如何在构造函数中初始化一个const字段？ [英] How to initialize a const field in constructor?

问题描述

假设我有一个C ++类Foo和一个类Bar，它必须用一个构造函数创建，其中传递一个Foo指针，这个指针意味着在Bar实例生命周期中保持不变。什么是正确的做法？

事实上，我认为我可以写下面的代码，但它不编译..

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

欢迎任何建议。

解决方案

您需要在初始化器列表中执行：

  Bar（Foo * _foo） foo（_foo）{
} 
  

（注意，我重命名了传入变量以避免混淆。）

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?

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;
};

Any suggestion is welcome.

解决方案

You need to do it in an initializer list:

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

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

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