C - pointer arithmetic

Question

In the following code:

int strlen(char *s){
 char *p = s;
 while(*p++ != '0円');
 return p - s;
}

Why does the above evaluate differently than this:

int strlen(char *s){
 char *p = s;
 while(*p != '0円') p++;
 return p - s;
}

It is my understanding that the expression will evaluate first, and then increment.

• Oliver Charlesworth: Do you understand that there's a difference between i++ and ++i?
• sherrellbc: Both increment the variable by 1, and result in the same answer if not in an expression. I can correctly use the the two in practice, but it just appeared to me that similar use with pointers yielded different results. As in: if(i++ > 5), the original value of i would be used, and incremented after the evaluation.
• Oliver Charlesworth: Because there's a difference between the two, but it's nothing to do with pointers. Try int i = 3, j = 3; printf("%d, %d\n", ++i, j++);.
• sherrellbc: I know, but the same concept when used on pointers does not seems to yield similar results. As in the use of *(++ptr), and *(ptr++). My question was, shouldn't the expression internal to the parenthesis be evaluated first since they have the highest precedence? If this is not the case, then it makes sense otherwise.
• sherrellbc: I just thought that, by use of the expression *(ptr++), than it would evaluate to the ptr++ first, and then dereference it. But it appears that I have assumed incorrectly.

Answer 1

in the first code p is incremented regardless of if the while() condition was true or false.

In the second snippet of code, p is incremented ONLY if while condition was true.

• Grijesh Chauhan: Bhai : first can be rectified as while(*p++); return (p- s -1);
• Aniket Inge: Surely bhai @GrijeshChauhan

Answer 2

Consider the last step in while loop when *p = '0円'.

In 1^st code:

while(*p++ != '0円');

p still get one increment, and pointer to the element behind '0円'.

In 2^nd code:

while(*p != '0円') p++;

*p != '0円' is not true, so while loop end, p get no increment. p pointer to '0円'.

Answer 3

In the first case:-

while(*p++ != '0円');

p will be incremented just after the evaluation of an expression irrespective of whether the condition is true or false because ++ is a part of conditional expression.

Whereas, in the second one:-

while(*p != '0円') p++;

First the condition will be checked and if it is true then only p will be incremented.

• Sanjay Manohar: No, I think the comparison happens before the increment in both cases. ++p would do the increment before the comparison. But in the first case, the increment is executed even if the string is zero length.
• Vivek Sadh: @Sanjay, Oh... yes actually the increment will take place just after the condition is checked. No matter whether its true or false(First Case) . I forgot that's a post increment. ;)
• Aniket Inge: the condition is first checked if *p contains 0円 or not. If it does, p++ is incremented in the second snippet. Whereas, p will be incremented regardless of if the condition where true or false in the first snippet.