Expression evaluation order

Question

I recently got confused by the following c++ snippet:

#include <cstdio>

int lol(int *k){
  *k +=5;
   return *k;
}

int main(int argc, const char *argv[]){
  int k = 0;
  int w = k + lol(&k);
  printf("%d\n", w);
  return 0;
}

Take a look at line:

int w = k + lol(&k);

Until now I thought that this expression would be evaluated from left to right: take current value of k (which before calll to lol function is 0) and then add it to the result of lol function. But compiler proves me I'm wrong, the value of w is 10. Even if I switch places to make it

int w = lol(&k) + k;

the result would be still 10. What am I doing wrong?

Tomek

Answer 1

This is because the parameters in an expression are not specified to be evaluated in any particular order.

The compiler is free to execute either parameter k or lol(&k) first. There are no sequence points in that expression. This means that the side-effects of the parameters can be executed in any order.

So in short, it's not specified whether the code prints 5 or 10. Both are valid outputs.

The exception to this is short-circuiting in boolean expressions because && and || are sequence points. (see comments)