Thread: Program about pointers

Hi, everyone. I was studying a C program that applies the ++ and * operatores on pointers, but I don't understand the output. The code is:

Code:

#include <stdio.h>
int data[2] = {100, 200};
int moredata[2] = {300, 400};

int main(void)
{
    int * p1, * p2, * p3; //declaration of three pointers

    p1 = p2 = data; //p1 = p2 = &data[0]
    p3 = moredata; //p3 = &moredata[0]
    printf("  *p1 = %d,   *p2 = %d,     *p3 = %d\n", *p1, *p2, *p3);
    printf("*p1++ = %d, *++p2 = %d, (*p3)++ = %d\n", *p1++, *++p2, (*p3)++);
    printf("  *p1 = %d,   *p2 = %d,     *p3 = %d\n", *p1, *p2, *p3);

    return 0;
}

The output is:



How come *p1++ is still 100? The way I see it, first p1 is incremented, poiting to data[1], and then the * operator should get 200 instead of 100. But this value is only obtained in the following printf line. And why is (*p3)++ 300 and not 301?

Thanks in advance!

Read this: C Operator Precedence Table. Pay particular attention to note #2 at the bottom. The postfix happens after the values are passed to printf.

Thank you! So, in

Code:

*p1++

,

the * operator has higher precedence and even though I use the ++ on the same statement, its effect is only obtained after that statement?

In the part

Code:

*++p2

,

how come it already displays 200 without the delay?

I think I am understanding it now. The point is that the postfix operator ++ has, in general, higher precedence than some operators bellow it on the table, except when there is the dereferencing operator in the statement - in that case, * goes first, right?

Originally Posted by stdq

Code:

*p1++

,

the * operator has higher precedence and even though I use the ++ on the same statement, its effect is only obtained after that statement?

No, reread the table in the link I provided. The ++ has higher precedence than the *. In this case, it's a post-fix increment, so the increment happens after. That means that all the other stuff is evaluated before the increment happens. p1 is dereferenced, giving you 100, then that value (the 100, not the address) is passed to printf, and lastly p1 is incremented to point to the next spot.

Code:

*++p2

,

how come it already displays 200 without the delay?

The ++ here also has higher precedence than the *, but is just lower than postfix ++. In this case, you're using a pre-fix increment, so the increment happens before the other stuff is evaluated. p2 is incremented to the second spot in the array, then the * dereferences it and passes 200 to printf.

Yes, the ++ operator has higher precedence. The difference is when the increment happens. If the ++ is on the left of something, it is incremented, and that new value used for the expression. If it's on the right, the old value is used, and the increment happens at the very end. It might help you to do some reading on sequence points: Sequence point - Wikipedia, the free encyclopedia and Question 3.8.

Originally Posted by stdq

How come *p1++ is still 100? The way I see it, first p1 is incremented, poiting to data[1], and then the * operator should get 200 instead of 100. But this value is only obtained in the following printf line. And why is (*p3)++ 300 and not 301?

*p1++ will first retrieve the value and then increment the pointer... if you want it to increment first use *(++p1)

(*p3)++ will be incremented AFTER retrieving the value ... If you want it to increment before retrieving the value use ++(*p3)

And yes, it's enough to make you pull your hair out on a good day....

Thank you, everyone!