Explicit Instantiation of Template

问题

This thread has been really helpful, but I still have questions about this process that don't appear to be answered.

To what extent must I explicitly instantiate a template? For example, if in my definitions file I am using templates on every function, friend class, operator overload, etc, must I instantiate each in a template instantiation file (the current method that I am using)?

Based off of my trial and error, the answer appears to be no, and that a simple

template class Class<type>;

will work for all of the class's members. However, I have read code that suggests otherwise, and a concrete answer would be greatly appreciated.

回答1:

In general you don't need to explicitly instantiate a template, but just define it in a header file and include that header file. However, a common application of explicit template instantiation is when you want to "hide" the definition of a template. Imagine the following situation, in which for simplicity we hide the implementation of a template function:

header.h

template<class X> void f(); // declaration

header.cpp

#include "header.h"

template<class X> void f(){ /* definition */ }
template void f<int>();    // explicit instantiation for int, force the compiler to generate code
template void f<double>(); // explicit instantiation for double, same

main.cpp

#include "header.h"

int main()
{
    f<int>(); // OK
    f<double>(); // also OK
    f<char>(); // linker error
}

As you can see, the function f is defined in the header.cpp file (and not in header.h), and therefore the implementation is hidden from the user. Because of the explicit instantiations for int and double, the compiler will be able to find the code of f<int>() and f<double>(); when compiling main.cpp. However, when trying to find the code for f<char>(); when compiling main.cpp, we get a linker error. That's because the compilation is done independently, and when the compiler compiles header.cpp, the compiler only generates code for f<int> and f<double>, and doesn't know we will invoke f<char> so it does not generate code for f<char>.

The only catch is that in order to make use of such a code hiding, we must explicitly instantiate the function for all the types for which we want to use it, otherwise we get linker errors.

来源：https://stackoverflow.com/questions/32707522/explicit-instantiation-of-template

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c++

templates