为什么const变量必须立即初始化？ [英] Why do const variables have to be initialized right away?

问题描述

这是一个一般的程序设计问题。我正在学习C ++，我已经知道任何const变量，例如： const int i 或 int * const ptr ，必须立即初始化。

这也是引用地址必须立即初始化的根本原因，因为地址 const 。

但是我找不到为什么必须这样做的原因/

任何人都可以为我解释这个问题吗？

解决方案

可以初始化它，或在稍后分配一个值。

  const int size; //没有初始化（错误）
 
 size = 100; //错误 - 你不能分配一个const变量。 
  

现在如果一个变量既不具有任何有意义的值，也不允许它有值，因为它是一个 const 变量，那么这个变量的点是什么？这是完全无用的。

但是，这只适用于内置和POD类型：

  struct A {}; // POD type 
 struct B {B（）{}}; //非POD类型，因为它有用户定义的构造函数！ 
 
 const int i; // error  -  built-in type 
 const A a; //错误 -  POD类型
 const B b; // ok  -  Non POD type 
 
 //同样
 const std :: string s; // ok  -  std :: string是一个非POD 
 const std :: vector< std :: string> v; // ok  -  std :: vector是非POD 
  

实际上是一个NON-POD类型 b> b

现在考虑这个结构，

  struct C 
 {
 const int i ; 
 C（）{} 
}; 
  

C 绝对是非POD类型，因为它有用户定义的构造函数。另外请注意，在构造函数中，它不会初始化 i 这是 int 声明为 const 。因为这个未初始化的 const i，以下会产生错误：

  const C c; //错误 -  
  

可能会认为错误是因为 const 在上面的变量 c 中声明。但这是短视和不是真的。即使你删除 const ，它也会给出错误：

  C c ; //错误 - 相同的错误
  

错误是因为 C :: i

演示： =http://ideone.com/NJT8L> http://ideone.com/NJT8L

---

此分析还表明，即使内置类型是非POD类型的成员，内置类型也会不自动初始化 。

内置类型（和POD类型）的默认初始化语法，是这样的：

  struct C 
 {
 const int i; 
 C（）：i（）{} //注意语法 - 它被称为成员初始化列表
}; 
  

现在允许这样：

  C x; // ok 
 const C y; // ok 
  

演示： http://ideone.com/84vD9

---

至于使用struct / class POD，请参阅this topic：

• Can not C ++ POD type have any constructor？

This is a general programming question. I'm learning about C++ and I've learned that any const variables, ie: const int i, or int *const ptr, have to be initialized right away.

This is also the underlying reason that references to addresses must be initialized right away, because the addresses are const.

But I can't find the reason why this must be done / why this rule is imposed.

Can anyone explain this for me please?

解决方案

Because there is no way you can initialize it, or assigned with a value, later on.

const int size; //no initialization (error)

size = 100; //error - you cannot assign a const variable.

Now if a variable which is neither having any meaningful value, nor are you allowed to make it to have value later on because it is a const variable, then what is the point of such a variable? It is completely useless.

However, this is true for only built-in and POD types:

struct A{}; //POD type
struct B{ B(){} }; //Non POD type because it has user-defined constructor!

const int i; //error - built-in type
const A a;   //error - POD type
const B b;   //ok -    Non POD type

//likewise
const std::string s; //ok - std::string is a non-POD
const std::vector<std::string> v; //ok - std::vector is a non-POD

Actually a NON-POD type cannot remain uninitialized, because the default constructor will be called, and the object would get initialized.

---

Now consider this struct,

struct C
{
   const int i;
   C() {}
};

C is definitely a non-POD type, because it has user-defined constructor. Also note that in the constructor, it doesn't initialize i which is int, declared as const. Because of this uninitialized const i, the following would give error:

const C c; //error - 

One might think the error is because of const in the above declaration of variable c. But that is short-sightedness and is not true. Even if you remove const, it would give error:

C c; //error - same error

The error is because of C::i which is declared const but has not been initialized.

Demo : http://ideone.com/NJT8L

---

This analysis also demonstrates that built-in types do not get initialized automatically even if they're members of non-POD types. This is true of non-POD class types as well.

And the syntax to default initialization for built-in types (and POD types) is this:

struct C
{
    const int i;
    C() : i() {} //note the syntax - it is called member-initialization list
};

Now this is allowed :

C x; //ok
const C y; //ok

Demo : http://ideone.com/84vD9

---

As for what makes a struct/class POD, see this topic:

• Can't C++ POD type have any constructor?

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