在编译时/模板检查变量的值

Question

So, we have this:

template<typename T, typename = std::enable_if_t<std::is_integral<T>::value>>
void fun(const T& val)
{
    std::cout << "val >= 0";
}

int main()
{
    fun(34);
}

Imagine we have other overloads of the function. How would I get the above function overload to only compile when the value of val is larger than 0 ?

On http://en.cppreference.com/w/cpp/types/is_integral I see that operator() is overloaded for std::is_integral and it returns the value so I tried this:

template<typename T, typename = std::enable_if_t<std::is_integral<T>::value() > 0>>

Of course, it looks wrong, and it is wrong as the compiler graciously lets me know.

How do I check the value of the variable at compile time?

Answer 1

Short answer: You can't.

Functions input parameters value is determined at run-time. Thus, SFINAE won't help at this neither will any else compile time trickery.

What you can do is attack the problem at runtime and define two independent functions that are going to be evoked accordingly:

template<typename T, typename = std::enable_if_t<std::is_integral<T>::value>>
void fun(const T& val)
{
    (val < 0)? lower_than_zero(val) : greater_equal_than_zero(val);
}

But probably you knew that already. If you're still in for compile time evaluation and you're sure that your variable is a compile time beast. Then you could pass it as a template non-type argument:

template<int N>
std::enable_if_t<N >= 0> fun() {
  std::cout << "N >= 0" << std::endl;
}

template<int N>
std::enable_if_t<N < 0> fun() {
  std::cout << "N < 0" << std::endl;
}

int main() {
  fun<42>();
  fun<-42>();
}

Answer 2

As @101010 answered there is no way to do this in general way. But if you need to check only one condition >= 0 , then you can do this:

void ff(unsigned int val) {
    val = 42;
}

int main()
{
    ff(34);
    ff(-34);
}

and compile like this:

g++ 1.cpp -Werror -Wsign-conversion

But it's just a hack for one special case.