= 和 {} 之间的区别，同时使用相同的构造函数参数名称初始化成员变量

Question

根据这个 answer 对成员变量和构造函数参数使用相同的名称是可以的。所以在做了一些调试之后，我想通了为什么下面的代码不起作用（class 的操作成员从未被初始化导致 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));
}

在构造函数中用 operation{operation} 替换行 operation = operation 解决了这个问题，但我仍然不明白为什么 operation = operation 与 operation{operation} 相比产生了错误的结果，什么是两者在构造函数初始化上下文中的区别；

Answer 1

两个标识符相同的特殊情况下赋值和成员初始化的区别：

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}更可取。

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

这是有效的，因为在成员初始化列表中，作用域（还）不包含参数列表（class 作用域）但是成员初始值设定项中的表达式在作用域中用参数解析（member功能范围）。

经过我草率的解释，我从cppreference.com:

中回忆起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

即对于初始值设定项

只有 class 标识符本身（对于委托构造函数调用）
base class 标识符（对于 base class 构造函数调用）
和数据成员（对于成员初始值设定项）

被考虑但不考虑参数（任何名称）。

在成员初始值设定项（圆括号或大括号）内，成员函数的常用名称解析适用，因此，参数可能会覆盖同名成员。

= 和 {} 之间的区别，同时使用相同的构造函数参数名称初始化成员变量

difference between = and {} while initialising member variables using the same name for constructor arguments

c++

constructor

initialization