Postfix对指针的一元增量操作

Question

I newly started working with C and Pointers . There is a concept that confuses me regarding to unary increment operation on pointers.

int num1, *pnum1
num1 = 2;
pnum1 = &num1;
printf("%d \n before: " , *pnum1);
num1 = (*pnum1)++;
printf("%d \n after: " , *pnum1);
return 0;

Since the unary increment operator (++) has greater precedence than the dereference operator (*), I put *pnum1 inside braces. What I expect was to see below result:

after: 3

But it doesn't increment the value of num1. Why would it be the case? Wasn't it suppose to increment the value of num1?

Answer 1

This is undefined behaviour. You're incrementing num1 (via (*pnum1)++ ), then assigning the result back to num1 . The order that incrementing and assigning happen is undefined in this case, so it could get the old value of num1 , increment num1 , and then assign the old value back to num1 , which seems to be what your compiler has chosen to do.

If you try num1 = num1++ instead, your compiler will probably warn you about this.

The solution is not to do stuff like this, since it's undefined behaviour.

Look up "sequence points" for more information.