[英]How to specialize a variadic template function in c++?
Let the function foo be given with the following "signature":让函数 foo 带有以下“签名”:
template<typename ...Ts>
void foo(Ts ...args)
This is a bit overdone since I need foo to proccess double
s only.这有点过头了,因为我只需要 foo 来处理
double
s。 How can I modify foo so it accepts double
s only?如何修改 foo 使其仅接受
double
s?
The original code has been Godbolted:原始代码已被Godbolted:
#include <tuple>
#include <iostream>
template<typename ...Ts>
void foo(Ts ...args)
{
std::tuple<Ts...> tu(args...);
std::apply([](Ts const&... tupleArgs)
{
((std::cout << tupleArgs << " "), ...);
}, tu);
}
int main()
{
foo(-2.44, -5.66, 78.99);
}
The output reads:输出如下:
-2.44 -5.66 78.99 -2.44 -5.66 78.99
You can constrain your template like this:你可以像这样约束你的模板:
template<typename ...Ts>
requires std::conjunction_v<std::is_same<double, Ts>...>
void foo(Ts ...args)
{
// ...
}
As pointed out by HolyBlackCat, you can write the template much more conveniently like this:正如 HolyBlackCat 所指出的,您可以像这样更方便地编写模板:
void foo(std::same_as<double> auto ...args)
{
(std::cout << ... << args);
}
Note also that you can use a fold expression to print the arguments.另请注意,您可以使用折叠表达式来打印参数。
Those of us who are still stuck with C++17 will have to do with a slightly uglier syntax :(我们这些仍然坚持使用 C++17 的人将不得不使用稍微丑陋的语法:(
template<typename ...Ts>
std::enable_if_t<std::conjunction_v<std::is_same<Ts, double>...>> foo(Ts ...args)
{
(std::cout << ... << args);
}
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