Exercise 1-4

Part 1: Is the following program valid? if so what does it do? If not, why not?

Part 2: What if we add a semi-colon between the second-last and third-last (right) curly braces?

#include <iostream>
#include <string>

int main()
{ 
    {
        const std::string s = "a string";
        std::cout << s << std::endl;
        {  
            const std::string s = "another string";
            std::cout << s << std::endl;
        }
    }

}

Solution (Part 1)

Yes. The program is valid.

Like the previous exercise (1-3), the key to this question is to understanding the term scope. Each pair of curly braces {} form a scope. It is okay to have scopes nested within a scope.

For clarity let me add some comments to the code to visualise these scopes.

#include <iostream>
#include <string>

int main()
{//scope main starts

    {   //scope main-1 start

        const std::string s = "a string";
        std::cout << s << std::endl;

        {   //scope main-1-1 starts

            const std::string s = "another string";
            std::cout << s << std::endl;

        }  //scope main-1-1 ends 

    }  //scope main-1 ends 

} //scope main ends

The const std::string variable s in scope main-1 is not the same const std::string variable s in scope main-1-1 (which is nested inside scope main-1). Even though scope main-1-1 is nested inside scope main-1, all local variables inside scope main-1-1 is hidden from view of main-1.

For completeness, let’s run the program at the top to confirm it runs okay.

a string
another string

Process returned 0 (0x0)   execution time : 0.317 s
Press any key to continue.

The program runs okay as expected.

Memory Treatment

Another thing to bear in mind is the fact that all local variables within a scope only has a life time within the scope. i.e. once the implementation reaches the closing curly brace, }, (in other words, the end of the scope) the local variables of that scope are destroyed. The memory that was taken up by the local variables are now freed up and returned back to the system.

In this case, we expect to see the followings memory allocations / de-allocations.

• When scope main starts: memory allocated for any local variables of scope main (in this case, none).
• When scope main-1 starts: memory allocated for the std::string s variable of scope main-1.
• When scope main-1-1 starts: memory allocated for the std::string s variable of scope main-1-1.
• When scope main-1-1 ends: The std::string s variable of scope main-1-1 is destroyed. Memory is freed up and returned to the system.
• When scope main-1 ends: The std::string s variable of scope main-1 is destroyed. Memory is freed up and returned to the system.
• When scope main ends: Any scope main local variables are destroyed (in this case none). Memory is freed up and returned to the system. (in this case none)

Notice how this differs to the case in exercise 1.3?

Solution (Part 2)

Adding a semi-colon (;) between the second last and third last right curly braces will still constitute a valid program. For clarity, this is what the program would look like with that semi-colon.

#include <iostream>
#include <string>

int main()
{//scope main starts

    {   //scope main-1 start

        const std::string s = "a string";
        std::cout << s << std::endl;

        {   //scope main-1-1 starts

            const std::string s = "another string";
            std::cout << s << std::endl;

        }  //scope main-1-1 ends 

        ;  // the additional semi-colon

    }  //scope main-1 ends 

} //scope main ends

The additional semi-colon essentially creates a null-statement within the main-1 scope. i.e. it has no effect to the code. I think the reason the authors ask this question is to solidify our understanding on scope. i.e. which scope does that semi-colon belong to? Writing a C++ code in the above manner helps us visualise this easier.

Submitting the program with this semi-colon would yield the same output.

Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000