c++ - 可变参数模板包扩展

Question

我正在尝试学习可变参数模板和函数。我不明白为什么这段代码不能编译:

template<typename T>
static void bar(T t) {}

template<typename... Args>
static void foo2(Args... args)
{
    (bar(args)...);
}

int main()
{
    foo2(1, 2, 3, "3");
    return 0;    
}

当我编译失败并出现错误:

Error C3520: 'args': parameter pack must be expanded in this context

(在函数 foo2 中)。

Answer 1

可能发生包扩展的地方之一是 braced-init-list。您可以通过将扩展放在虚拟数组的初始化列表中来利用这一点:

template<typename... Args>
static void foo2(Args &&... args)
{
    int dummy[] = { 0, ( (void) bar(std::forward<Args>(args)), 0) ... };
}

更详细地解释初始化器的内容:

{ 0, ( (void) bar(std::forward<Args>(args)), 0) ... };
  |       |       |                        |     |
  |       |       |                        |     --- pack expand the whole thing 
  |       |       |                        |   
  |       |       --perfect forwarding     --- comma operator
  |       |
  |       -- cast to void to ensure that regardless of bar()'s return type
  |          the built-in comma operator is used rather than an overloaded one
  |
  ---ensure that the array has at least one element so that we don't try to make an
     illegal 0-length array when args is empty

Demo .

在 {} 中扩展的一个重要优势是它保证了从左到右的评估。

---

使用 C++17 fold expressions , 你可以写

((void) bar(std::forward<Args>(args)), ...);