以默认参数为模板类型的函数 [英] Function with default parameter as template type

问题描述

我正在尝试使用带有默认参数的函数作为函数指针模板参数:

I am trying to use a function with a default argument as a function pointer template parameter:

template <void (*F)()>
class A {};

void foo1(int a = 0) {}
void foo2() {}

int main() 
{
    //A<foo1> a1;   <-- doesn't work
    A<foo2> a2;
}

编译错误是:

main.cpp:7:7: 错误:无法将模板参数‘foo1’转换为‘void (*)()’

main.cpp:7:7: error: could not convert template argument ‘foo1’ to ‘void (*)()’

是否有特定的语法可以使其工作?还是特定的语言限制?否则，替代方法是使用两个单独的函数而不是默认参数:

Is there specific syntax for this to work? Or a specific language limitation? Otherwise, the alternative is to have two separate functions instead of a default parameter:

void foo1(int a) {}
void foo1() { foo1(0); }

更新我知道签名是不同的，但我想知道是否有一种方法可以方便地完成这项工作，而无需修改所有带有默认参数的函数?

Update I understand that the signatures are different, but I'm wondering if there is a way to make this work conveniently without needing to modify all the functions with default parameters?

推荐答案

根据 C++ 标准第 8.3.6 节，

According to section 8.3.6 of the C++ standard,

如果在参数声明中指定了表达式，则该表达式将用作默认参数.在缺少尾随参数的调用中将使用默认参数.

If an expression is specified in a parameter declaration this expression is used as a default argument. Default arguments will be used in calls where trailing arguments are missing.

由于A 不是函数调用，默认参数被忽略.事实上，除了函数的调用，它们在所有上下文中都被忽略，例如

Since A<foo1> is not a call of the function, default arguments are ignored. In fact, they are ignored in all contexts except the calls of the function, for example

typedef void (*FFF)();
FFF x = foo1;

不会编译，并产生与您尝试使用 foo1 作为模板参数时得到的相同消息:

will not compile, and produce the same message that you get when trying to use foo1 as a template parameter:

error: invalid conversion from ‘void (*)(int)’ to ‘void (*)()’

这是有道理的，因为评估默认参数是调用中的一个单独步骤:

This makes sense, because evaluating default arguments is a separate step in the invocation:

8.3.6.9:每次调用函数时都会评估默认参数.

8.3.6.9: Default arguments will be evaluated each time the function is called.

默认参数的存在不会改变函数的签名.例如，您不能使用带有默认参数的单参数函数来覆盖无参数虚拟成员函数.

The presence of default arguments does not alter the signature of your function. For example, you cannot use a single-argument function with a default argument to override a no-argument virtual member function.

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