c++ - 函数参数默认值为 "passing_parametr"*2

Question

我写了以下内容:

#include <stdio.h>

int foo(int x, int y=2*x)
{
    return y;
}

int main()
{
    printf("%d\n",foo(5));
}

但是我有编译时错误error: local variable ‘x’ may not appear in this context但我希望它会好起来，因为 3.3.4/1:

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.

结束最近的封闭函数声明符是，声明点紧跟在int x之后。那么为什么它不起作用？

Answer 1

这不是因为范围。 8.3.6/7 表示

Local variables shall not be used in a default argument.

和8.3.6/9:

Default arguments are evaluated each time the function is called. The order of evaluation of function arguments is unspecified. Consequently, parameters of a function shall not be used in a default argument, even if they are not evaluated. Parameters of a function declared before a default argument are in scope and can hide namespace and class member names.

禁止这样做是有道理的，因为函数参数的计算顺序是未指定的。如果 2*x 在 x 的参数之前计算，y 的值是多少？