分配给 std::tie 和引用元组有什么区别? [英] What is the difference between assigning to std::tie and tuple of references?

问题描述

对下面的元组业务有点疑惑:

I am a bit puzzled by the following tuple business:

int testint = 1;
float testfloat = .1f;
std::tie( testint, testfloat ) = std::make_tuple( testint, testfloat );
std::tuple<int&, float&> test = std::make_tuple( testint, testfloat );

使用 std::tie 它可以工作，但是直接分配给引用元组不会编译，给出

With std::tie it works, but assigning directly to the tuple of references doesn't compile, giving

错误:从‘std::tuple’转换为非标量类型‘std::tuple’请求"

"error: conversion from ‘std::tuple<int, float>’ to non-scalar type ‘std::tuple<int&, float&>’ requested"

或

"没有合适的用户定义的从 std::tuple 到 std::tuple"的转换

"no suitable user-defined conversion from std::tuple<int, float> to std::tuple<int&, float&>"

为什么?我仔细检查了编译器是否真的与通过执行此操作分配给的类型相同:

Why? I double checked with the compiler if it's really the same type that is being assigned to by doing this:

static_assert( std::is_same<decltype( std::tie( testint, testfloat ) ), std::tuple<int&, float&>>::value, "??" );

评估结果为真.

我还检查了 online 是否可能是 msvc 的错误，但所有编译器给出的结果都是一样的.

I also checked online to see if it maybe was the fault of msvc, but all compilers give the same result.

推荐答案

std::tie()函数实际上是初始化std::tuple 的引用，其中 std::tuple<T&...> 不能由临时 std::tuple.std::tie() 所做的操作和初始化相应对象的表达式如下:

The std::tie() function actually initializes the members of the std::tuple<T&...> of references where is the std::tuple<T&...> can't be initialized by a templatory std::tuple<T...>. The operation std::tie() does and initializing a corresponding object would be expressed like this:

std::tuple<int&, float&> test = 
    std::tuple<int&, float&>(testint, testfloat) = std::make_tuple(testint, testfloat);

(显然，您通常会使用与已绑定变量不同的值).

(obviously, you would normally use different values than those of the already bound variables).

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