put_money 是按值还是按引用来保存其参数? [英] Does put_money hold its argument by value or reference?

问题描述

以下是否会调用未定义的行为?

Does the following invoke undefined behavior?

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <experimental/iterator>

int main() {
    long double values[] = {1, 2, 3};
    std::transform(
        std::begin(values), std::end(values),
        std::experimental::make_ostream_joiner(std::cout, ", "),
        [](long double v) {
            return std::put_money(v + 1);
        }
    );
    return 0;
}

我担心的是 return std::put_money(v + 1) 返回对临时 v + 1 的引用.

My worry is that return std::put_money(v + 1) returns a reference to the temporary v + 1.

推荐答案

标准 ([ext.manip]/6) 只定义了这个特定的表达式:

The standard ([ext.manip]/6) only defines this specific expression:

out << put_­money(mon, intl);

同时mon是如何存储的还没有具体说明，它绝对有可能成为一个悬空引用并成为UB.

It is unspecified how mon is stored in the mean time, and it is definitely possible for it to become a dangling reference and be UB.

一个简单"的解决方法是让你自己的类知道你存储了值:

An "easy" fix is making your own class to know you store the value:

struct money_putter {
    long double value;

    template<class charT, class traits>
    friend std::basic_ostream<charT, traits>& operator<<(std::basic_ostream<charT, traits>& os, const money_putter& mon) {
        return os << std::put_money(mon.value);
    }
};


int main() {
    int values[] = {1, 2, 3};
    std::transform(
        std::begin(values), std::end(values),
        std::experimental::make_ostream_joiner(std::cout, ", "),
        [](int i)  {
            return money_putter{i};  // or i + 1
        }
    );
    return 0;
}

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