问题描述
我正在尝试创建专门用于某些给定类型的全局函数模板.看起来像这样:
I am trying to create a global function template specialized for some given types. It looks something like that:
A.h (主要模板，模板专业化，外部)
A.h (primary template, template specialization, extern)
template <typename T> void foo() { std::cout << "default stuff" << std::endl; }
template<> void foo<int>() { std::cout << "int stuff" << std::endl; }
extern template void foo<int>();
A.cpp (显式实例化)
template void foo<int>();
B.h
void bar();
B.cpp (包括A.h)
void bar() { foo<int>(); }
main.cpp
foo<int>();
bar();
编译器崩溃了:'void foo()'的多个定义.我认为应该由extern来处理.B编译单元不应实例化foo，而应在链接上使用A实例化时间，不?我在这里弄错了什么?
The compiler crashes on me: "multiple definitions of 'void foo()'. I thought that the extern was supposed to take care of this. The B compile unit should not instantiate foo, and instead use the A instantiation at link time, no? What am I getting wrong here?
请注意，如果我不专门研究foo，代码可以很好地编译.函数专门化和实例化之间是否存在某种冲突?
Note that if I do not specialize foo, the code compiles just fine. Is there some kind of conflict between function specialization and instantiation?
推荐答案
此处不需要extern来抑制实例化.通过声明显式专业化，您已经在告诉任何调用foo<int>的代码以使用显式专业化而不是主模板.相反，您只想在A.h中声明专业化，然后在A.cpp中定义它:
You don't need extern here to suppress instantiation. By declaring an explicit specialization, you're already telling any code that calls foo<int> to use the explicit specialization rather than the primary template. Instead, you simply want to declare the specialization in A.h and then define it in A.cpp:
// A.h
template <typename T> void foo() { std::cout << "default stuff" << std::endl; }
template <> void foo<int>();
// A.cpp
template <> void foo<int>() { std::cout << "int stuff" << std::endl; }
如果要在某个翻译单元中提供主模板的显式实例化，而不是的显式专业化，则可以使用extern.
Your use of extern would be appropriate if you wanted to provide an explicit instantiation of the primary template in some translation unit, and not an explicit specialization.
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