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, Jul 17 '17 at 06:22
Possible duplicate of [difference between \*y++ and ++\*y?](https://stackoverflow.com/questions/17978286/difference-between-y-and-y) — msc, Jul 17 '17 at 06:23
@Evert Yes, it is for an exercise. I'm trying to understand the output. — WCKennedays, Jul 17 '17 at 06:24
@AjayBrahmakshatriya as in `printf("x=%d, y=%p, z=%p\n", x, y, z);` ? — WCKennedays, Jul 17 '17 at 06:25
`*y++=*z++` is also undefined behavior since you are performing pointer arithmetic on non array pointers — Ajay Brahmakshatriya, Jul 17 '17 at 06:25
`y` is not 500, it is the address of `x`. Also, it most probably isn't 500 or anything such, use `%p` for pointers. — Antti Haapala -- Слава Україні, Jul 17 '17 at 06:25
@JosefHoppe Apologies. It is assumed. Edited the question (added comment) `/* Assume address of x is 500 and integer is 4 byte size */` — WCKennedays, Jul 17 '17 at 06:27
@AjayBrahmakshatriya how's that dereferencing an invalid pointer? — Antti Haapala -- Слава Україні, Jul 17 '17 at 06:27
in any case `%d` doesn't work with pointers, you must use `%p` with a `(void*)` cast. — Antti Haapala -- Слава Україні, Jul 17 '17 at 06:27
@AnttiHaapala I misread the expression. I edited my comment and removed the second point. — Ajay Brahmakshatriya, Jul 17 '17 at 06:38
What does [**C Operator Precedence**](http://en.cppreference.com/w/c/language/operator_precedence) tell you should happen? — David C. Rankin, Jul 17 '17 at 06:50

CIsForCookies · Accepted Answer · 2017-07-17T11:43:36.013

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 '17 at 11:43

answered Jul 17 '17 at 06:27

CIsForCookies

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The value 500 is assumed, as this is a practice exercise. `/* Assume address of x is 500 and integer is 4 byte size */` – WCKennedays Jul 17 '17 at 06:31
What if it's a prefix? Like `*++*y`. Any changes to the output? – WCKennedays Jul 17 '17 at 06:40
1

@WCKennedays, you mean `*++y`? then it is UB because the pointer `y+1` is not necessarily valid. – Ajay Brahmakshatriya Jul 17 '17 at 06:41
@AjayBrahmakshatriya sorry, I mean `++*y` – WCKennedays Jul 17 '17 at 06:45
1

@WCKennedays then you mean `++*y = ++*z`? that is again UB because you are writing to the same pointer twice between 2 sequence points. – Ajay Brahmakshatriya Jul 17 '17 at 06:49
1

[**C11 Standard (draft n1570)**](http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf) *§ 6.5 (2) of Expressions* is your reference – David C. Rankin Jul 17 '17 at 07:24
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@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++);` – 0___________ Jul 17 '17 at 08:07
@PeterJ but http://en.cppreference.com/w/c/language/operator_precedence states that postfix precedes indirection. Is it wrong? – CIsForCookies Jul 17 '17 at 08:26
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@CIsForCookies but expressions are evaluated from the left to the right - so the first is *p then p++. – 0___________ Jul 17 '17 at 08:31
@PeterJ OK, My mistake. Thx!! – CIsForCookies Jul 17 '17 at 08:37
1

Thanks, everyone. Big help. – WCKennedays Jul 19 '17 at 18:18

C Programming - Pointers (with increments)

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