C ++ 11可变参数模板和以逗号分隔的表达式对等 [英] C++11 variadic templates and comma-separated expressions equivalence

问题描述

在可变参数模板中，...运算符将参数包扩展为一系列逗号分隔的参数（以最简单的形式）。我的问题是：为什么用逗号分隔的多个参数调用some_function（）并用...运算符调用它不会呢？

In a variadic template the ... operator expands a parameter pack into a series of comma-separated arguments (in the simplest form). My question is: how come that calling some_function() for multiple arguments comma-separated works and calling it with the ... operator doesn't?

我在说关于此代码：

template<typename... Args> inline void expand(Args&&... args) 
{
   some_function(22),some_function(32); // Works
   some_function(args)...; // Doesn't work - ERROR
}

这两行不会产生

推荐答案

因为在第一种情况下，您没有用逗号分隔的参数，而是使用了逗号-

Because in the first case you don't have comma-separated arguments but you are instead using the comma-operator, a totally different beast.

您可以递归实现 expand 函数：

inline void expand() {}

template<typename T, typename... Args>
inline void expand(T&& head, Args&&... tail)
{
    some_function(head);
    expand(tail...);
}

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