CODEWITHSUNDEEP
c++
stamaimer
stamaimer

Reputation: 6475

how the local variable organized in stack when function call?

I run the follow code:

    void func()
    {
        int i;
        int array[10];

        cout << &i << endl;

        for(int i = 0; i < 10; ++i)
        {
            cout << &array[i] << '\n';
        }
    }

the output is:

    0x28fe98
    0x28fe70
    0x28fe74
    0x28fe78
    0x28fe7c
    0x28fe80
    0x28fe84
    0x28fe88
    0x28fe8c
    0x28fe90
    0x28fe94

why the address of i is 0x28fe98?
i think the address of i must be &array[0] - 4.
why the address of i is &array[9] + 4 in fact?

Upvotes: 2

Views: 164

Answers (1)

Drax
Drax

Reputation: 13288

why the address of i is 0x28fe98?

It is located just after the end of your array

i think the address of i must be &array[0] - 4.

Why should it be ?

why the address of i is &array[9] + 4 in fact?

Because i is after your array :)

Typically elements are pushed on the stack and their adresses decrease while the heap limit increases but no code should rely on that except if you're on a low level (asm etc..)

Upvotes: 5

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