Returning pointer to local variable

Question

What is the problem associated with returning a pointer to a local variable?

And (i don't know if its legal to do this) What is the problem with returning a reference to pointer in main? ex:

int *p;
p=abc();

where abc would be returning int&

P.S. Sorry if not clear. I am confused too :P

If the return type of `abc` is `int&` – that is, *reference* to `int` – then the above assignment is a type error. Did you probably meant `int *`, aka *pointer* to `int`? — 5gon12eder, Sep 11 '14 at 22:35

score 1 · Answer 1 · answered Sep 11 '14 at 22:15

1

You can syntactically return a pointer to a local variable, but that variable will no longer be valid after the function has returned.

answered Sep 11 '14 at 22:15

Mark Harrison

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So instead of returning the correct value, I would get some garbage value? – Raghav Sharma Sep 11 '14 at 22:17
It will get a garbage value, and that garbage value may change at unpredictable times. – Brendan Sep 11 '14 at 22:18
@RaghavSharma, correct. If you are unlucky, the memory would not be modified yet, and it would *appear* to be correct. – Mark Harrison Sep 11 '14 at 22:22

score 1 · Answer 2 · answered Sep 11 '14 at 22:23

1

If your code is:

int *abc() {
    int n = 3;
    return &n;
}

void foo() {
    int m = 4;
}

int main() {
    int *p;
    p=abc();
    foo();
    printf("%d", *p);
}

then m will probably overwrite n and the result will be 4.

answered Sep 11 '14 at 22:23

Elad

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score 0 · Answer 3 · answered Sep 11 '14 at 22:34

0

In simple terms, local variables are available only as long as it is in that function is in the function stack, once the function is done, its popped out of the function stack.

answered Sep 11 '14 at 22:34

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Returning pointer to local variable

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