std::tie 与 std::make_tuple [英] std::tie vs std::make_tuple

问题描述

此代码编译，但我想知道应该首选哪个版本:

This code compiles but I'm wondering which version should be preferred:

#include <iostream>
#include <tuple>
using namespace std;

tuple<int, int, int> return_tuple1() {
    int a = 33;
    int b = 22;
    int c = 31;
    return tie(a, b, c);
}

tuple<int, int, int> return_tuple2() {
    int a = 33;
    int b = 22;
    int c = 31;
    return make_tuple(a, b, c);
}

int main() {
    auto a = return_tuple1();
    auto b = return_tuple2();
    return 0;
}

既然函数是按值返回一个元组，那么使用 std::tie 应该没有任何问题吧?(即没有悬空引用)

since the function is returning a tuple by value there shouldn't be any problem in using std::tie right? (i.e. no dangling references)

推荐答案

使用 std::tie 时要非常小心.返回一个 tie 在逻辑上等同于返回一个引用，以及随之而来的所有警告.

Be very careful with std::tie. Returning a tie is logically equivalent to returning a reference, with all of the caveats that come with it.

逻辑上，这三个是等价的:

Logically, these three are equivalent:

int& foo();
std::reference_wrapper<int> foo();
std::tuple<int&> foo();

还有这个:

int a = 10;
return std::tie(a);

相当于:

int a = 10;
return std::ref(a);

因为它产生以下之一:

std::tuple<int&>

在您的示例中，您可以通过返回值的隐式转换来保存.但是，将返回类型替换为 auto 会显示逻辑错误:

In your example you are saved by the return value's implicit conversion. However, replacing the return type with auto reveals the logic error:

#include <iostream>
#include <tuple>
using namespace std;

auto return_tuple1() {  // function name is now lying
    int a = 33;         // it should be return_chaos()
    int b = 22;
    int c = 31;
    return tie(a, b, c);
}

auto return_tuple2() {
    int a = 33;
    int b = 22;
    int c = 31;
    return make_tuple(a, b, c);
}

int main() {
    auto a = return_tuple1(); // uh-oh...

    auto b = return_tuple2();

    std::get<0>(a); // undefined behaviour - if you're lucky you'll get a segfault at some point.
    std::get<0>(b); // perfectly ok
    return 0;
}

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