Thread: about constant functions and variables

Hi guys,

I wanted to know that if we have a function like:

const int GetRoll();

Does this constant function returns a type "int" which is fixed?

Is it necessary that with a constant function, the variable should also be constant?

eg:

const int RollNo;

const int GetRollNo();

Thanks

No, adding const to a function return value of a built-in type (i.e. int, float, etc.) serves no useful purpose at all. In fact, some compilers will give you a warning saying so.

Note that adding const to a return value type of a non-built-in type may make sense if you want to prevent the user of a function from accidentally doing something silly like the following:

Code:

#include <string>

std::string returnSomeString()
{
    std::string result = "blah blah";
    // ...

    return result;
}

int main()
{
    // This makes no sense! returnSomeString() returns a temporary that will
    // cease to exist right after the append() method invoked on it returns. Hence
    // there is no point in changing the temporary.
    returnSomeString().append( "meep meep!" );
}

The following would prevent that from even compiling.

Code:

#include <string>

const std::string returnSomeString()
{
    std::string result = "blah blah";
    // ...

    return result;
}

int main()
{
    // error! this won't compile ... can't modify a constant object
    returnSomeString().append( "meep meep!" );
}

Originally Posted by R41D3N

returnSomeString() still is temporary and hence even if you append it or you dont(w/o const RT), hardly makes any difference in the code. Right?

The point is that unlike the first example (without the const), which will compile and do the pointless append() without giving you any indication that you're doing something silly, the second example (with the const) won't compile at all, which should give you a pretty good hint that you might be doing something wrong.