重载解析，模板和继承 [英] Overload resolution, templates and inheritance

问题描述

#include <iostream>

struct A {};
struct B : public A {};

template<typename T>
void foo(const T &x) { std::cout << "Called template" << std::endl; }

void foo(const A &a) { std::cout << "Called A" << std::endl; }

int main()
{
    foo(A());
    foo(B());
    return 0;
}

打印：

Called A
Called template

在一个合适的非模板函数将总是在模板函数上被选择的印象。有人可以向我解释导致这个令人惊讶的结果的解决步骤吗？

I was under the impression that a suitable non-template function would always be chosen over a template function. Can someone explain to me the resolution steps that lead to this somewhat surprising result?

推荐答案

在模板函数下总是选择合适的非模板函数。

I was under the impression that a suitable non-template function would always be chosen over a template function.

这只适用于模板和非模板是同样好的候选人。这就是为什么为 foo（A（））选择非模板的原因。

This only holds if the template and the non-template are equally good candidates. That's why the non-template is chosen for foo(A()).

foo（B（）），使用非模板需要一个派生到基础的转换。

However, in the case of foo(B()), using the non-template requires a derived-to-base conversion. So the function template is strictly better, and hence it's chosen.

foo 模板实例化为 void foo（const B&）。考虑没有模板的样子：

The foo template instantiates into void foo(const B&). Consider what it would look like without templates:

void foo(const B &x) { std::cout << "Called template" << std::endl; }

void foo(const A &a) { std::cout << "Called A" << std::endl; }

我相信你会同意调用 foo / code>应明确选择第一个。这就是为什么选择模板。

I believe you'll agree calling foo(B()) should unambiguously pick the first one. That's exactly why the template is chosen.

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