c++ - 我们如何在函数声明中使用函数参数？

Question

标准N4296::3.3.4/1 [basic.scope.proto]:

In a function declaration, or in any function declarator except the declarator of a function definition (8.4), names of parameters (if supplied) have function prototype scope, which terminates at the end of the nearest enclosing function declarator.

我尝试了以下示例:

1.

template<const int a>
class A{ };

const int a = 4;

A<a> b; // OK

void foo(const int a = 4, A<a>); //non-type template argument is 
                                 //not a constant expression

DEMO

2.

void foo(const int a = 4, int b = a); //default argument references parameter 'a'

DEMO

我们如何使用这个示波器的功能？它是为了什么而引入的？

Answer 1

这是一个人为的例子:

void foo(overly::long_type::which_should_be_a_typedef_anyway a, decltype(a) b); // fine

decltype(a) c; // oops, a not declared