C Programming - Pointers (with increments)

Question

int main()
{
    int x = 30, *y, *z;
    y = &x; 
    /* Assume address of x is 500 and integer is 4 byte size */
    z = y;
    *y++ = *z++;
    x++;
    printf("x=%d, y=%d, z=%d\n", x, y, z);
    return 0;
}

The above is the code. The output is: x=31, y=504, z=504

Please correct me if I'm wrong:

From what I understand: y=&x; assigns the address of x to y. So, y now holds the value of 500.

z=y; Since y = 500, this assigns 500 to z.

What really confuses me is this part *y++=*z++;, I don't exactly know what this means as there's many things going on at the same time. z gets incremented and pointed to somewhere (AND where is it pointing actually? There is no address assigned to it like y i.e y=&x;. Then *y also gets incremented at the same time (are you even allowed to do that?).

Another thing that confuses me is that: in my opinion, since y points to x, when y++ happens, x should be incremented to 31, and then when going down the code block x++ happens, x should now be 32.

So, question is, how did we get that output of x=31, y=504, z=504? Thank you.

printf("x=%d, y=%d, z=%d\n", x, y, z); is undefined behavior. If you absolutely have to, use %p to print pointers. — Ajay Brahmakshatriya
– Ajay Brahmakshatriya, Commented Jul 17, 2017 at 6:22
@Evert Yes, it is for an exercise. I'm trying to understand the output. — WCKennedays
– WCKennedays, Commented Jul 17, 2017 at 6:24
@AjayBrahmakshatriya as in printf("x=%d, y=%p, z=%p\n", x, y, z); ? — WCKennedays
– WCKennedays, Commented Jul 17, 2017 at 6:25

CIsForCookies · Accepted Answer · 2017-07-17 11:43:36Z

2

Let's break it down.

int x=30, *y, *z; y,z are defined and x is initialized.
y=&x; z=y; both of the pointers are assigned with the address of x (i.e. &x)
*y++=*z++; This one contains a few things, so I'll split it up:

According to this the postfix++ takes precedence over indirection* so we can look at *y++ as *(y++)
y++ means that the value of y will be incremented by 1 and since y is int* it will actually add sizeof(int) to it's value [4 in your case]. But the value of the expression y++ is the value of y before the increment, which leads us to...
indirection: after the postfix++ is done, the indirection is evaluated and since y++'s value is the value of "old" y you get the value of x

Conclusion: y was incremented by 1 but the indirection (dereference) led us to x.

The same applies to z. Eventually what happens is that both y,z are incremented, but before that the value in x is assigned to x (obsolete :) )

Next you'll have x++; which increments the value of x by 1. And that's it.

You can think of *y++=*z++; as:

*y = *z;
y++, z++;

edited Jul 17, 2017 at 11:43

answered Jul 17, 2017 at 6:27

CIsForCookies

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11 Comments

WCKennedays Over a year ago

The value 500 is assumed, as this is a practice exercise. /* Assume address of x is 500 and integer is 4 byte size */

Ajay Brahmakshatriya Over a year ago

@WCKennedays, you mean *++y? then it is UB because the pointer y+1 is not necessarily valid.

Ajay Brahmakshatriya Over a year ago

@WCKennedays then you mean ++*y = ++*z? that is again UB because you are writing to the same pointer twice between 2 sequence points.

David C. Rankin Over a year ago

C11 Standard (draft n1570) § 6.5 (2) of Expressions is your reference

0___________ Over a year ago

@CIsForCookies 'Since postfix++ takes precedence over *' It does not - first value of the *p is taken, then p is incremented so you can : while(*p) putch(*p++);

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