Pointer points to uninitialized variable

Question

In the following program, ptr points to uninitialized variable x. Before printing ptr, I have assigned 10 to ptr and print it.

#include <stdio.h>

int main()
{
    int *ptr;
    int x;

    ptr = &x;
    *ptr = 10;

    printf(" x = %d\n", x);
    printf(" *ptr = %d\n", *ptr);
}

Both ptr and x print the correct value. But, I have doubt, Is it defined behavior?

Please read about it. Why do you think it would be defined or undefined? Where is your research effort? What did you find which you did not understand? Adding those makes it a good question, no offense. — Sourav Ghosh
– Sourav Ghosh, Commented Sep 11, 2017 at 6:09
You only invoke potentially poorly-defined behavior when you access the value of a variable that has not been initialized. — Lundin
– Lundin, Commented Sep 11, 2017 at 6:48
Could you please explain what you mean with "Before dereferenced ptr, i have assigned 10 to ptr and dereferenced it." ? Assigning 10to *ptr(not ptris dereferencing. You dereference it before you dereference it? — Gerhardh
– Gerhardh, Commented Sep 11, 2017 at 7:25

StoryTeller - Unslander Monica · Accepted Answer · 2017-09-11 05:38:04Z

13

Yes, it is. You assign a valid value to ptr and then use indirection to assign a valid value to x.

The address of a variable like x and its value are separate things. After storage is allocated, taking the address is always well defined, regardless of the value in the variable.

answered Sep 11, 2017 at 5:38

StoryTeller - Unslander Monica

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2 Comments

Antti Haapala Over a year ago

(though the compiler may still optimize these as if, as the pointer is never passed around anywhere)

StoryTeller - Unslander Monica Over a year ago

@AnttiHaapala - Yeah, it certainly may. The behavior of the code still remains will defined either way, however.

WorkaroundNewbie · Accepted Answer · 2017-09-11 06:12:00Z

2

Yes , because when you declare x the placeholder / memory will become available for you .

ptr = &x; *ptr = 10; code effectively means

x =10

answered Sep 11, 2017 at 6:12

WorkaroundNewbie

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Comments

Codor · Accepted Answer · 2017-09-11 05:37:41Z

1

To my understanding it is defined behaviour, as it it not necessary to have memory initialized before writing to it.

answered Sep 11, 2017 at 5:37

Codor

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