Check Equality of Types

Question

I have a struct and template class, within which there is a function that should check if T is equal to the struct and if so, do something.

The struct:

struct mystruct
{
    int x;
    int y;
    int z;    
};

the template class:

template <typename T>
class myclass
{
    public:
    void myfunc()
    {
        // this condition is ignored..
        if(std::is_same<T,mystruct>::value==0)
        {
             cout << "T is not mystruct type" << '\n';

        }
        else
        {
            T ms;
            ms.x = 5;
            ms.y = 4;
            ms.z = 3;
        }
    }
};

in main function, if T == mystruct everything goes through fine:

int main()
{
    // no issues
   myclass<mystruct> x;
   x.myfunc();
}

but if T != mystruct:

int main()
{
    //tries to unsuccessfuly convert int to mystruct
   myclass<int> x;
   x.myfunc();
}

execution fails with the below error:

error: request for member 'x' in 'ms', which is of non-class type 'int'
             ms.x = 5;

does anyone have an idea why the if-else statement not working as expected? Thanks!

Which version of C++ are you targetting? If C++17 is targeted, you can use a `if constexpr`. Else you must use SFINAE — Antoine Morrier, Jul 28 '19 at 19:58
Possible duplicate [Difference between if constexpr vs if](https://stackoverflow.com/questions/43434491/difference-between-if-constexpr-vs-if) — n. m. could be an AI, Jul 28 '19 at 20:03
Thank you @Antoine Morrier! I will give `if constexpr` a try. — SAR, Jul 28 '19 at 20:26

score 2 · Accepted Answer · answered Jul 28 '19 at 20:04

Even if the if condition evaluates to false, for a particular template instantiation, the entire template still must consist of valid C++ code.

If the template's parameter is int, for example, then the else part of the if statement becomes equivalent to:

    else
    {
        int ms;
        ms.x = 5;
        ms.y = 4;
        ms.z = 3;
    }

It should be obvious why this won't compile. Because the entire template becomes, equivalent to:

    if (true)
    {
        cout << "T is not mystruct type" << '\n';
    }
    else
    {
        int ms;
        ms.x = 5;
        ms.y = 4;
        ms.z = 3;
    }

Even though else never gets executed, it stlil must be valid C++ code. Templates are no different.

C++17 introduced if constexpr which requires that the evaluated if expression is constant, and only the appropriate part of the if statement ends up being compiled, with the rest being effectively discarded. So, with C++17, you should be able to change the if statement to if constexpr, and get the expected results.

Thank you very much @Sam Varshavchik. I was going nuts over this one..I will try the `if constexpr` and see how it goes. much appreciated! — SAR, Jul 28 '19 at 20:25

Check Equality of Types

1 Answers1