c++ - 为什么默认参数不能依赖于非默认参数？

Question

考虑以下构造函数:

class MyClass {
    MyClass(unsigned int dimension, std::vector vector=unitaryVector(dimension));
};

其中 unitaryVector(d) 是一个返回 d 维随机 std::vector 的函数。

这会产生以下编译器错误:

error: default argument references parameter 'dimension'
    MyClass(unsigned int dimension, std::vector vector=unitaryVector(dimension));

为什么这个习语在 C++11 中无效？这似乎很明显:如果提供了 vector 参数，则将 vector 初始化为参数的拷贝，否则，调用函数并将其初始化为返回值的拷贝值(value)。为什么编译器不能理解这一点？

Answer 1

C++ 标准禁止这样做。

dcl.fct.default

9 default argument is evaluated each time the function is called with no argument for the corresponding parameter. A parameter shall not appear as a potentially-evaluated expression in a default argument. Parameters of a function declared before a default argument are in scope and can hide namespace and class member names.

[ Example:

int a;
int f(int a, int b = a);            // error: parameter a
                                    // used as default argument
typedef int I;
int g(float I, int b = I(2));       // error: parameter I found
int h(int a, int b = sizeof(a));    // OK, unevaluated operand

— end example ]

请注意，如果未提供默认参数，则在调用站点替换

Intro.execution (强调我的)

11: [ Note: The evaluation of a full-expression can include the evaluation of subexpressions that are not lexically part of the full-expression. For example, subexpressions involved in evaluating default arguments ([dcl.fct.default]) are considered to be created in the expression that calls the function, not the expression that defines the default argument. — end note ]

---

您可以简单地重载构造函数并委托(delegate)它:

class MyClass {
    explicit MyClass(unsigned int dimension) 
        : MyClass(dimension, unitaryVector(dimension))  //delegation
    {  }
    MyClass(unsigned int dimension, std::vector vector);
};

脚注:使单参数构造函数显式是一件好事