编译错误：未解析的重载函数类型 [英] Compile error: unresolved overloaded function type

问题描述

我尝试使用g ++ 4.7.2编译以下内容：

I try to compile the following with g++ 4.7.2:

template <typename T>
struct A {
    struct B {
        T t;

        template<T B::*M>
        T get() {
            return this->*M;
        }
    };

    B b;

    T get() {
        return b.get<&B::t>();
    }
};


int main() {
    A<int> a;
    a.get();
}

它给我

test.cpp: In member function ‘T A<T>::get()’:
test.cpp:15:23: error: expected primary-expression before ‘)’ token
test.cpp: In instantiation of ‘T A<T>::get() [with T = int]’:
test.cpp:22:8:   required from here
test.cpp:15:23: error: invalid operands of types ‘<unresolved overloaded function type>’ and ‘int A<int>::B::*’ to binary ‘operator<’

为什么？

谢谢。

推荐答案

您需要使用模板 disambiguator：

You need to use the template disambiguator:

return b.template get<&B::t>();

没有它，解析表达式时：

Without it, when parsing the expression:

b.get<&B::t>();

编译器无法判断是否应该解释 get 作为成员变量的名称，后跟< 符号（小于），或作为成员函数模板的实例化 get 。

The compiler can't tell whether it should interpret get as the name of a member variable followed by a < sign (less-than), or as the instantiation of a member function template called get.

虽然我们知道我们的表达式的含义是什么，编译器不能，至少不会在实例化之前发生，并且执行句法解析即使您的函数从未实例化。

Although we know what is the intended meaning of our expression, the compiler cannot, at least not before instantiation occurs - and syntactic parsing is performed even though your function is never instantiated.

这篇关于编译错误：未解析的重载函数类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！