c++ - 如何断点多个 C++ 模板方法，就像非模板一样？

Question

如果我这样做:

(gdb) b nmspace::TestClass::compareFOO

然后下面的方法签名将作为一个断点被签名:

nmspace::TestClass::compareFOO(blah::Foo const&, blah::Foo const&, unsigned int)
nmspace::TestClass::compareFOO(blah::Foo const&, blah::FooField const&, unsigned int)
nmspace::TestClass::compareFOO(blah::FooField const&, blah::Foo const&, unsigned int)
nmspace::TestClass::compareFOO(blah::FooField const&, blah::FooField const&, unsigned int)

是不是有类似下面的，或者说模板方法每次都要写这四个？多态性不也适用于C++模板吗？

nmspace::TestClass::compareFOOES<blah::Foo, blah::Foo>
nmspace::TestClass::compareFOOES<blah::Foo, blah::FooField>
nmspace::TestClass::compareFOOES<blah::FooField, blah::Foo>
nmspace::TestClass::compareFOOES<blah::FooField, blah::FooField>

我试过了 nmspace::TestClass::compareFOOES , nmspace::TestClass::compareFOOES* , nmspace::TestClass::compareFOOES<>()等

Answer 1

如果所有函数的主体都相同，您可以使用 std::enable_if 或 boost::enable_if 编写一个可以接受多种类型参数的全局函数:

template< class T >
struct is_valid_field
    : boost::or_<boost::is_same<T, blah::Foo>, boost::is_same<T, blah::FooField>>
{
};

template< class T, class Q >
nmspace::TestClass::compareFOO(T const&, Q const&, unsigned int,
    typename boost::enable_if<
        boost::and_<is_valid_field<T>, is_valid_field<Q>
    >::type* = 0)
{
    // implementation
}