hide function template, declare specializations

问题

This is a followup to C++ templates: prevent instantiation of base template

I use templates to achieve function overloading without the mess of implicit type conversions: declare the function template, define desired specializations (overloads). all is well except wrong code does not produce errors until the link phase:

lib.hpp:

template<class T> T f(T v);

lib.cpp:

#include "lib.hpp"

template<> long f(long v) { return -v; }
template<> bool f(bool v) { return !v; }

main.cpp:

#include <iostream>
#include "lib.hpp"

int main()
{
    std::cout
        << f(123L) << ", "
        << f(true) << ", "
        << f(234) << "\n"
    ;
}

gcc output:

c++ -O2 -pipe -c main.cpp
c++ -O2 -pipe -c lib.cpp
c++ main.o lib.o -o main
main.o(.text+0x94): In function `main':
: undefined reference to `int get<int>(int)'

I'd like to have it fail during compilation of main.cpp. Can I somehow declare only specializations actually implemented?

What are my options? The target is C++03, and I'm mainly interested in gcc-4.x and VC9.

回答1:

It seems to produce a linker error even if you don't put it in the separate file.

However, to produce a compiler error for other instantiations, implement the function and use a compile-time assertion, e.g

#include <boost/static_assert.hpp>

template <class T> T f(T)
{
    //assert some type-dependent "always-false" condition,
    //so it won't be triggered unless this function is instantiated
    BOOST_STATIC_ASSERT(sizeof(T) == 0 && "Only long or bool are available");
}

template<> long f(long v) { return -v; }
template<> bool f(bool v) { return !v; }

int main()
{
    //f(100);
    f(100L);
    f(false);
}

And just for general information, C++0x has a much more elegant way to deal with it:

template <class T> T f(T) = delete;

template<> long f(long v) { return -v; }
template<> bool f(bool v) { return !v; }

回答2:

The best way is to implement that basic template with something invalid (not illegal) C++ code. For example,

template<class T> T f(T v) { return v.Default_Implementation_Not_Available; }

This error will be compile time; and it's generated only when you instantiate any version other than 'long' & 'bool'. If you don't instantiate 'int' version, compilation will go fine.

回答3:

I don't believe it's possible to do what you want. See these FAQs for more info:

How can I avoid linker errors with my template functions?

How can I avoid linker errors with my template classes?

回答4:

When compiling main.cpp, there's no way for the compiler to know which template specializations may exist in some other compilation unit -- so there's no way to flag this error at compile time, you have to wait until link time.

来源：https://stackoverflow.com/questions/2155491/hide-function-template-declare-specializations

标签

c++

templates

declaration

template-specialization