returning address of a local variable

[manojg]

Hi,
I know that this code is wrong because it returns the address of local variable which no longer exist after function:

int *f1()
{
    int x = 2;
    return &x;
}

However, why this one does not show problem and runs fine, here the pointer p is also a local variable:

int *f2()
{
    int x = 2;
    int *p;
    p = &x;
    return p;
}

[Cubbi]

Define "runs fine". You can execute a function call to f2() just as you can execute a function call to f1(), that's not a problem. The behavior of a program that dereferences the pointer returned by either of those function calls is undefined.

[manojg]

f2() returns pointer to 2 which is supposed to be. Compiler does not show any error.

int main()
{
    int *x = f2();
    cout << *x << endl;
}

output is 2

[Cubbi]

the output of "2" is just one of the many possible results of executing an undefined program. See http://stackoverflow.com/questions/6441218/can-a-local-variables-memory-be-accessed-outside-its-scope/6445794#6445794

[Lachlan Easton]

There's no guarantee that it will work correctly as it is undefined behavior. The two examples are functionally the same. The address returned will still contain the location of where x was, x's value (2) will probably still be there after the function returns. But it could be re-purposed at any time so it becomes unsafe to use.

[manojg]

Funny but very illustrative link. I will read it all. Thanks for explaining it. By-the-way, when a function returns a value, that is also related with the local memory defined within the function because 2 is stored somewhere in the memory. Then, why does the "value return" has no problem?

[Cubbi]

When a function returns a value, it is not related to memory: it has no address. Try:

#include <iostream>
int f3() {
    return 2;
}
int main() 
{
   std::cout << "The address of the value returned by f3() is " << &f3() << '\n';
}

that won't compile