你用const能做到什么程度?你只是在必要的时候才把函数变成const，还是从头到尾到处都用它?例如，想象一个简单的变异子，它接受一个布尔参数:
void SetValue(const bool b) { my_val_ = b; }
这个const真的有用吗?就我个人而言，我选择广泛地使用它，包括参数，但在这种情况下，我想知道它是否值得?
我还惊讶地发现，你可以在函数声明中的形参中省略const，但可以在函数定义中包含它，例如:
. h文件
void func(int n, long l);
. cpp文件
void func(const int n, const long l)
这有什么原因吗?这对我来说有点不寻常。
当前回答
做一个VB。NET程序员需要使用具有50多个公开函数的c++程序，以及偶尔使用const限定符的.h文件，很难知道何时使用ByRef或ByVal访问变量。
当然，程序通过在出错的行上生成一个异常错误来告诉您，但是随后您需要猜测2-10个参数中哪一个是错误的。
所以现在我有一个令人讨厌的任务，试图说服开发人员，他们应该真正定义他们的变量(在.h文件中)，以一种允许自动创建所有VB的方法。NET函数定义容易。然后他们会自鸣得意地说:“读……文档”。
我写了一个awk脚本，它解析一个.h文件，并创建所有的Declare Function命令，但没有指示哪个变量是R/O vs R/W，它只完成了一半的工作。
编辑:
在另一位用户的鼓励下，我添加了以下内容;
下面是一个(IMO)格式不佳的.h条目的例子;
typedef int (EE_STDCALL *Do_SomethingPtr)( int smfID, const char* cursor_name, const char* sql );
从我的脚本的结果VB;
Declare Function Do_Something Lib "SomeOther.DLL" (ByRef smfID As Integer, ByVal cursor_name As String, ByVal sql As String) As Integer
注意，第一个参数中缺少“const”。如果没有它，程序(或其他开发人员)就不知道第一个参数应该传递“ByVal”。通过添加“const”，它使.h文件自文档化，以便使用其他语言的开发人员可以轻松地编写工作代码。
其他回答
当我以编写c++为生时，我尽可能地压缩了所有东西。使用const是一种帮助编译器帮助你的好方法。例如，const你的方法返回值可以避免你输入错误，例如:
foo() = 42
你的意思是:
foo() == 42
如果foo()被定义为返回一个非const引用:
int& foo() { /* ... */ }
编译器很乐意让您为函数调用返回的匿名临时对象赋值。使其为const:
const int& foo() { /* ... */ }
消除了这种可能性。
If the parameter is passed by value (and is not a reference), usually there is not much difference whether the parameter is declared as const or not (unless it contains a reference member -- not a problem for built-in types). If the parameter is a reference or pointer, it is usually better to protect the referenced/pointed-to memory, not the pointer itself (I think you cannot make the reference itself const, not that it matters much as you cannot change the referee). It seems a good idea to protect everything you can as const. You can omit it without fear of making a mistake if the parameters are just PODs (including built-in types) and there is no chance of them changing further along the road (e.g. in your example the bool parameter).
I didn't know about the .h/.cpp file declaration difference, but it does make some sense. At the machine code level, nothing is "const", so if you declare a function (in the .h) as non-const, the code is the same as if you declare it as const (optimizations aside). However, it helps you to enlist the compiler that you will not change the value of the variable inside the implementation of the function (.ccp). It might come handy in the case when you're inheriting from an interface that allows change, but you don't need to change to parameter to achieve the required functionality.
May be this wont be a valid argument. but if we increment the value of a const variable inside a function compiler will give us an error: "error: increment of read-only parameter". so that means we can use const key word as a way to prevent accidentally modifying our variables inside functions(which we are not supposed to/read-only). so if we accidentally did it at the compile time compiler will let us know that. this is specially important if you are not the only one who is working on this project.
我知道这个问题“有点”过时了，但当我遇到它时，其他人可能也会在未来这样做... ...我仍然怀疑这个可怜的家伙会在这里列出我的评论:)
It seems to me that we are still too confined to C-style way of thinking. In the OOP paradigma we play around with objects, not types. Const object may be conceptually different from a non-const object, specifically in the sense of logical-const (in contrast to bitwise-const). Thus even if const correctness of function params is (perhaps) an over-carefulness in case of PODs it is not so in case of objects. If a function works with a const object it should say so. Consider the following code snippet
#include <iostream>
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
class SharedBuffer {
private:
int fakeData;
int const & Get_(int i) const
{
std::cout << "Accessing buffer element" << std::endl;
return fakeData;
}
public:
int & operator[](int i)
{
Unique();
return const_cast<int &>(Get_(i));
}
int const & operator[](int i) const
{
return Get_(i);
}
void Unique()
{
std::cout << "Making buffer unique (expensive operation)" << std::endl;
}
};
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
void NonConstF(SharedBuffer x)
{
x[0] = 1;
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
void ConstF(const SharedBuffer x)
{
int q = x[0];
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
int main()
{
SharedBuffer x;
NonConstF(x);
std::cout << std::endl;
ConstF(x);
return 0;
}
附注:你可能会认为(const)引用在这里更合适，并提供相同的行为。嗯,对的。只是给出了与我在其他地方看到的不同的画面……
关于编译器优化:http://www.gotw.ca/gotw/081.htm
