英文:
How recursive variadic templates is work?
问题
为什么 count 等于 1？我预期 count 应该是 8。sum 模板函数只被调用了一次吗？
不好意思，我只能回答您提出的翻译请求。如果您有其他问题，可以提出并请求翻译。
英文:
With this code:
static unsigned count = 0;
template<typename... T>
auto sum(T... args)
{
++count;
return (... + args);
}
int main (void)
{
std::cout << sum(12, 32, 32, 12, 4, 3, 43, 432) << std::endl;
std::cout << "TIME: " << count << std::endl;
}
Output is:
$> ./program.out
570
TIME: 1
Why is count equals to 1? I expected count to be 8. Is sum template function called only once?
答案1
得分: 2
是的，它不会被递归调用。相反，该表达式会为折叠表达式展开。
折叠表达式的实例化将表达式e展开如下：
...
2) 一元左折叠*（... op E）* 变为 (((E1 op E2) op ...) op EN)
...
（其中N是包扩展中元素的数量）
您可以将++count放入折叠表达式中，例如：
template<typename... T>
auto sum(T... args)
{
return (... + (++count, args));
}
正如@Xatyrian所指出的，它的值与包扩展中元素的数量相同，也可以通过sizeof...获取。
英文:
> Is sum template function call once ?
Yes, it won't be called recursively. Instead, the expression is expanded for fold expression.
> The instantiation of a fold expression expands the expression e as
> follows:
>
> ...
> 2) Unary left fold (... op E) becomes (((E<sub>1</sub> op
> E<sub>2</sub>) op ...) op E<sub>N</sub>)
> ...
>
> (where N is the number of elements in the pack expansion)
You might want to put ++count into the fold expression, e.g.
template<typename... T>
auto sum(T... args)
{
return (... + (++count, args));
}
As @Xatyrian pointed, its value is just same as the number of elements in the pack expansion, which could be taken by sizeof... too.
答案2
得分: 0
如果您想多次调用sum函数，您可以使用递归方式实现：
static unsigned count = 0;
template <typename T>
auto sum(T t)
{
++count;
return t;
}
template <typename T, typename... Ts>
auto sum(T t, Ts... ts)
{
++count;
return t + sum(ts...);
}
英文:
If you wanted to call sum multiple times, you could do so recursively:
static unsigned count = 0;
template <typename T>
auto sum(T t)
{
++count;
return t;
}
template <typename T, typename... Ts>
auto sum(T t, Ts... ts)
{
++count;
return t + sum(ts...);
}
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