Question

following code we always use and is alright,

while(int c=getchar() != 'q');

but if change the ints to pointers like below, compile will rise an error.

#include <stdio.h>
int main(){
  int* c=0;
  if(int* a =c !=0)
    printf("ok");
}

error: cannot convert `bool' to `int*' in assignment

what had happened here? seems the priority was changed. can anyone give me a hint. and if I change it to this, it will work.

#include <stdio.h>
int main(){
  int* c=0;
  int* a;
  if((a =c) !=0)
    printf("ok");
}

Answer 1

No, operator precedence hasn’t changed between the two code snippets. The following:

while(int c=getchar() != 'q')

is equivalent to:

while(int c = (getchar() != 'q'))

Here, getchar() is called, its return value is compared to 'q', and the result of the comparison is stored in the int variable c.

In other words, c doesn’t contain the character that’s just been read; it contains a boolean indicator saying whether the most recent character was 'q'.

Since c is never looked at, the code works. However, it is probably not what was intended.