在模板函数和const quailifier中键入deduction [英] Type deduction in templated functions and const quailifier

问题描述

我有以下的c ++ 11代码：

I have the following piece of c++11-code:

#include <iostream>

struct object {
    void talk(const char* text) const { std::cout << "talk " << text << std::endl; }
};

void makeItTalk(object& obj) { obj.talk("non-const"); }
void makeItTalk(const object& obj) { obj.talk("const"); }

template<typename P> void f(P&& p) { 
    makeItTalk(std::forward<P>(p));
}


int main() {

    const object obj;
    f(obj);

    return 0;
}

运行时我得到 talk const 这是它应该是，但我想知道它是如何工作。从我到目前为止读取的 const -qualifier在模板中被忽略。由于 obj 是 const object& 类型，我们有一个 P&& code>作为 f 中的参数我希望模板参数解析为 object& c $ c>& &&& =& 函数 f 应成为

When running I get talk const which is what it should be, but I'm wondering how it works. From what I read so far the const-qualifier is ignored in template deduction. Since obj is of type const object& and we have a P&& as parameter in f I would expect that the template parameter resolves to object& and since & && = & the function f should become

void f(object& p) { makeItTalk(std::forward<object&>(p)); }

但是这个函数甚至不能被调用 obj 。所以我想知道我是否错了，说 const 被忽略了？

But this function is not even allowed to be called for obj. So I'm wondering if I am wrong by saying the const is ignored?

推荐答案

根据我的理解，当函数模板接受指针或引用参数时，类型推导不会忽略const限定符。

As I understand it, type deduction does not ignore the const qualifier when the function template takes a pointer or reference parameter. The top-level consts are removed but not the constness of what is pointed to or referenced.

可以在这里找到一个更广泛的参数： http://cpp-next.com/archive/2011/04/appearing -and-disappearing-consts-in-c /

A more extensive argument can be found here: http://cpp-next.com/archive/2011/04/appearing-and-disappearing-consts-in-c/

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