C++ scopes and allocation

Question

If I understand correctly, variables that are not dynamically allocated are supposed to be deleted at the end of their scope. However, I wanted to try it out with this code (which I think is not correct as I am supposed to use dynamic allocation) :

int* function()
{
    int a = 3;
    return &a;
}

int main(int argc, char const *argv[]) {
    int* a(function());
    std::cout << *a << std::endl; // prints 3
}

Why can I still access the value of the variable a by using the pointer returned by the function when it is supposed not to exist anymore ?

score 0 · Answer 1 · answered Jul 17 '16 at 15:26

0

a and hence the return value from function has gone out of scope. You are just lucky.

Just be careful and compile with all the warnings enables - and take heed of those warnings.

answered Jul 17 '16 at 15:26

Ed Heal

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Actually, He is **very unlucky** that the code didn't crash immediately. – WhiZTiM Jul 17 '16 at 15:27
... Depends on how you see the world – Ed Heal Jul 17 '16 at 15:27
What do you mean by lucky ? It seems to work every time – MaxV37 Jul 17 '16 at 15:30
1

That may be the case in this example. But in the future when using this methodology it will break. – Ed Heal Jul 17 '16 at 15:32
1

@MaxV37 try it with a different optimization level, with a different compiler (or compiler version) or on a different platform. It is *not guaranteed to work* so that it does is just pure luck and may break at any time. – Jesper Juhl Jul 17 '16 at 15:33

score 0 · Answer 2 · answered Jul 17 '16 at 15:27

0

The fact that you can access the value is pure luck. The code has undefined behaviour (since the variable is destroyed) and the compiler can generate whatever it wants - but you cannot rely on it.

answered Jul 17 '16 at 15:27

Jesper Juhl

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C++ scopes and allocation

2 Answers2