Advanced Calculator

Question

You must write a program that evaluates a string that would be entered into an advanced calculator.

The program must accept input using stdin and output the correct answer. For languages that do no have functions to accept stdin, you may assume the functions readLine and print to handle these tasks.

Requirements:

• Does not use any kind of "eval" functions
• Can handle floating point and negative numbers
• Supports at least the +, -, *, /, and ^ operators
• Supports brackets and parenthesis for overriding the normal order
• Can handle input containing one or more spaces between the operators and numbers
• Evaluates the input using the standard order of operations
  
  

Test Cases

Input

10 - 3 + 2

Output

9

---

Input

8 + 6 / 3 - 7 + -5 / 2.5

Output

1

---

Input

4 + [ ( -3 + 5 ) * 3.5 ] ^ 2 - 12

Output

41

Answer 1

C++, 640 583

string k="[]()+-*/^";stack<double> m;stack<char> n;
#define C(o,x,y) ('^'==o?x<y:x<=y)
#define Q(a) double a=m.top();m.pop();
#define R(o) {Q(b)Q(a)m.push(o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:o=='/'?a/b:o=='^'?pow(a,b):0);n.pop();}
while(!cin.eof()){string s;getline(cin,s,' ');if(s.empty())continue;if('\n'==*--s.end())s.erase(--s.end());(s.size()==1&&s.npos!=k.find(s[0]))?({char c=s[0]=='['?'(':s[0]==']'?')':s[0];while(!n.empty()&&'('!= c&&C(c,k.find(c),k.find(n.top())))R(n.top());')'==c?n.pop():n.push(c);}):m.push(strtod(s.c_str(),0));}while(!n.empty())R(n.top());cout<<m.top()<<endl;

Indented

string k="[]()+-*/^";
stack<double> m;
stack<char> n;
#define C(o,x,y) ('^'==o?x<y:x<=y)
#define Q(a) double a=m.top();m.pop();
#define R(o) {Q(b)Q(a)m.push(o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:o=='/'?a/b:o=='^'?pow(a,b):0);n.pop();}
while(!cin.eof())
{
    string s;
    getline(cin,s,' ');
    if(s.empty())continue;
    if('\n'==*--s.end())s.erase(--s.end());
    (s.size()==1&&s.npos!=k.find(s[0]))?({
        char c=s[0]=='['?'(':s[0]==']'?')':s[0];
        while(!n.empty()&&'('!= c&&C(c,k.find(c),k.find(n.top())))
            R(n.top());
        ')'==c?n.pop():n.push(c);
    }):m.push(strtod(s.c_str(),0));
}
while(!n.empty())
    R(n.top());
cout<<m.top()<<endl;

My first code golf, so looking forward to comments & criticism!

Answer 2

PHP - 394 354 312 characters

<?=e(!$s=preg_split('#\s+#',`cat`,-1,1),$s);function e($P,&$s){$S='array_shift';if(($a=$S($s))=='('|$a=='['){$a=e(0,$s);$S($s);}while($s&&($p=strpos(' +-*/^',$o=$s[0]))&&$p>=$P){$b=e($p+($S($s)!='^'),$s);if($o=='+')$a+=$b;if($o=='-')$a-=$b;if($o=='*')$a*=$b;if($o=='/')$a/=$b;if($o=='^')$a=pow($a,$b);}return$a;}

Indented:

<?
preg_match_all('#\d+(\.\d+)?|\S#',`cat`,$m);
$s=$m[0];
function e($P) {
        global $s;
        if (strpos(" ([",$s[0])){
                array_shift($s);
                $a=e(0);
                array_shift($s);
        } else {
                $a=array_shift($s);
                if ($a=='-')$a.=array_shift($s);
        }
        while ($s && ($p=strpos(' +-*/^',$o=$s[0])) && $p >= $P) {
                array_shift($s);
                $b = e($p+($o!='^'));
                switch($o){
                case'+':$a+=$b;break;
                case'-':$a-=$b;break;
                case'*':$a*=$b;break;
                case'/':$a/=$b;break;
                case'^':$a=pow($a,$b);
                }
        }
        return $a;
}
echo e(0);

Answer 3

Postscript, 446

This uses the shunting yard algorithm.

[/*[/p
2/e{mul}>>/d[/p
2/e{div}>>/+[/p
1/e{add}>>/-[/p
1/e{sub}>>/o[/p
9/e{}>>/c[/p
-1/e{}>>/^[/p
3/e{exp}>>/p
0>>begin/s(%stdin)(r)file 999 string readline pop def
0 1 s length 1 sub{s exch[0 1 255{}for]dup[(\(o)([o)(\)c)(]c)(/d)]{{}forall
put dup}forall
pop
3 copy pop
get
get
put}for{s token not{exit}if
exch/s exch store{cvr}stopped{load
dup/p get
p
le{currentdict end
exch begin/e get exec}{begin}ifelse}if}loop{{e end}stopped{exit}if}loop
=

Un-golfed and commented:

% We associate the operators with their precedence /p and the executed commend /e
[
  (*)[/p  2 /e{mul}>>
  (d)[/p  2 /e{div}>> % This is division
  (+)[/p  1 /e{add}>>
  (-)[/p  1 /e{sub}>>
  (o)[/p  9 /e{   }>> % This is open bracket
  (c)[/p -1 /e{   }>> % This is close bracket
  (^)[/p  3 /e{exp}>>
  /p 0
>>begin

% Let's read the input string
/s(%stdin)(r)file 999 string readline pop def

% If we want to use the token operator, we have to replace (, [, ), ] and / to get meaningful results
% We use kind of an encoding array (familiar to PostScripters) to map those codes to o, c, and d.
0 1 s length 1 sub{        % index
  s exch                   % string index
  [0 1 255{}for] dup       % string index translationArray translationArray
  [(\(o)  ([o)  (\)c)  (]c)  (/d)] % string index translationArray translationArray reencodeArray
  {                        % string index translationArray translationArray translationString
    {}forall               % string index translationArray translationArray charCode newCharCode
    put dup                % string index translationArray translationArray
  }forall                  % string index translationArray translationArray
  pop                      % string index translationArray
  3 copy pop               % string index translationArray string index
  get                      % string index translationArray charCode
  get                      % string index translatedCharCode
  put                      % -/-
}for

% Now we can actually start interpreting the string
% We use the stack for storing numbers we read and the dictionary stack for operators that are "waiting"
{                          % number*
  s token not{exit}if      % number* string token
  exch /s exch store       % number* token
  % We try to interpret the token as a number
  {cvr}stopped{            % number* token
    % If interpretation as number fails, we have an operator
    load                   % number* opDict
    % Compare operator precedence with last operator on dictstack
    dup /p get             % number* opDict opPrec
    p                      % number* opDict opPrec prevOpPrec
    le {                   % number* opDict
      % If the last operator on the stack has at least the same precedence, execute it
      currentdict end      % number* opDict prevOpDict
      exch begin           % number* prevOpDict
      /e get exec          % number*
    }{                     % number* opDict
      % If last operator doesn't have higher precedence, put the new operator on the dictstack as well
      begin
    }ifelse
  }if
}loop
% If we're finished with interpreting the string, execute all operators that are left on the dictstack
{{e end}stopped{exit}if}loop
=

TODO: Right-associativity of exponentation

Answer 4

Python 2,339 335 bytes

import re
x,s=input(),re.sub
def f(y):
 y,r=s('- ','+ -',y).split(),['^','*','/','+','-']
 for c in r:
  while c in y:d=y.index(c)-1;a,b=map(float,[y[d],y[d+2]]);y=y[:d]+[((a,-a)[a<0]**b,a*b,a/b,a+b,a-b)[r.index(c)]]+y[d+3:]
 return`y[0]`
w=lambda b:s("[([]+[\d+\-*/^ .]*[)\]]",lambda m:f(m.group()[1:]),s(' +',' ',b))
print f(w(w(x)))

Try it online!

• -4 bytes by changing str(x) with backticks ``!

Answer 5

Postscript, 1000 695 665 494

Stole ideas from ThomasW. Added feature: accepts strings with or without spaces around operators.[feature removed]

---

Using ARGUMENTS is shorter than %stdin, and easier to test, to boot!

---

Simplified the substitution to just replace brackets with parens.

575(1)10:36 PM:ps 0> gsnd -q -- calc2bg.ps '10 - 3 + 2'
9
576(1)10:37 PM:ps 0> gsnd -q -- calc2bg.ps '8 + 6 / 3 - 7 + -5 / 2.5'
1.0
577(1)10:37 PM:ps 0> gsnd -q -- calc2bg.ps '4 + [ ( -3 + 5 ) * 3.5 ] ^ 2 - 12'
41.0

Code:

/T[/^[/C{exp}/P 4/X{le}>>/*[/C{mul}/P 3/X{lt}>>/[/C{div}/P
3/X{lt}>>/+[/C{add}/P 2/X{lt}>>/-[/C{sub}/P
2/X{lt}>>>>def[/integertype{}/realtype{}/stringtype{V}/nametype{cvlit/N
exch store{P T N get dup/P get exch/X get exec{exit}if C end}loop T N get
begin}91 40 93 41>>begin/V{0 1 2 index length 1 sub{2 copy get
dup where{exch get}if 3 copy put pop pop}for[/N 0/R 0/P 0/C{}>>begin{token
not{exit}if exch/R exch store dup type exec R}loop{P 0 eq{end exit}if C
end}loop}def ARGUMENTS{V ==}forall

Ungolfed and commented:

%!
%Shunting-Yard Algorithm using dictstack for operators
%invoke with %gsnd -q -- calc2bg.ps [ 'expr1' ]*

%The operator table. C:code P:precedence X:test(implements associativity)
/T[
    /^[/C{exp}/P 4/X{le}>>
    /*[/C{mul}/P 3/X{lt}>>
    /[/C{div}/P 3/X{lt}>>
    /+[/C{add}/P 2/X{lt}>>
    /-[/C{sub}/P 2/X{lt}>>
>>def

%The type-dispatch dictionary
%numbers: do nothing
%string: recurse
%name: process op
[%/integertype{}/realtype{} %now uses `where` below
/stringtype{V}/nametype{
pstack()=
    cvlit/N exch store %stash cur-op
    {
        P %prec(tos)
        T N get %prec(tos) cur-op-dict
        dup/P get %prec(tos) cur-op-dict prec(cur-op)
        exch/X get %prec(tos) prec(cur-op) test(cur-op)
        exec{exit}if %exit if prec(tos) < || <= prec(cur-op)
/C load ==
        C %pop-and-apply
        end
pstack()=
    } loop
    T N get begin %push cur-op
}>>begin

%substitutions
[91 40 93 41>>begin %replace brackets with parens
/V {
    %pre-process
    0 1 2 index length 1 sub {
        2 copy get
        dup where { exch get } if
        3 copy put pop pop
    } for
dup ==

    [/N 0/R 0/P 0/C{}>>begin %dummy base operator and storage
    { token not{exit}if exch /R exch store %extract token, stash Remainder
pstack(>)=
        %dispatch type procedure
        dup type dup where { pop exec }{ pop } ifelse
    R }loop
pstack()=
    {
        P 0 eq{end exit}if %hit dummy op: exit
/C load ==
        C end %pop and apply
    } loop
} def

ARGUMENTS{V ==}forall %iterate through the command-line arguments