模板方法未实例化
为什么我在这个程序上遇到链接错误（使用 gcc 4.6.2）：
#include <iostream>
// prints something;
// the template doesn't add anything
template <typename T>
struct Printer
{
void print()
{
std::cout << "Printer::print" << std::endl;
}
};
// this is an actual template
// calls the method indicated by the second template argument
// belonging to the class indicated by the first template argument
template < typename U, void(U::*func)()>
struct Caller
{
void call(U obj)
{
(obj.*func)();
}
};
// just a wrapper
template<typename V>
struct Wrapper
{
void call_caller_on_printer()
{
Printer<int> a_printer;
Caller<Printer<int>, &Printer<int>::print> caller;
caller.call(a_printer);
}
};
int main()
{
Wrapper<int> the_wrapper;
the_wrapper.call_caller_on_printer();
return 0;
}
链接器抱怨 Printer::print 是一个未定义的引用。但是，如果您将 Wrapper 设置为非模板（模板不会在其中添加任何内容），那么它就可以工作。 Printer的打印方法似乎没有被实例化。这是为什么？
Why am I getting link errors on this program (with gcc 4.6.2):
#include <iostream>
// prints something;
// the template doesn't add anything
template <typename T>
struct Printer
{
void print()
{
std::cout << "Printer::print" << std::endl;
}
};
// this is an actual template
// calls the method indicated by the second template argument
// belonging to the class indicated by the first template argument
template < typename U, void(U::*func)()>
struct Caller
{
void call(U obj)
{
(obj.*func)();
}
};
// just a wrapper
template<typename V>
struct Wrapper
{
void call_caller_on_printer()
{
Printer<int> a_printer;
Caller<Printer<int>, &Printer<int>::print> caller;
caller.call(a_printer);
}
};
int main()
{
Wrapper<int> the_wrapper;
the_wrapper.call_caller_on_printer();
return 0;
}
The linker complains that Printer::print is an undefined reference. However, if you make Wrapper a non-template (the template doesn't add anything there), it works. The print method of Printer does not seem to be instantiated. Why is that?
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我在 GCC 4.5.1 上遇到了一个看起来类似的问题（是的，它看起来确实像回归）。
就我而言，它有助于将指针显式转换为所需类型，以使 GCC 4.5.1 吞下此代码。尝试在这里做同样的事情。即
（未经测试；顺便说一句，这里的强制转换在语法上是否有效？否则元函数可能会有所帮助。）
I’ve had a problem that looks similar on GCC 4.5.1 (and yes, it does look like a regression).
In my case, it helped to explicitly cast the pointer to the desired type to make GCC 4.5.1 swallow this code. Try doing the same here. I.e.
(Untested; incidentally, is a cast even syntactically valid here? Otherwise a metafunction might help.)