C++0x 将参数传递给可变参数模板函数

发布于 2024-09-09 19:34:03 字数 718 浏览 3 评论 0原文

通过引用获取可变数量的参数是什么意思？这是否意味着每个参数都是通过引用传递的？

例如，考虑以下函数，它们对其每个参数执行一些处理：

void f()  // base case for recursion
{
}


template <typename Head, typename ... Tail>
void f(Head& head, Tail&... tail)
{
    // Do processing on head
    process(head);

    // Now recurse on rest of arguments
    f(tail...);
}

现在，如果我有：

int a, b, c;
...
f(a, b, c);

这会导致 f(int&, int&, int&)、f(int&, int&) 和最后 f 的实例化（整数&）？

如果我将 f() 的第二个参数更改为“Tail...”而不是“Tail&...”怎么样？现在实例化将是 f(int&, int, int)、f(int&, int) 和最后的 f(int&)，这意味着例如 'c' 将通过前两次调用进行复制，最后一次调用将被复制修改副本而不是原始文件？

有人可以指出一个很好的参考来解释可变参数模板到底是如何工作的吗？

原文

What does it mean to take a variable number of arguments by reference? Does it mean that each of the arguments are passed by reference?

Consider for example the following functions which performs some processing on each its arguments:

void f()  // base case for recursion
{
}


template <typename Head, typename ... Tail>
void f(Head& head, Tail&... tail)
{
    // Do processing on head
    process(head);

    // Now recurse on rest of arguments
    f(tail...);
}

Now if I have:

int a, b, c;
...
f(a, b, c);

Will this result in instantiations of f(int&, int&, int&), f(int&, int&), and finally f(int&)?

How about if I change the second parameter of f() to be "Tail..." instead of "Tail&...". Will the instantiations now be f(int&, int, int), f(int&, int), and finally f(int&), meaning that e.g. 'c' will be copied through the first two calls and the last call will be modifying a copy instead of the original?

Could someone point to a good reference that explains how exactly variadic templates work?

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