Pointer in C++ / Static Scope

Question

I have this code:

int foo() {
    int r = 100;
    int *s = &r;
    return *s;
}

void main() {
    cout << foo();
    system("pause");
}

I thought the result should be junk value, not 100.

But when I test it in Windows 8.1 and Mac OS X, the result is 100.

Can someone explain it to me?

It seems fairly straightforward to me that it should return 100. why do you think it should be junk? — Fabio, Sep 04 '15 at 17:14
Because when `return *s` , it will return the memory of `r` , but when `foo` was finished , the memory of `r` would be deleted . So I think the result should be junk value — Forrest, Sep 04 '15 at 17:18
It doesn't return the memory, doing `return *s` instead of `return s` means that you're returning the value at the address instead of the address. Your function returns `int`, not `int*`. — porglezomp, Sep 04 '15 at 17:20
ya, I mean it will return the value of `r` , but when foo was finished , the address/memory of `r` would be released, so why it still returns 100 there? — Forrest, Sep 04 '15 at 17:27
@DươngAnhKhoa At return the return value is passed (constructed) before local variables go out of scope. — , Sep 04 '15 at 17:31
ok,thanks This is a problem that my teacher had asked me today — Forrest, Sep 04 '15 at 17:34

score 2 · Answer 1 · answered Sep 04 '15 at 17:20

2

No, there shouldn't be a "junk value". s points to an object within the same scope, both s and r are destroyed when the function exits. You safely dereference s before r is destroyed.

Also void main() is incorrect.

answered Sep 04 '15 at 17:20

user5301649

21
1

`void main()` should instead be `int main()` Most compilers will return an error when you use void instead of int. – Quiver Sep 04 '15 at 17:30
To add to my previous comment. The reason you don't use `void main()` is because it is the standard to use `int main()` - You can read more about it here: http://stackoverflow.com/questions/4207134/what-is-the-proper-declaration-of-main – Quiver Sep 04 '15 at 17:38

Fabio · Accepted Answer · 2015-09-04T18:41:12.330

That's because *s is simply the value that the pointer is pointing to.

if you have:

int a = 2;
int *b = &a;
printf("%d\n", *b);
a = 3;
printf("%d", *b);

it's gonna print:

2
3

but if you have:

int a = 2;
int b = 3;
int *c = &b;
printf("%d\n", a);
a = *c;
printf("%d\n", a);
*c = 5;
printf("%d\n", a);

it's gonna print

2
3
3

and not

2
3
5

because when you do a = *c; it just copies the value of c into a, but establishes no further relation between c and a, which means you can change c or the value that it's pointing to whatever you want, and a is gonna continue holding that value that c had when you did that assignment. They're 2 different spaces in memory and what a = *c does is merely copy the value that c is pointing to to a

What I mean by all that is that when you return *s you're not returning the address to the memory s is pointing to, but the value that it contains

Pointer in C++ / Static Scope

2 Answers2