如果我这样做:
(gdb) b nmspace::TestClass::compareFOO
然后将以下方法签名作为断点签名:
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)
是否有类似于以下的内容,或者我们必须每次都为模板方法编写四个?多态性是否也适用于C ++模板?
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>
我已尝试nmspace::TestClass::compareFOOES
,nmspace::TestClass::compareFOOES*
,nmspace::TestClass::compareFOOES<>()
等。
答案 0 :(得分:1)
如果所有函数的主体相同,您可以编写一个可以使用std::enable_if
或boost::enable_if
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
}