如何在构造函数中初始化 const 字段? [英] How to initialize a const field in constructor?

问题描述

想象一下，我有一个 C++ 类 Foo 和一个类 Bar，它们必须用一个构造函数来创建，在该构造函数中传递了一个 Foo 指针，并且这个指针在 Bar 实例生命周期中保持不变.正确的做法是什么?

Imagine I have a C++ class Foo and a class Bar which has to be created with a constructor in which a Foo pointer is passed, and this pointer is meant to remain immutable in the Bar instance lifecycle. What is the correct way of doing it?

事实上，我以为我可以像下面的代码那样写，但它不能编译..

In fact, I thought I could write like the code below but it does not compile..

class Foo;

class Bar {
public:
    Foo * const foo;
    Bar(Foo* foo) {
        this->foo = foo;
    }
};

class Foo {
public:
  int a;
};

欢迎提出任何建议.

推荐答案

你需要在初始化列表中进行:

You need to do it in an initializer list:

Bar(Foo* _foo) : foo(_foo) {
}

(请注意，我重命名了传入变量以避免混淆.)

(Note that I renamed the incoming variable to avoid confusion.)

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