带有指定模板参数的 C++11 make_pair 无法编译

问题描述

我只是在玩启用了 -std=c++11 的 g++ 4.7(后来的快照之一).我试图编译一些我现有的代码库，一个失败的案例让我有些困惑.

I was just playing around with g++ 4.7 (one of the later snapshots) with -std=c++11 enabled. I tried to compile some of my existing code base and one case that failed somewhat confuses me.

如果有人能解释发生了什么，我将不胜感激.

I would appreciate if someone can explain what is going on.

代码如下:

#include <utility>
#include <iostream>
#include <vector>
#include <string>

int main ( )
{
    std::string s = "abc";

    // 1 ok
    std::pair < std::string, int > a = std::make_pair ( s, 7 );

    // 2 error on the next line
    std::pair < std::string, int > b = std::make_pair < std::string, int > ( s, 7 );

    // 3 ok
    std::pair < std::string, int > d = std::pair < std::string, int > ( s, 7 );

    return 0;
}

我知道 make_pair 意味着用作 (1) 情况(如果我指定类型，那么我不妨使用 (3))，但我不明白为什么在这种情况下它失败了.

I understand that make_pair is meant to be used as the (1) case (if I specify the types, then I might as well use (3)), but I don't understand why it's failing in this case.

确切的错误是:

test.cpp: In function ‘int main()’:
    test.cpp:11:83: error: no matching function for call to ‘make_pair(std::string&, int)’
    test.cpp:11:83: note: candidate is:
    In file included from /gcc4.7/usr/local/lib/gcc/i686-pc-linux-gnu/4.7.0/../../../../include/c++/4.7.0/utility:72:0,
                 from test.cpp:1:
    /gcc4.7/usr/local/lib/gcc/i686-pc-linux-gnu/4.7.0/../../../../include/c++/4.7.0/bits/stl_pair.h:274:5:
note: template<class _T1, class _T2> constexpr std::pair<typename std::__decay_and_strip<_T1>::__type, typename std::__decay_and_strip<_T2>::__type> std::make_pair(_T1&&, _T2&&)
    /gcc4.7/usr/local/lib/gcc/i686-pc-linux-gnu/4.7.0/../../../../include/c++/4.7.0/bits/stl_pair.h:274:5:
note:   template argument deduction/substitution failed:
    test.cpp:11:83: note:   cannot convert ‘s’ (type ‘std::string {aka std::basic_string<char>}’) to type ‘std::basic_string<char>&&’

同样，这里的问题只是发生了什么?"我知道我可以通过删除模板规范来解决这个问题，但我只是想知道这里到底发生了什么问题.

Again, the question here is just "what's going on?" I know that I can fix the problem by removing the template specification, but I just want to know what's failing here under the covers.

• g++ 4.4 编译此代码没有问题.
• 删除 -std=c++11 也可以毫无问题地编译代码.

推荐答案

这不是 std::make_pair 的用途；您不应该显式指定模板参数.

This is not how std::make_pair is intended to be used; you are not supposed to explicitly specify the template arguments.

C++11 std::make_pair 接受两个参数，类型为 T&& 和 U&&，其中T 和 U 是模板类型参数.实际上，它看起来像这样(忽略返回类型):

The C++11 std::make_pair takes two arguments, of type T&& and U&&, where T and U are template type parameters. Effectively, it looks like this (ignoring the return type):

template <typename T, typename U>
[return type] make_pair(T&& argT, U&& argU);

当您调用 std::make_pair 并明确指定模板类型参数时，不会发生参数推导.相反，类型参数被直接替换到模板声明中，产生:

When you call std::make_pair and explicitly specify the template type arguments, no argument deduction takes place. Instead, the type arguments are substituted directly into the template declaration, yielding:

[return type] make_pair(std::string&& argT, int&& argU);

请注意，这两种参数类型都是右值引用.因此，它们只能绑定到右值.这对于您传递的第二个参数 7 来说不是问题，因为这是一个右值表达式.然而，s 是一个左值表达式(它不是临时的，也不会被移动).这意味着函数模板与您的参数不匹配，这就是您收到错误的原因.

Note that both of these parameter types are rvalue references. Thus, they can only bind to rvalues. This isn't a problem for the second argument that you pass, 7, because that is an rvalue expression. s, however, is an lvalue expression (it isn't a temporary and it isn't being moved). This means the function template is not a match for your arguments, which is why you get the error.

那么，为什么当您没有明确指定模板参数列表中的 T 和 U 是什么时它会起作用?简而言之，右值引用参数在模板中是特殊的.部分由于称为 reference collapsing 的语言功能，一个右值引用A&& 类型的参数，其中A 是模板类型参数，可以绑定到任何类型的A.

So, why does it work when you don't explicitly specify what T and U are in the template argument list? In short, rvalue reference parameters are special in templates. Due in part to a language feature called reference collapsing, an rvalue reference parameter of type A&&, where A is a template type parameter, can bind to any kind of A.

无论 A 是左值、右值、const 限定、volatile 限定还是非限定，A&& 都可以绑定到那个对象(同样，当且仅当 A 本身是一个模板参数).

It doesn't matter whether the A is an lvalue, an rvalue, const-qualified, volatile-qualified, or unqualified, an A&& can bind to that object (again, if and only if A is itself a template parameter).

在您的示例中，我们调用:

In your example, we make the call:

make_pair(s, 7)

这里，s 是 std::string 类型的左值，7 是 int 类型的右值>.由于您没有为函数模板指定模板参数，因此执行模板参数推导以找出参数是什么.

Here, s is an lvalue of type std::string and 7 is an rvalue of type int. Since you do not specify the template arguments for the function template, template argument deduction is performed to figure out what the arguments are.

将左值s绑定到T&&，编译器推导出T为std::string&，产生一个 std::string& 类型的参数&&.但是，没有对引用的引用，所以这个双重引用"不存在.折叠成 std::string&.s 是匹配项.

To bind s, an lvalue, to T&&, the compiler deduces T to be std::string&, yielding an argument of type std::string& &&. There are no references to references, though, so this "double reference" collapses to become std::string&. s is a match.

将7 绑定到U&& 很简单:编译器可以推导出U 为int，产生一个 int&& 类型的参数，它成功绑定到 7 因为它是一个右值.

It's simple to bind 7 to U&&: the compiler can deduce U to be int, yielding a parameter of type int&&, which binds successfully to 7 because it is an rvalue.

这些新的语言功能有很多微妙之处，但如果您遵循一个简单的规则，那就很容易了:

There are lots of subtleties with these new language features, but if you follow one simple rule, it's pretty easy:

如果一个模板参数可以从函数参数中推导出来，就让它推导出来.除非绝对必须，否则不要明确提供参数.

If a template argument can be deduced from the function arguments, let it be deduced. Don't explicitly provide the argument unless you absolutely must.

让编译器来完成繁重的工作，无论如何，99.9% 的时间它都会是您想要的.当它不是您想要的时，您通常会收到一个易于识别和修复的编译错误.

Let the compiler do the hard work, and 99.9% of the time it'll be exactly what you wanted anyway. When it isn't what you wanted, you'll usually get a compilation error which is easy to identify and fix.

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