Cout中的C ++ 17折叠表达式 [英] C++17 fold expression in cout

问题描述

我正在学习新的c ++ 17 fold表达式，并且从

其中 E 是 pack表达式和 I 是初始化表达式。

---

没有匹配您的二进制折叠（std :: cout<< args< ...），其形式为（I op E op .. ）。

I am learning the new c++17 fold expression and I saw this code from c++17 fold expression. I would like to know why this code work :

template<typename ...Args>
void printer(Args&&... args) {
    (std::cout << ... << args) << '\n';
}

but not this one :

template<typename ...Args>
void printer(Args&&... args) {
    (std::cout << args << ...) << '\n';
}

which could seems logic too and would reverse the print order in my opinion.

解决方案

As seen on cppreference, binary folds can have the following two forms:

Where E is the pack expression and I is the initialization expression.

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There is no binary fold that matches your (std::cout << args << ...), which has the form of (I op E op ...).

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