what is the output and explain how?

Question

I am confused about the output of following 2 programs. Could someone explain the precedence and associativity rules?

Program 1:

char arr[] = "geeksforgeeks";
char *p = arr;
*p++;
printf(" %c", *p);

Program 2:

char arr[]  = "geeksforgeeks";
char *ptr  = arr;

while(*ptr != '\0')
    ++*ptr++;
printf("%s %s", arr, ptr);

Please don't write code like that. You _can_ does not mean you _have to_. — Sourav Ghosh, Jun 07 '16 at 14:09
here in first program *p++ is evaluated as *(p++) because post fix has higher precedence over pointer * — skvatss, Jun 07 '16 at 14:11
i am not able to sort out how the second program is evaluating the expression ++*ptr++ — skvatss, Jun 07 '16 at 14:12
It's evaluated as `++(*(ptr++))`. Postfix `++` has higher precedence than prefix `++` and dereference `*`. The latter two have the same precedence so *associativity* comes in to play, which is right to left for those operators. — Bathsheba, Jun 07 '16 at 14:14
@Bathsheba then how is the output coming as h ,it should be f according to your evaluation. — skvatss, Jun 07 '16 at 14:15
Don't forget that the postfix increment is *delayed* until the next sequencing point. — Bathsheba, Jun 07 '16 at 14:17
"I am confused about the output of following 2 programs." That's by design. Programs like 2 are not supposed to be wtitten. — n. m. could be an AI, Jun 07 '16 at 15:56

score 1 · Answer 1 · answered Jun 07 '16 at 16:07

First let's simplify your program into something that's comparable:

char arr[] = "geeksforgeeks";
char* p = arr;

*p++;
printf("%s %c\n", arr, *p);

char* ptr = arr;

++*ptr++;
printf("%s %c\n", arr, *ptr);

Next lets look at the operator precedence and hypothesize what we expect to happen:

Postfix increment (Happens 1^st because it has the highest precedence of the three operators)
Dereference (Happens 2^nd because operators of precedence 2 associate Right-to-left)
Prefix increment (Happens last because it is the leftmost operator of the lowest precedence)

So we expect *p++ to have no effect, other than advancing p to the 2^nd position because:

And we expect ++*ptr++ to increment the current character and advance ptr to the 2^nd position because:

And by looking at this Live Example you can see that our hypothesis was correct, we even get a warning that:

Value computed is not used [-Wunused-value] *p++

Our results are:

geeksforgeeks e
heeksforgeeks e

1 Answers1