在C++中,命名参数的范围是否包含其默认值的表达式？

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 -std=c++14喜欢它,但clang ++ 3.8 -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

来自basic.scope.pdecl/1:

声明的声明就在其完整的声明者之后和初始化者之前(如果有的话),除非如下所述.

例:

unsigned char x = 12;
{ unsigned char x = x; }

这里第二个x用自己的(不确定的)值初始化.

Clang与标准的这一部分一致.海湾合作委员会似乎只适用于它{block}.

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();
 }

Answer 2

它似乎不合法:参见C++ 14规范(第3.3.2节 - 声明点).

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

在你的情况下,声明d是完整的,所以当你使用d()你指的是变量,而不是函数.