $s[0];

Previously entertained techniques:

<?do{$_GET[a]=preg_replace_callback('~(-?[\d.]+) +(-?[\d.]+) +([*/+-])~',fn($m)=>($m[3]=='+'?$m[1]+$m[2]:($m[3]=='-'?$m[1]-$m[2]:($m[3]=='*'?$m[1]*$m[2]:$m[1]/$m[2]))),$_GET[a],1,$c);}while($c);echo $_GET[a]; // 208

<?$v=$_GET[a];do{$v=preg_replace_callback('~(-?[\d.]+) +(-?[\d.]+) +([*/+-])~',fn($m)=>($m[3]=='+'?$m[1]+$m[2]:($m[3]=='-'?$m[1]-$m[2]:($m[3]=='*'?$m[1]*$m[2]:$m[1]/$m[2]))),$v,1,$c);}while($c);echo $v; // 202

<?while(preg_match('~(.*?)(-?[\d.]+) (-?[\d.]+) ([*/+-])(.*)~',$_GET[a],$m))$_GET[a]=$m[1].($m[4]=='+'?$m[2]+$m[3]:($m[4]=='-'?$m[2]-$m[3]:($m[4]=='*'?$m[2]*$m[3]:$m[2]/$m[3]))).$m[5];echo $_GET[a]; // 198

<?$v=$_GET[a];while(preg_match('~(.*?)(-?[\d.]+) +(-?[\d.]+) +([*/+-])(.*)~',$v,$m))$v=$m[1].($m[4]=='+'?$m[2]+$m[3]:($m[4]=='-'?$m[2]-$m[3]:($m[4]=='*'?$m[2]*$m[3]:$m[2]/$m[3]))).$m[5];echo $v;  // 194

Answer 42

Python 2.7 (205 bytes)

Edit: Version 2. After consideration I was able to shave off a lot by using the operator module:

from operator import *
f={'*':mul,'/':div,'+':add,'-':sub}
def e(r):
    l=[]
    for i in r:
        try:i=float(i);l.append(i)
        except:l.append(f[i](*l[-2:]));del l[-3:-1]
    return l
print e(raw_input().split())[0]

Answer 43

Java 243

Input comes from the command line. Element 0 in the Float[ ] isn't actually used for storing the inputs, however it is needed for the first assignment of the temp value. The Float constructor takes care of parsing for negative values and values with a decimal place.

interface R{static void main(String[]a){Float t,f[]=new Float[a.length];int i=0;for(String s:a){t=f[i];switch(s){case"-":t=-t;case"+":f[--i]+=t;break;case"/":t=1/t;case"*":f[--i]*=t;break;default:f[++i]=new Float(s);}}System.out.print(f[i]);}}

Ungolfed:

interface ReversePolishNotation{
    static void main(String[]arg){
        Float temp, fVals[]=new Float[arg.length];
        int i=0;
        for(String str:arg){
            temp=fVals[i];
            switch(str){
                case"-":
                    temp=-temp;
                case"+":
                    fVals[--i]+=temp; break;
                case"/":
                    temp=1/temp;
                case"*":
                    fVals[--i]*=temp; break;
                default:
                    fVals[++i]=new Float(str);
            }
        }
        System.out.print(fVals[i]);
    }
}