在 C++ 中，命名参数的范围是否包括其默认值的表达式？

Question

示例：这是合法的 C++14 吗？

#include <iostream>

static int d() {
    return 42;
}

static int e(int d = d()) {
    return d;
}

int main() {
    std::cout << e() << " " << e(-1) << std::endl;
}

g++ 5.4 with -std=c++14 喜欢它，但是 clang++ 3.8 with -std=c++14 抱怨：

samename.cxx:3:23: error: called object type 'int' is not a function or function pointer
static int e(int d = d()) {return d;}
                     ~^

Answer 1

似乎不合法：请参阅 C++14 规范 (section 3.3.2 - Point of declaration)。

int x = 5;
{ int x = x; } // second x is initialized with its own undetermined state
{ int x[x]; }  // declares an array with 5 elements, since the declaration of 
               // second x is not complete when the first x is used

在你的例子中，d 的声明是完整的，所以当你使用 d() 时，你指的是变量，而不是函数。

Answer 2

来自basic.scope.pdecl/1：

The point of declaration for a name is immediately after its complete declarator and before its initializer (if any), except as noted below.

Example:
unsigned char x = 12;
{ unsigned char x = x; }
Here the second x is initialized with its own (indeterminate) value.

Clang 与标准的这一部分一致。 GCC 似乎只在 {block}.

中应用它

鉴于：

constexpr int d() {return 42;}

以下在 Clang 中失败，但在 GCC 中有效：

static int e(int d = d()) {return d;}
             // This ^^d should refer to the new name that shadows ::d()

以下在 Clang 和 GCC 中都失败了：

void func(){
    int d = d();
         // ^^d here refers to the new name that shadowed ::d();
 }

在 C++ 中，命名参数的范围是否包括其默认值的表达式？

In C++, does the scope of a named parameter include the expression for its default value?

c++

scope

overloading

default-value

c++14