Weird string calculation

Question

String is given in form of a variable. Let's call that variable s. It is known that the string looks like a calculation expression using +, -, *, /. There's no parentheses.

Calculate the expression inside the string and store the result in variable r (you can assume it's already declared). All the divisions can be performed without remainders (4-6*3/8 is invalid because 8 is not divisible by 3). Length of string is not limited. String can't contain 0.

Challenge is not supposed to give advantage for some languages, so I decided to change some math rules:

Operators precedence is a little bit different than it is in math and programming language expressions: multiplication/division is still higher than addition/substraction. However, if there're some multiplications/divisions in a row, it is performed right-to-left. Addition/substraction is preformed left-to-right.

Sample s and r:

2+5-6/9*8+7 -> 2

1/3*6-6 -> 12

The shortest code (in bytes) wins.

Answer 1

Mathematica (41)

s = "2+5-6/9*8+7";
ToExpression@StringReplace[s,"/"->"^-1 "]

2

Correct me if I'm wrong

Answer 2

Ruby, 24

 r=eval s.gsub ?/,'**-1*'

Thank you to @ybeltukov for the Mathematica technique. I thought there might be a substitution that could be used to leverage eval and yours was much simpler than the crazy regexes I was trying!

This shortcut effectively changes the operator precedence in s to what the question requires. It has the side effect of sometimes resulting in a Rational object type (e.g. 2/1 == 2). The results are correct for the cases I've tried.

Answer 3

Perl 5, 27 20

s/(.)\//1\/$1*/g;print eval

Edit:

s/\//**-1*/g;$r=eval

---

Assuming the input of:

$_ = "2+5-6/9*8+7";
> $r = 2

and

$_ = "1/3*6-6";
> $r = 12

This regex is based in the mathematical theory of @ybeltukov, I'm new to regex so it probably could be much better.

Answer 4

JavaScript (213 characters compacted, 133 if I'm willing to sacrifice my soul)

For clarity, this is the code before I compacted it down (JS-Fiddle of it):

function math(match, leftDigit, operator, rightDigit, offset, string) {
    var L = parseInt(leftDigit)
    var R = parseInt(rightDigit)
    switch (operator)
    {
        case '*': return R*L;
        case '\/': return R/L;
        case '+': return L+R;
        case '-': return L-R;
    }
};

multAndDivRegex = /(\-?\d+)([\*\/])(\-?\d+)(?=[^\*\/]*$)/; //Right to left
addAndSubRegex = /(\-?\d+)([\+\-])(\-?\d+)/; // Left to right

while(multAndDivRegex.test(str)) {
    str = str.replace(multAndDivRegex, math);
}

while(addAndSubRegex.test(str)) {
    str = str.replace(addAndSubRegex, math);
}

Here it is compacted down to 213 characters (containing some extra help from Toni Almeida) (JS-Fiddle of it):

function m(x,l,o,r){
    L=~~l;R=~~r;
    return o=='*'?R*L:o=='/'?R/L:o=='+'?L+R:L-R
}

for(M=/(\-?\d+)([\*\/])(\-?\d+)(?=[^\*\/]*$)/;M.test(s);s=s.replace(M,m));
for(A=/(\-?\d+)([\+\-])(\-?\d+)/;A.test(r=s);s=s.replace(A,m));

Since I'm ending lines with semi-colons, all removable whitespace was ignored for character counting, but left in for clarity.

I can get it down to 133 characters if I use eval(), but that's not how I roll... (JS-Fiddle)

function m(x,l,o,r){return o=='*'?~~r*~~l:~~r/~~l}for(M=/(\-?\d+)([\*\/])(\-?\d+)(?=[^\*\/]*$)/;M.test(s);s=s.replace(M,m));r=eval(s)

This solution would mean that people can enter ANY JavaScript code they want! The 303 character version does not execute any JavaScript; it only does math.

Answer 5

Python 3, 35

r=int(eval(s.replace("/","**-1*")))

Python 3, 30

If you are not scared of floating point value, try this one.

r=eval(s.replace("/","**-1*"))

Both of them evaluate the string s and store the result in variable r

Answer 6

GTB, 6

x?_)→R

6 chars and 6 bytes

EDIT - now that the original question has changed, here's some bloated code:

@iS_,"*")\\iS_,"/$_4;I,1,l?A;)-1:s;A;,2I,1)+A;&s;A;,1I→_&x?_)→R

I think I did a pretty good job; I made it only 63 characters ;) more than 1,000% bloat