What will this code print? #includeusing namespace std;int x = 7;int main(){int x = 2;{int x = 3;cout << ::x << endl;}return 0;}

Question

What will this code print?

#include &lt;iostream&gt;
using namespace std;

int x = 7;

int main() {
    int x = 2;
    {
        int x = 3;
        cout &lt;&lt; ::x &lt;&lt; endl;
    }
    return 0;
}

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Solution

This code will print "7".

Here's the step-by-step explanation:

The code starts with the global variable int x = 7;. This variable is accessible throughout the entire program.
Then, in the main() function, a local variable int x = 2; is declared. This variable is only accessible within the main() function and it hides the global variable x.
Inside the main() function, there's another block of code {int x = 3;cout << ::x << endl;}. Here, another local variable int x = 3; is declared. This variable is only accessible within this block of code and it hides the x variable of the main() function.
However, in the line cout << ::x << endl;, the scope resolution operator :: is used before x. This operator allows access to the global variable x, not the local ones.
Therefore, the value of the global variable x, which is 7, is printed.
The program then returns 0 and ends.

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What will this code print? #includeusing namespace std;int x = 7;int main(){int x = 2;{int x = 3;cout << ::x << endl;}return 0;}

Question

What will this code print?

Solution

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