问题描述
我知道我们可以不要使用可变参数扩展，就像它是一个逗号运算符链。在该问题中，示例如下：
I know that we can't use variadic expansions as if it is a chain of comma operators. In that question the sample is like this:
template<typename... Args>
inline void increment_all(Args&... args)
{
++args...;
}
这可能是不明确的，无论是增加还是先扩大括号， ：
It might be ambiguous either to increment or expand first so parentheses won't hurt:
template<typename... Args>
inline void increment_all(Args&... args)
{
(++args)...;
}
或类似这样：
template<typename... Args>
void cout_all(Args&&... args)
{
(std::cout << std::forward<Args>(args))...;
}
我知道我们可以使用一些递归技巧来获得我们想要的， a href =http://stackoverflow.com/questions/12342633/how-do-i-play-out-the-arguments-of-a-function-using-a-variadic-template>喜欢这样。我不知道为什么标准没有描述这样的行为？
I know that we can use some recursion tricks to get what we want, like this. What I don't know is why does not the standard describe such behavior? I mean, what is the reason behind it?
推荐答案
允许扩展包的其他上下文是列表，其中逗号是
The other contexts where a pack expansion is allowed are lists where a comma is a separator between list elements, not an operator.
例如， f（args ...）展开为函数参数列表 tuple< Args ...> 扩展为模板参数列表。
For example, f(args...) expands to a function argument list, tuple<Args...> expands to a template argument list.
示例pack扩展形成语句，并且语句的子表达式之间的逗号是逗号运算符，可以重载，导致任意复杂的代码，并且与内置逗号运算符不同，不是强制左右对照评价。如果你的（std :: cout<< std :: forward< Args>（args））...; 例子写出了args未指定的顺序，因为参数包中的一个类型重载运算符< <和运算符评论。
In your examples the pack expansion forms a statement, and commas between sub-expressions of a statement are the comma operator, which could be overloaded, leading to arbitrarily complicated code, and unlike the builtin comma operator, not forcing left-to-right evaluation. You'd be surprised if your (std::cout << std::forward<Args>(args))...; example wrote out the args in unspecified order because one of the types in the parameter pack overloaded operator<< and operator, and broke the order of evaluation.
这不会是对当前规则的简单扩展，它将是一个完全不同的上下文，具有非常不同的效果。
Doing this would not be a simple extension to the current rules, it would be a completely different context with very different effects.
不，它不会含糊不清。可以使用 f（++ args ...），它是明确和明确的。你的建议的困难不是如何解析 ++ args ... 这是将它扩展为包含逗号运算符的语句后会发生什么。
No, it wouldn't be ambiguous. It's OK to use f(++args...) and it's clear and unambiguous. The difficulty with your suggestion is not how to parse ++args... it's what happens after you expand it to a statement containing comma operators.
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