When is a variable in a nested scope destroyed from the memory?

Question

void main() {
    int x=5;
    {
        int x= 6; 
        cout<<x;
    }
    cout<<x;
}

the output is 6, 5.

My question is: when Assign x to 6, it reserved a memory cell for it, and then entered the nested block here( the old x is destroyed!! Or how it can defined two variables for the same name?

I presume thus is C++ code. (If you don't say what the language, the question is unanswerable.) — Stephen C, Mar 23 '22 at 10:06
Are you sure your output is `56`, not `65`? BTW, `cout>>x;` is an error - it should be `cout< — ForceBru, Mar 23 '22 at 11:02
`void main()` is **not *standard C++***. You should change it to `int main()`. — Jason, Mar 23 '22 at 11:20
Think of those two `x`s as `x_inside_main` and `x_inside_block_inside_main`. They're two different variables. That's all there is to it. — Pete Becker, Mar 23 '22 at 13:14

score 3 · Answer 1 · answered Mar 23 '22 at 11:36

how it can defined two variables for the same name?

The two x in your example exists in different scopes. When you defined the inner x, it hides the x from outer scope. See the comments in the below given program.

int main()
{//---------------------------->scope1 started here
//------v---------------------->this variable x is in scope1
    int x=5;  
    {//------------------------>scope2 started here
//----------v------------------>this variable x is in scope2
        int x= 6;
//------------v---------------->this prints scope2's x value        
        cout<<x;
    }//------------------------>scope2 ends here and x from scope2 is destroyed
//--------v---------------->this prints scope1's x value   
    cout<<x;
}//---------------->scope1 ends here

Also, the inner scope x is destroyed when that scope ends. Moreover, the inner scope x and outer scope x occupy different addresses in memory.

The output of the above program is 65 which can be seen here.

Additionally, note that void main() is not standard C++, you should instead replace it with int main() as shown in my above snippet.

When is a variable in a nested scope destroyed from the memory?

1 Answers1