passing reference to function

Question

I have a code that does something like following example, and I'm not sure if this is correct because the executable runs as expected.

// source.cpp
void compute_x(int& ref)
{
    ref = 0;
}

void f(int x) 
{
    int local = x;
    local = 1;

    if (local)
    {
          return copute_x(local);
    }
    else return;
}

int main()
{
    f(2);
    return 0;
}

the code runs but, Is variable local valid once f returns?

Answer 1

The variable local goes out of scope after f returns.

After your edit : but the return value is returned from the function and subsequently returned from main .

Answer 2

否，变量local是无效的，因为其作用域位于f函数定义的括号之间...此后不存在