深入理解C/C++ 关键字const
1 常量函数则包含一个this的常量指针
这仅仅是编译器的限制，我们仍然可以绕过编译器的限制，去改变对象的状态，可以强制转换类型
Fred* pFred = (Fred*)this;
pFred->intValue = 50;
也可以构造另外一个指向同一块内存的指针去修改它的状态。
2 常量对象调用非常量函数，编译时会产生语法错误
3 C++也允许在数据成员的定义前面加上mutable，以允许该成员可以在常量函数中被修改
4 当存在重载函数时，常量对象调用常量成员；非常量对象调用非常量的成员。
| int x { 4 }; // initialize x with the value of 4 x = 5; // change value of x to 5 |
However, it’s sometimes useful to define variables with values that can not be changed. For example, consider the gravity of Earth (near the surface): 9.8 meters/second^2. This isn’t likely to change any time soon (and if it does, you’ve likely got bigger problems than learning C++). Defining this value as a constant helps ensure that this value isn’t accidentally changed.
To make a variable constant, simply put the const keyword either before or after the variable type, like so:
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const double gravity { 9.8 }; // preferred use of const before type int const sidesInSquare { 4 }; // okay, but not preferred |
Although C++ will accept const either before or after the type, we recommend using const before the type because it better follows standard English language convention where modifiers come before the object being modified (e.g. a “green ball”, not a “ball green”).
Const variables must be initialized when you define them, and then that value can not be changed via assignment.
Declaring a variable as const prevents us from inadvertently changing its value:
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const double gravity { 9.8 }; gravity = 9.9; // not allowed, this will cause a compile error |
Defining a const variable without initializing it will also cause a compile error:
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const double gravity; // compiler error, must be initialized upon definition |
Note that const variables can be initialized from other variables (including non-const ones):
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std::cout << "Enter your age: "; int age; std::cin >> age;
const int usersAge { age }; // usersAge can not be changed |
Const is often used with function parameters:
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void printInteger(const int myValue) { std::cout << myValue; } |
Making a function parameter const does two things. First, it tells the person calling the function that the function will not change the value of myValue. Second, it ensures that the function doesn’t change the value of myValue.
When arguments are passed by value, we generally don’t care if the function changes the value of the parameter (since it’s just a copy that will be destroyed at the end of the function anyway). For this reason, we usually don’t const parameters passed by value. But later on, we’ll talk about other kinds of function parameters (where changing the value of the parameter will change the value of the argument passed in). For these types of parameters, judicious use of const is important.
Runtime vs compile time constants
C++ actually has two different kinds of constants.
Runtime constants are those whose initialization values can only be resolved at runtime (when your program is running). Variables such as usersAge and myValue in the snippets above above are runtime constants, because the compiler can’t determine their initial values at compile time. usersAge relies on user input (which can only be given at runtime) and myValue depends on the value passed into the function (which is only known at runtime). However, once initialized, the value of these constants can’t be changed.
Compile-time constants are those whose initialization values can be resolved at compile-time (when your program is compiling). Variable gravity above is an example of a compile-time constant. Compile-time constants enable the compiler to perform optimizations that aren’t available with runtime constants. For example, whenever gravity is used, the compiler can simply substitute the identifier gravity for the literal double 9.8.
When you declare a const variable, the compiler will implicitly keep track of whether it’s a runtime or compile-time constant.
In most cases, this doesn’t matter, but there are a few odd cases where C++ requires a compile-time constant instead of a run-time constant (such as when defining the length of a fixed-size array -- we’ll cover this later).
constexpr
To help provide more specificity, C++11 introduced new keyword constexpr, which ensures that a constant must be a compile-time constant:
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constexpr double gravity { 9.8 }; // ok, the value of 9.8 can be resolved at compile-time constexpr int sum { 4 + 5 }; // ok, the value of 4 + 5 can be resolved at compile-time
std::cout << "Enter your age: "; int age; std::cin >> age; constexpr int myAge { age }; // not okay, age can not be resolved at compile-time |
Best practice
Any variable that should not be modifiable after initialization and whose initializer is known at compile-time should be declared as constexpr.
Any variable that should not be modifiable after initialization and whose initializer is not known at compile-time should be declared as const.
https://www.learncpp.com/cpp-tutorial/const-constexpr-and-symbolic-constants/
///
The const keyword allows you to specify whether or not a variable is modifiable. You can use const to prevent modifications to variables and const pointers and const references prevent changing the data pointed to (or referenced).
But why do you care?
Const gives you the ability to document your program more clearly and actually enforce that documentation. By enforcing your documentation, the const keyword provides guarantees to your users that allow you to make performance optimizations without the threat of damaging their data. For instance, const references allow you to specify that the data referred to won't be changed; this means that you can use const references as a simple and immediate way of improving performance for any function that currently takes objects by value without having to worry that your function might modify the data. Even if it does, the compiler will prevent the code from compiling and alert you to the problem. On the other hand, if you didn't use const references, you'd have no easy way to ensure that your data wasn't modified.
Documentation and Safety
The primary purpose of constness is to provide documentation and prevent programming mistakes. Const allows you to make it clear to yourself and others that something should not be changed. Moreover, it has the added benefit that anything that you declare const will in fact remain const short of the use of forceful methods (which we'll talk about later). It's particularly useful to declare reference parameters to functions as const references:
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Here, a myObj object is passed by reference into verifyObjectCorrectness. For safety's sake, const is used to ensure that verifyObjectCorrectness cannot change the object--after all, it's just supposed to make sure that the object is in a valid state. This can prevent silly programming mistakes that might otherwise result in damaging the object (for instance, by setting a field of the class for testing purposes, which might result in the field's never being reset). Moreover, by declaring the argument const, users of the function can be sure that their object will not be changed and not need to worry about the possible side effects of making the function call.
Syntax Note
When declaring a const variable, it is possible to put const either before or after the type: that is, both
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and
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result in x's being a constant integer. Note that in both cases, the value of the variable is specified in the declaration; there's no way to set it later!
Const Pointers
We've already seen const references demonstrated, and they're pretty natural: when you declare a const reference, you're only making the data referred to const. References, by their very nature, cannot change what they refer to. Pointers, on the other hand, have two ways that you can use them: you can change the data pointed to, or change the pointer itself. Consequently, there are two ways of declaring a const pointer: one that prevents you from changing what is pointed to, and one that prevents you from changing the data pointed to.
The syntax for declaring a pointer to constant data is natural enough:
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You can think of this as reading that *p_int is a "const int". So the pointer may be changeable, but you definitely can't touch what p_int points to. The key here is that the const appears before the *.
On the other hand, if you just want the address stored in the pointer itself to be const, then you have to put const after the *:
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Personally, I find this syntax kind of ugly; but there's not any other obviously better way to do it. The way to think about it is that "* const p_int" is a regular integer, and that the value stored in p_int itself cannot change--so you just can't change the address pointed to. Notice, by the way, that this pointer had to be initialized when it was declared: since the pointer itself is const, we can't change what it points to later on! Them's the rules.
Generally, the first type of pointer, where the data is immutable, is what I'll refer to as a "const pointer" (in part because it's the kind that comes up more often, so we should have a natural way of describing it).
Const Functions
The effects of declaring a variable to be const propagate throughout the program. Once you have a const object, it cannot be assigned to a non-const reference or use functions that are known to be capable of changing the state of the object. This is necessary to enforce the const-ness of the object, but it means you need a way to state that a function should not make changes to an object. In non-object-oriented code, this is as easy as using const references as demonstrated above.
In C++, however, there's the issue of classes with methods. If you have a const object, you don't want to call methods that can change the object, so you need a way of letting the compiler know which methods can be safely called. These methods are called "const functions", and are the only functions that can be called on a const object. Note, by the way, that only member methods make sense as const methods. Remember that in C++, every method of an object receives an implicit this pointer to the object; const methods effectively receive a const this pointer.
The way to declare that a function is safe for const objects is simply to mark it as const; the syntax for const functions is a little bit peculiar because there's only one place where you can really put the const: at the end of the function:
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For instance,
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Note that just because a function is declared const that doesn't prohibit non-const functions from using it; the rule is this:
- Const functions can always be called
- Non-const functions can only be called by non-const objects
That makes sense: if you have a const function, all that means is that it guarantees it won't change the object. So just because it is const doesn't mean that non-const objects can't use it.
As a matter of fact, const functions have a slightly stronger restriction than merely that they cannot modify the data. They must make it so that they cannot be used in a way that would allow you to use them to modify const data. This means that when const functions return references or pointers to members of the class, they must also be const.
Const Overloading
In large part because const functions cannot return non-const references to an objects' data, there are many times where it might seem appropriate to have both const and non-const versions of a function. For instance, if you are returning a reference to some member data (usually not a good thing to do, but there are exceptions), then you may want to have a non-const version of the function that returns a non-const reference:
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On the other hand, you don't want to prevent someone using a const version of your object,
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from getting the data. You just want to prevent that person from changing it by returning a const reference. But you probably don't want the name of the function to change just because you change whether the object is const or not--among other things, this would mean an awful lot of code might have to change just because you change how you declare a variable--going from a non-const to a const version of a variable would be a real headache.
Fortunately, C++ allows you to overload based on the const-ness of a method. So you can have both const and non-const methods, and the correct version will be chosen. If you wish to return a non-const reference in some cases, you merely need to declare a second, const version of the method that returns a const method:
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Const iterators
As we've already seen, in order to enforce const-ness, C++ requires that const functions return only const pointers and references. Since iterators can also be used to modify the underlying collection, when an STL collection is declared const, then any iterators used over the collection must be const iterators. They're just like normal iterators, except that they cannot be used to modify the underlying data. (Since iterators are a generalization of the idea of pointers, this makes sense.)
Const iterators in the STL are simple enough: just append "const_" to the type of iterator you desire. For instance, we could iterator over a vector as follows:
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Note that I used a const iterator to iterate over a non-const collection. Why do that? For the same reason that we normally use const: it prevents the possibility of silly programming mistakes ("oops, I meant to compare the two values, not assign them!") and it documents that we never intend to use the iterator to change the collection.
Const cast
Sometimes, you have a const variable and you really want to pass it into a function that you are certain won't modify it. But that function doesn't declare its argument as const. (This might happen, for instance, if a C library function like strlen were declared without using const.) Fortunately, if you know that you are safe in passing a const variable into a function that doesn't explicitly indicate that it will not change the data, then you can use a const_cast in order to temporarily strip away the const-ness of the object.
Const casts look like regular typecasts in C++, except that they can only be used for casting away constness (or volatile-ness) but not converting between types or casting down a class hierarchy.
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Note that you can also use const_cast to go the other way--to add const-ness--if you really wanted to.
Efficiency Gains? A note about Conceptual vs. Bitwise Constness
One common justification for const correctness is based on the misconception that constness can be used as the basis for optimizations. Unfortunately, this is generally not the case--even if a variable is declared const, it will not necessarily remain unchanged. First, it's possible to cast away constness using a const_cast. It might seem like a silly thing to do when you declare a parameter to a function as const, but it's possible. The second issue is that in classes, even const classes can be changed because of the mutable keyword.
Mutable Data in Const Classes
First, why would you ever want to have the ability to change data in a class that's declared const? This gets at the heart of what constness means, and there are two ways of thinking about it. One idea is that of "bitwise constness", which basically means that a const class should have exactly the same representation in memory at all times. Unfortunately (or fortunately), this is not the paradigm used by the C++ standard; instead, C++ uses "conceptual constness". Conceptual constness refers to the idea that the output of the const class should always be the same. This means that the underlying data might change as long as the fundamental behavior remains the same. (In essence, the "concept" is constant, but the representation may vary.)
Why have conceptual constness?
Why would you ever prefer conceptual constness to bitwise constness? One reason is efficiency: for instance, if your class has a function that relies on a value that takes a long time to calculate, it might be more efficient to calculate the value once and then store it for later requests. This won't change the behavior of the function--it will always return the same value. It will, however, change the representation of the class because it must have some place to cache the value.
C++ Support for Conceptual Constness
C++ provides for conceptual constness by using the mutable keyword: when declaring a class, you may specify that some of the fields are mutable:
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this will allow const functions to change the field regardless of whether or not the object itself was declared as const.
Other Ways of Achieving The Same Gains
If you were planning on using const to increase efficiency, think about what this would really mean--it would be akin to using the original data without making a copy of it. But if you wanted to do that, the simplest approach would just be to use references or pointers (preferably const references or pointers). This gives you a real efficiency gain without relying on compiler optimizations that probably aren't there.
Dangers of Too-much Constness
Beware of exploiting const too much; for instance, just because you can return a const reference doesn't mean that you should return a const reference. The most important example is that if you have local data in a function, you really ought not return a reference to it at all (unless it is static) since it will be a reference to memory that is no longer valid.
Another time when returning a const reference may not a good idea is when you are returning a reference to member data of an object. Although returning a const reference prevents anyone from changing the data by using it, it means that you have to have persistent data to back the reference--it has to actually be a field of the object and not temporary data created in the function. Once you make the reference part of the interface to the class, then, you fix the implementation details. This can be frustrating if you later wish to change your class's private data so the result of the function is computed when the function is invoked rather than actually be stored in the class at all times.
Summary
Don't look at const as a means of gaining efficiency so much as a way to document your code and ensure that some things cannot change. Remember that const-ness propagates throughout your program, so you must use const functions, const references, and const iterators to ensure that it would never be possible to modify data that was declared const.
https://www.cprogramming.com/tutorial/const_correctness.html
https://www.cnblogs.com/yc_sunniwell/archive/2010/07/14/1777416.html
为什么使用const？采用符号常量写出的代码更容易维护；指针常常是边读边移动，而不是边写边移动；许多函数参数是只读不写的。const最常见用途是作为数组的界和switch分情况标号(也可以用枚举符代替)，分类如下：
常变量： const 类型说明符 变量名
常引用： const 类型说明符 &引用名
常对象： 类名 const 对象名
常成员函数： 类名::fun(形参) const
常数组： 类型说明符 const 数组名[大小]
常指针： const 类型说明符* 指针名 ，类型说明符* const 指针名
首先提示的是：在常变量（const 类型说明符 变量名）、常引用（const 类型说明符 &引用名）、常对象（类名 const 对象名）、 常数组（类型说明符 const 数组名[大小]）， const” 与 “类型说明符”或“类名”（其实类名是一种自定义的类型说明符） 的位置可以互换。如：
const int a=5; 与 int const a=5; 等同
类名 const 对象名 与 const 类名 对象名 等同
用法1：常量
取代了C中的宏定义，声明时必须进行初始化(!c++类中则不然）。const限制了常量的使用方式，并没有描述常量应该如何分配。如果编译器知道了某const的所有使用，它甚至可以不为该const分配空间。最简单的常见情况就是常量的值在编译时已知，而且不需要分配存储。―《C++ Program Language》
用const声明的变量虽然增加了分配空间，但是可以保证类型安全。
C标准中，const定义的常量是全局的，C++中视声明位置而定。
用法2：指针和常量
使用指针时涉及到两个对象：该指针本身和被它所指的对象。将一个指针的声明用const“预先固定”将使那个对象而不是使这个指针成为常量。要将指针本身而不是被指对象声明为常量，必须使用声明运算符*const。
所以出现在 * 之前的const是作为基础类型的一部分：
char *const cp; //到char的const指针
char const *pc1; //到const char的指针
const char *pc2; //到const char的指针（后两个声明是等同的）
从右向左读的记忆方式：
cp is a const pointer to char. 故pc不能指向别的字符串，但可以修改其指向的字符串的内容
pc2 is a pointer to const char. 故*pc2的内容不可以改变，但pc2可以指向别的字符串
且注意：允许把非 const 对象的地址赋给指向 const 对象的指针,不允许把一个 const 对象的地址赋给一个普通的、非 const 对象的指针。
用法3：const修饰函数传入参数
将函数传入参数声明为const，以指明使用这种参数仅仅是为了效率的原因，而不是想让调用函数能够修改对象的值。同理，将指针参数声明为const，函数将不修改由这个参数所指的对象。
通常修饰指针参数和引用参数：
void Fun( const A *in); //修饰指针型传入参数
void Fun(const A &in); //修饰引用型传入参数
用法4：修饰函数返回值
可以阻止用户修改返回值。返回值也要相应的付给一个常量或常指针。
用法5：const修饰成员函数(c++特性)
const对象只能访问const成员函数，而非const对象可以访问任意的成员函数，包括const成员函数；
const对象的成员是不能修改的，而通过指针维护的对象确实可以修改的；
const成员函数不可以修改对象的数据，不管对象是否具有const性质。编译时以是否修改成员数据为依据进行检查。
具体展开来讲：
(一). 常量与指针
常量与指针放在一起很容易让人迷糊。对于常量指针和指针常量也不是所有的学习C/C++的人都能说清除。例如：
const int *m1 = new int(10);
int* const m2 = new int(20);
在上面的两个表达式中，最容易让人迷惑的是const到底是修饰指针还是指针指向的内存区域？其实，只要知道：const只对它左边的东西起作用，唯一的例外就是const本身就是最左边的修饰符，那么它才会对右边的东西起作用。根据这个规则来判断，m1应该是常量指针（即，不能通过m1来修改它所指向的内容。）；而m2应该是指针常量（即，不能让m2指向其他的内存模块）。由此可见：
1. 对于常量指针，不能通过该指针来改变所指的内容。即，下面的操作是错误的：
int i = 10;
const int *pi = &i;
*pi = 100;
因为你在试图通过pi改变它所指向的内容。但是，并不是说该内存块中的内容不能被修改。我们仍然可以通过其他方式去修改其中的值。例如：
// 1: 通过i直接修改。
i = 100;
// 2: 使用另外一个指针来修改。
int *p = (int*)pi;
*p = 100;
实际上，在将程序载入内存的时候，会有专门的一块内存区域来存放常量。但是，上面的i本身不是常量，是存放在栈或者堆中的。我们仍然可以修改它的值。而pi不能修改指向的值应该说是编译器的一个限制。
2. 根据上面const的规则，const int *m1 = new int(10);我们也可写作：
int const *m1 = new int(10);
这是，理由就不须作过多说明了。
3. 在函数参数中指针常量时表示不允许将该指针指向其他内容。
void func_02(int* const p)
{
int *pi = new int(100);
//错误！P是指针常量。不能对它赋值。
p = pi;
}
int main()
{
int* p = new int(10);
func_02(p);
delete p;
return 0;
}
4. 在函数参数中使用常量指针时表示在函数中不能改变指针所指向的内容。
void func(const int *pi)
{
//错误！不能通过pi去改变pi所指向的内容！
*pi = 100;
}
int main()
{
int* p = new int(10);
func(p);
delete p;
return 0;
}
我们可以使用这样的方法来防止函数调用者改变参数的值。但是，这样的限制是有限的，作为参数调用者，我们也不要试图去改变参数中的值。因此，下面的操作是在语法上是正确的，但是可能破还参数的值：
#include
#include
void func(const int *pi)
{
//这里相当于重新构建了一个指针，指向相同的内存区域。当然就可以通过该指针修改内存中的值了。
int* pp = (int*)pi;
*pp = 100;
}
int main()
{
using namespace std;
int* p = new int(10);
cout << "*p = " << *p << endl;
func(p);
cout << "*p = " << *p << endl;
delete p;
return 0;
}
(二)：常量与引用
常量与引用的关系稍微简单一点。因为引用就是另一个变量的别名，它本身就是一个常量。也就是说不能再让一个引用成为另外一个变量的别名, 那么他们只剩下代表的内存区域是否可变。即：
int i = 10;
// 正确：表示不能通过该引用去修改对应的内存的内容。
const int& ri = i;
// 错误！不能这样写。
int& const rci = i;
由此可见，如果我们不希望函数的调用者改变参数的值。最可靠的方法应该是使用引用。下面的操作会存在编译错误：
void func(const int& i)
{
// 错误！不能通过i去改变它所代表的内存区域。
i = 100;
}
int main()
{
int i = 10;
func(i);
return 0;
}
这里已经明白了常量与指针以及常量与引用的关系。但是，有必要深入的说明以下。在系统加载程序的时候，系统会将内存分为4个区域：堆区，栈区，全局区（静态）和代码区。从这里可以看出，对于常量来说，系统没有划定专门的区域来保护其中的数据不能被更改。也就是说，使用常量的方式对数据进行保护是通过编译器作语法限制来实现的。我们仍然可以绕过编译器的限制去修改被定义为“常量”的内存区域。看下面的代码：
const int i = 10;
// 这里i已经被定义为常量，但是我们仍然可以通过另外的方式去修改它的值。
// 这说明把i定义为常量，实际上是防止通过i去修改所代表的内存。
int *pi = (int*) &i;
局部常量存在哪个段呢？ 栈段？
(三)：常量函数
常量函数是C++对常量的一个扩展，它很好的确保了C++中类的封装性。在C++中，为了防止类的数据成员被非法访问，将类的成员函数分成了两类，一类是常量成员函数（也被称为观察着）；另一类是非常量成员函数（也被成为变异者）。在一个函数的签名后面加上关键字const后该函数就成了常量函数。对于常量函数，最关键的不同是编译器不允许其修改类的数据成员。例如：
class Test
{
public:
void func() const;
private:
int intValue;
};
void Test::func() const
{
intValue = 100;
}
上面的代码中，常量函数func函数内试图去改变数据成员intValue的值，因此将在编译的时候引发异常。
当然，对于非常量的成员函数，我们可以根据需要读取或修改数据成员的值。但是，这要依赖调用函数的对象是否是常量。通常，如果我们把一个类定义为常量，我们的本意是希望他的状态（数据成员）不会被改变。那么，如果一个常量的对象调用它的非常量函数会产生什么后果呢？看下面的代码：
class Fred{
public:
void inspect() const;
void mutate();
};
void UserCode(Fred& changeable, const Fred& unChangeable)
{
changeable.inspect(); // 正确，非常量对象可以调用常量函数。
changeable.mutate(); // 正确，非常量对象也允许修改调用非常量成员函数修改数据成员。
unChangeable.inspect(); // 正确，常量对象只能调用常理函数。因为不希望修改对象状态。
unChangeable.mutate(); // 错误！常量对象的状态不能被修改，而非常量函数存在修改对象状态的可能
}
从上面的代码可以看出，由于常量对象的状态不允许被修改，因此，通过常量对象调用非常量函数时将会产生语法错误。实际上，我们知道每个成员函数都有一个隐含的指向对象本身的this指针。而常量函数则包含一个this的常量指针。如下：
void inspect(const Fred* this) const;
void mutate(Fred* this);
也就是说对于常量函数，我们不能通过this指针去修改对象对应的内存块。但是，在上面我们已经知道，这仅仅是编译器的限制，我们仍然可以绕过编译器的限制，去改变对象的状态。看下面的代码：
class Fred{
public:
void inspect() const;
private:
int intValue;
};
void Fred::inspect() const
{
cout << "At the beginning. intValue = "<< intValue << endl;
// 这里，我们根据this指针重新定义了一个指向同一块内存地址的指针。
// 通过这个新定义的指针，我们仍然可以修改对象的状态。
Fred* pFred = (Fred*)this;
pFred->intValue = 50;
cout << "Fred::inspect() called. intValue = "<< intValue << endl;
}
int main()
{
Fred fred;
fred.inspect();
return 0;
}
上面的代码说明，只要我们愿意，我们还是可以通过常量函数修改对象的状态。同理，对于常量对象，我们也可以构造另外一个指向同一块内存的指针去修改它的状态。这里就不作过多描述了。
另外，也有这样的情况，虽然我们可以绕过编译器的错误去修改类的数据成员。但是C++也允许我们在数据成员的定义前面加上mutable，以允许该成员可以在常量函数中被修改。例如：
class Fred{
public:
void inspect() const;
private:
mutable int intValue;
};
void Fred::inspect() const
{
intValue = 100;
}
但是，并不是所有的编译器都支持mutable关键字。这个时候我们上面的歪门邪道就有用了。
关于常量函数，还有一个问题是重载。
#include
#include
using namespace std;
class Fred{
public:
void func() const;
void func();
};
void Fred::func() const
{
cout << "const function is called."<< endl;
}
void Fred::func()
{
cout << "non-const function is called."<< endl;
}
void UserCode(Fred& fred, const Fred& cFred)
{
cout << "fred is non-const object, and the result of fred.func() is:" << endl;
fred.func();
cout << "cFred is const object, and the result of cFred.func() is:" << endl;
cFred.func();
}
int main()
{
Fred fred;
UserCode(fred, fred);
return 0;
}
输出结果为：
fred is non-const object, and the result of fred.func() is:
non-const function is called.
cFred is const object, and the result of cFred.func() is:
const function is called.
从上面的输出结果，我们可以看出。当存在同名同参数和返回值的常量函数和非常量函数时，具体调用哪个函数是根据调用对象是常量对像还是非常量对象来决定的。常量对象调用常量成员；非常量对象调用非常量的成员。
总之，我们需要明白常量函数是为了最大程度的保证对象的安全。通过使用常量函数，我们可以只允许必要的操作去改变对象的状态，从而防止误操作对对象状态的破坏。但是，就像上面看见的一样，这样的保护其实是有限的。关键还是在于我们开发人员要严格的遵守使用规则。另外需要注意的是常量对象不允许调用非常量的函数。这样的规定虽然很武断，但如果我们都根据原则去编写或使用类的话这样的规定也就完全可以理解了。
(四)：常量返回值
很多时候，我们的函数中会返回一个地址或者引用。调用这得到这个返回的地址或者引用后就可以修改所指向或者代表的对象。这个时候如果我们不希望这个函数的调用这修改这个返回的内容，就应该返回一个常量。这应该很好理解，大家可以去试试。
+++++++++++++++++++++++++++++++++++++++
c++ 中const
+++++++++++++++++++++++++++++++++++++++
1. const常量，如const int max = 100;
优点：const常量有数据类型，而宏常量没有数据类型。编译器可以对前者进行类型安全检查，而对后者只进行字符替换，没有类型安全检查，并且在字符替换时可能会产生意料不到的错误（边际效应）
2. const 修饰类的数据成员。如：
class A
{
const int size;
…
}
const数据成员只在某个对象生存期内是常量，而对于整个类而言却是可变的。因为类可以创建多个对象，不同的对象其const数据成员的值可以不同。所以不能在类声明中初始化const数据成员，因为类的对象未被创建时，编译器不知道const 数据成员的值是什么。如
class A
{
const int size = 100; //错误
int array[size]; //错误，未知的size
}
const数据成员的初始化只能在类的构造函数的初始化表中进行。要想建立在整个类中都恒定的常量，应该用类中的枚举常量来实现。如
class A
{…
enum {size1=100, size2 = 200 };
int array1[size1];
int array2[size2];
}
枚举常量不会占用对象的存储空间，他们在编译时被全部求值。但是枚举常量的隐含数据类型是整数，其最大值有限，且不能表示浮点数。
3. const修饰指针的情况，见下式：
int b = 500;
const int* a = & [1]
int const *a = & [2]
int* const a = & [3]
const int* const a = & [4]
如果你能区分出上述四种情况，那么，恭喜你，你已经迈出了可喜的一步。不知道，也没关系，我们可以参考《Effectivec++》Item21上的做法，如果const位于星号的左侧，则const就是用来修饰指针所指向的变量，即指针指向为常量；如果const位于星号的右侧，const就是修饰指针本身，即指针本身是常量。因此，[1]和[2]的情况相同，都是指针所指向的内容为常量（const放在变量声明符的位置无关），这种情况下不允许对内容进行更改操作，如不能*a = 3；[3]为指针本身是常量，而指针所指向的内容不是常量，这种情况下不能对指针本身进行更改操作，如a++是错误的；[4]为指针本身和指向的内容均为常量。
4. const的初始化
先看一下const变量初始化的情况
1) 非指针const常量初始化的情况：A b;
const A a = b;
2) 指针const常量初始化的情况：
A* d = new A();
const A* c = d;
或者：const A* c = new A();
3）引用const常量初始化的情况：
A f;
const A& e = f; // 这样作e只能访问声明为const的函数，而不能访问一般的成员函数；
[思考1]： 以下的这种赋值方法正确吗？
const A* c=new A();
A* e = c;
[思考2]： 以下的这种赋值方法正确吗？
A* const c = new A();
A* b = c;
5. 另外const 的一些强大的功能在于它在函数声明中的应用。在一个函数声明中，const可以修饰函数的返回值，或某个参数；对于成员函数，还可以修饰是整个函数。有如下几种情况，以下会逐渐的说明用法：A&operator=(const A& a);
void fun0(const A* a );
void fun1( ) const; // fun1( ) 为类成员函数
const A fun2( );
1) 修饰参数的const，如 void fun0(const A* a ); void fun1(const A& a);
调用函数的时候，用相应的变量初始化const常量，则在函数体中，按照const所修饰的部分进行常量化，如形参为const A*a，则不能对传递进来的指针的内容进行改变，保护了原指针所指向的内容；如形参为const A&a，则不能对传递进来的引用对象进行改变，保护了原对象的属性。
[注意]：参数const通常用于参数为指针或引用的情况，且只能修饰输入参数;若输入参数采用“值传递”方式，由于函数将自动产生临时变量用于复制该参数，该参数本就不需要保护，所以不用const修饰。
[总结]对于非内部数据类型的输入参数，因该将“值传递”的方式改为“const引用传递”，目的是为了提高效率。例如，将void Func(A a)改为void Func(const A &a)
对于内部数据类型的输入参数，不要将“值传递”的方式改为“const引用传递”。否则既达不到提高效率的目的，又降低了函数的可理解性。例如void Func(int x)不应该改为void Func(const int &x)
2) 修饰返回值的const，如const A fun2( ); const A* fun3( );
这样声明了返回值后，const按照"修饰原则"进行修饰，起到相应的保护作用。const Rational operator*(const Rational& lhs, const Rational& rhs)
{
return Rational(lhs.numerator() * rhs.numerator(),
lhs.denominator() * rhs.denominator());
}
返回值用const修饰可以防止允许这样的操作发生:Rational a,b;
Radional c;
(a*b) = c;
一般用const修饰返回值为对象本身（非引用和指针）的情况多用于二目操作符重载函数并产生新对象的时候。
[总结]
1. 一般情况下，函数的返回值为某个对象时，如果将其声明为const时，多用于操作符的重载。通常，不建议用const修饰函数的返回值类型为某个对象或对某个对象引用的情况。原因如下：如果返回值为某个对象为const（const A test = A实例）或某个对象的引用为const（const A& test = A实例），则返回值具有const属性，则返回实例只能访问类A中的公有（保护）数据成员和const成员函数，并且不允许对其进行赋值操作，这在一般情况下很少用到。
2. 如果给采用“指针传递”方式的函数返回值加const修饰，那么函数返回值（即指针）的内容不能被修改，该返回值只能被赋给加const 修饰的同类型指针。如：
const char * GetString(void);
如下语句将出现编译错误：
char *str=GetString();
正确的用法是：
const char *str=GetString();
3. 函数返回值采用“引用传递”的场合不多，这种方式一般只出现在类的赙值函数中，目的是为了实现链式表达。如：
class A
{…
A &operate = (const A &other); //负值函数
}
A a,b,c; //a,b,c为A的对象
…
a=b=c; //正常
(a=b)=c; //不正常，但是合法
若负值函数的返回值加const修饰，那么该返回值的内容不允许修改，上例中a=b=c依然正确。(a=b)=c就不正确了。
[思考3]： 这样定义赋值操作符重载函数可以吗？
const A& operator=(const A& a);
6. 类成员函数中const的使用
一般放在函数体后，形如：void fun() const;
任何不会修改数据成员的函数都因该声明为const类型。如果在编写const成员函数时，不慎修改了数据成员，或者调用了其他非const成员函数，编译器将报错，这大大提高了程序的健壮性。如：
class Stack
{
public:
void Push(int elem);
int Pop(void);
int GetCount(void) const; //const 成员函数
private:
int m_num;
int m_data[100];
};
int Stack::GetCount(void) const
{
++m_num; //编译错误，企图修改数据成员m_num
Pop(); //编译错误，企图调用非const函数
Return m_num;
}
7. 使用const的一些建议
1) 要大胆的使用const，这将给你带来无尽的益处，但前提是你必须搞清楚原委；
2) 要避免最一般的赋值操作错误，如将const变量赋值，具体可见思考题；
3) 在参数中使用const应该使用引用或指针，而不是一般的对象实例，原因同上；
4) const在成员函数中的三种用法（参数、返回值、函数）要很好的使用；
5) 不要轻易的将函数的返回值类型定为const;
6) 除了重载操作符外一般不要将返回值类型定为对某个对象的const引用;
[思考题答案]
1) 这种方法不正确，因为声明指针的目的是为了对其指向的内容进行改变，而声明的指针e指向的是一个常量，所以不正确；
2) 这种方法正确，因为声明指针所指向的内容可变；
3) 这种做法不正确；
在const A::operator=(const A& a)中，参数列表中的const的用法正确，而当这样连续赋值的时侯，问题就出现了：
A a,b,c:
(a=b)=c;
因为a.operator=(b)的返回值是对a的const引用，不能再将c赋值给const常量。
++++++++++++++++++++++++++++++++++++++++
const 在c和c++中的区别
http://tech.e800.com.cn/articles/2009/722/1248229886744_1.html
++++++++++++++++++++++++++++++++++++++++
1. C++中的const正常情况下是看成编译期的常量,编译器并不为const分配空间,只是在编译的时候将期值保存在名字表中,并在适当的时候折合在代码中.所以,以下代码:
using namespace std;
int main()
{
const int a = 1;
const int b = 2;
int array[ a + b ] = {0};
for (int i = 0; i < sizeof array / sizeof *array; i++)
{
cout << array << endl;
}
}
在可以通过编译,并且正常运行.但稍加修改后,放在C编译器中,便会出现错误:
int main()
{
int i;
const int a = 1;
const int b = 2;
int array[ a + b ] = {0};
for (i = 0; i < sizeof array / sizeof *array; i++)
{
printf("%d",array);
}
}
错误消息:
c:\test1\te.c(8): error C2057: 应输入常数表达式
c:\test1\te.c(8): error C2466: 不能分配常数大小为 0 的数组
出现这种情况的原因是:在C中,const是一个不能被改变的普通变量,既然是变量,就要占用存储空间,所以编译器不知道编译时的值.而且,数组定义时的下标必须为常量.
2. 在C语言中: const int size; 这个语句是正确的，因为它被C编译器看作一个声明,指明在别的地方分配存储空间.但在C++中这样写是不正确的.C++中const默认是内部连接,如果想在C++中达到以上的效果,必须要用extern关键字.即C++中,const默认使用内部连接.而C中使用外部连接.
(1) 内连接:编译器只对正被编译的文件创建存储空间,别的文件可以使用相同的表示符或全局变量.C/C++中内连接使用static关键字指定.
(2) 外连接:所有被编译过的文件创建一片单独存储空间.一旦空间被创建,连接器必须解决对这片存储空间的引用.全局变量和函数使用外部连接.通过extern关键字声明,可以从其他文件访问相应的变量和函数.
/* C++代码 header.h */
const int test = 1;
/* C++代码 test1.cpp */
#include "header.h"
using namespace std;
int main() { cout << "in test1 :" << test << endl; }
/* C++代码 test2.cpp */
#include "header.h"
using namespace std;
void print() { cout << "in test2:" << test << endl;}
以上代码编译连接完全不会出问题,但如果把header.h改为:
extern const int test = 1;
在连接的时候,便会出现以下错误信息:
test2 error LNK2005: "int const test" (?test@@3HB) 已经在 test1.obj 中定义
因为extern关键字告诉C++编译器test会在其他地方引用,所以,C++编译器就会为test创建存储空间,不再是简单的存储在名字表里面.所以,当两个文件同时包含header.h的时候,会发生名字上的冲突.
此种情况和C中const含义相似:
/* C代码 header.h */
const int test = 1;
/* C代码 test1.c */
#include "header.h"
int main() { printf("in test1:%d\n",test); }
/* C代码 test2.c */
#include "header.h"
void print() { printf("in test2:%d\n",test); }
错误消息:
test3 fatal error LNK1169: 找到一个或多个多重定义的符号
test3 error LNK2005: _test 已经在 test1.obj 中定义
也就是说：在c++ 中const 对象默认为文件的局部变量。与其他变量不同，除非特别说明，在全局作用域声明的 const 变量是定义该对象的文件的局部变量。此变量只存在于那个文件中，不能被其他文件访问。通过指定 const 变更为 extern，就可以在整个程序中访问 const 对象：
// file_1.cc
// defines and initializes a const that is accessible to other files
extern const int bufSize = fcn();
// file_2.cc
extern const int bufSize; // uses bufSize from file_1
// uses bufSize defined in file_1
for (int index = 0; index != bufSize; ++index)
// ...
3. C++中,是否为const分配空间要看具体情况.如果加上关键字extern或者取const变量地址,则编译器就要为const分配存储空间.
4. C++中定义常量的时候不再采用define,因为define只做简单的宏替换，并不提供类型检查.