调用可变参数模板函数而无args失败 [英] Calling variadic template function with no args failing

问题描述

给出以下代码：

#include <iostream>

template <typename... Args>
void foo(Args&&... bargs, Args&&... aargs)
{
    std::cout << "Hello" << std::endl;
}

int main()
{
    foo<int, double>(1, 2.0, 3, 4.0); //OK
    foo<char>('c', 'd'); // OK
    foo(); //FAIL
}

我得到以下编译器错误：

I get the following compiler error:

In function 'int main()':
15:9: error: no matching function for call to 'foo()'
15:9: note: candidate is:
6:6: note: template<class ... Args> void foo(Args&& ..., Args&& ...)
6:6: note:   template argument deduction/substitution failed:
15:9: note:   candidate expects 1 argument, 0 provided

是否可以在不使用args的情况下调用该函数？可以将函数更改为支持零个或多个args吗？

Is it possible to call such a function with no args? Can the function be changed to support zero or more args?

推荐答案

您必须指定不带args的函数版本：

You have to specify a version of the function without args:

#include <iostream>

template <typename... Args>
void foo(Args&&... bargs, Args&&... aargs)
{
    std::cout << "Hello with args" << std::endl;
}

void foo()
{
    std::cout << "Hello without args" << std::endl;
}

int main()
{
    foo<int, double>(1, 2.0, 3, 4.0); //OK
    foo<char>('c', 'd'); // OK
    foo(); // Also OK now
}

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