c++ - 在使用相同名称的构造函数参数初始化成员变量时，=和{}之间的区别

Question

按照这个answer，为成员变量和构造函数参数使用相同的名称是可以的。因此，在进行了一些调试之后，我发现了以下代码为何不起作用的原因(该类的操作成员从未初始化过，从而导致switch语句失败)。

#include <cstdio>
#include <cstdlib>

enum class Operator {
    Add,
    Subtract,
    Multiply,
    Divide
};

class Calculator {
    Operator operation;
    public:
        Calculator(Operator operation) {
            operation = operation;
        }

        int calculate(int a, int b) {
            switch (operation) {
                case Operator::Add: {
                    return a+b;
                }
                case Operator::Subtract: {
                    return  abs(a-b);
                }
                case Operator::Multiply: {
                    return a*b;
                }
                case Operator::Divide: {
                    return a/b;
                }

            }
        }
};

int main() {
    Calculator calculator{Operator::Add};
    int a = 100;
    int b = 20;
    printf("%d + %d = %d\n", a, b, calculator.calculate(a, b));
}

用构造函数内部的 operation = operation替换 operation{operation}行解决了该问题，但是我仍然不明白为什么 operation = operation与 operation{operation}相比产生错误的结果，并且在构造函数初始化的上下文中两者之间有什么区别；

Answer 1

在两个标识符相同的特殊情况下，赋值和成员初始化之间的区别:
operation = operation;是从参数operation到参数operation的赋值。参数operation使成员operation黯然失色。

要解决此问题，您必须编写this->operation = operation;。

class Calculator {
    Operator operation;
    public:
        Calculator(Operator operation) {
            this->operation = operation;
        }
};

除此之外，成员初始化 operation(operation)或 operation{operation}是可取的。

class Calculator {
    Operator operation;
    public:
        Calculator(Operator operation):
          operation{operation}
        {
        }
};

之所以可行，是因为在成员初始化列表中，作用域还没有包含参数列表(类作用域)，但是成员初始化器中的表达式在作用域中使用参数(成员函数作用域)进行了解析。

经过草率的解释后，我从 cppreference.com 撤回了 Constructors and member initializer lists:

The body of a function definition of any constructor, before the opening brace of the compound statement, may include the member initializer list, whose syntax is the colon character :, followed by the comma-separated list of one or more member-initializers, each of which has the following syntax

---

class-or-identifier ( expression-list(optional) ) (1)

---

class-or-identifier brace-init-list (2) (since C++11)

---

parameter-pack ... (3) (since C++11)

---

1) Initializes the base or member named by class-or-identifier using direct initialization or, if expression-list is empty, value-initialization

2) Initializes the base or member named by class-or-identifier using list-initialization (which becomes value-initialization if the list is empty and aggregate-initialization when initializing an aggregate)

3) Initializes multiple bases using a pack expansion

• class-or-identifier - any identifier, class name, or decltype expression that names a non-static data member, a direct or virtual base, or (for delegating constructors) the class itself

• expression-list - possibly empty, comma-separated list of the parameters to pass to the constructor of the base or member

• braced-init-list - brace-enclosed list of comma-separated initializers and nested braced-init-lists

• parameter-pack - name of a variadic template parameter pack

即用于初始化

仅类标识符本身(用于委派构造函数调用)

基类标识符(用于基类构造函数调用)

和数据成员(用于成员初始化程序)

被考虑但不考虑参数(具有任何名称)。

在成员初始化器(括号或花括号)内，使用成员函数的常用名称解析，因此，参数可能会使具有相同名称的成员黯然失色。