Thread: Strange behaviour of printf

Hello every one I dont understand why my output is different
here is my code

Code:

#include <stdio.h>
#include <stdlib.h>

struct {
	int len ;   //cannot initialize to a value directly
	char *str;
	int a;
} *p;
//size of pointer is always constant = 4

main(){
	p = malloc(sizeof(p));
	p->len = 0;
	printf("%d\t%d",p->len,p->len++);
}

Expected output
0 0

original output
1 0

and when write almost similar code like this one

Code:

#include <stdio.h>
#include <stdlib.h>

struct {
	int len ;   //cannot initialize to a value directly
	char *str;
	int a;
} *p;
//size of pointer is always constant = 4

main(){
	p = malloc(sizeof(p));
	p->len = 0;
	printf("%d\t%d",p->len,p->len+1);
}

I get the out put as
0 1
which is expected

can any one clarify my doubt, is it some thing related to precedence of '++' then why does printf print the first argument as 1 and latter as 0

You have no guarantee when ++ happens, except "before the end of the line". It might happen before the other p->len argument to printf is evaluated, or it might not, depending on how the people who write the compiler want to do it.

Also, you get extremely lucky as your malloc does not allocate nearly enough storage for an entire struct -- it only gets enough memory for a pointer to the struct. You should malloc sizeof(*p) instead.

i have analyzed the cases with

Code:

	printf("%d\t%d\t%d\n",p->len++,p->len++,p->len++);

the out put i get is
2 1 0

and even with

Code:

p->str = "what is it";
printf("%c\t%c\t",*p->str,*p->str++);

out put
h w
i get the reverse way in all the cases

yeah the sizeof(*p) was my mistake sorry

thank you...