sizeof in preprocessor command doesn't compile with error C1017

Question

I want to use preprocessor command to control code executive path. Because in this way can save runtime time.

#if (sizeof(T)==1 doesn\'t comple with error:

Answer 1

The preprocessor runs before the C++ compiler. It knows nothing of C++ types; only preprocessor tokens.

While I would expect any decent compiler to optimize away an if (sizeof(T) == 1), you can be explicit about it in C++17 with the new if constexpr:

template
class String
{
public:
    static void showSize()
    {
        if constexpr (sizeof(T) == 1) {
            std::cout << "char\n";
        } else {
            std::cout << "wchar_t\n";
        }
    }
};

Live Demo

Pre C++17 it's a bit less straightforward. You could use some partial-specialization shenanigans. It's not particularly pretty, and I don't think it will even be more efficient in this case, but the same pattern could be applied in other situations:

template 
struct size_shower
{
    static void showSize()
    {
        std::cout << "wchar_t\n";
    }
};

template 
struct size_shower
{
    static void showSize()
    {
        std::cout << "char\n";
    }
};

template
class String
{
public:
    static void showSize()
    {
        size_shower::showSize();
    }
};

Live Demo

In this case you could directly specialize String, but I'm assuming in your real situation it has other members that you don't want to have to repeat.