CODEWITHSUNDEEP
c++compiler-errorsassignment-operator
dongjk
dongjk

Reputation: 343

an assignment has two Equal sign

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");
}

Upvotes: 1

Views: 183

Answers (3)

NPE
NPE

Reputation: 500753

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.

Upvotes: 4

Ray Toal
Ray Toal

Reputation: 88438

You are indeed seeing precedence in action.

!= has higher precedence than =. So

int* a =c !=0

parses as

int* a = (c !=0)

which is an assignment of a bool to an int*

Upvotes: 3

cnicutar
cnicutar

Reputation: 182684

Because of precedence, while(int c=getchar() != 'q'); is wrong. It will always store the result of comparing the value returned by getchar() and q. So c will always be 1 or 0.

You likely want:

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

Or maybe

while((c = getchar()) != EOF && c != 'q');

Upvotes: 6

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