问题描述
示例代码:
MyItemType a;
MyItemType b;
a.someNumber = 5;
b = a;
cout << a.someNumber << endl;
cout << b.someNumber << endl;
b.someNumber = 10;
cout << a.someNumber << endl;
cout << b.someNumber << endl;
输出:
5
5
5
10
如果a和b是引用类型，则最后两行应该是10和10，而不是我猜的5和10.
If a and b were reference types, the last 2 lines would have been 10 and 10 instead of 5 and 10 I guess.
这是否意味着您进行如下声明:
Does this mean when you do a declaration like this:
AClassType anInstance;
它被视为值类型吗?
------这里是MyItemType.h ------------
------Here is MyItemType.h------------
#ifndef MYITEMTYPE_H
#define MYITEMTYPE_H
class MyItemType{
public:
int someNumber;
MyItemType();
};
MyItemType::MyItemType(){
}
#endif /* MYITEMTYPE_H */
推荐答案
实际上，它不是像值类型那样对待.
It is not treated like a value type, in fact it is.
尽管在Java对象变量中存储了对对象的引用，但在C ++中，对象与其引用之间存在重要区别.默认情况下，分配实际上是按值分配的.
While in Java object variables store references to objects, in C++ there is an important difference between an object and its reference. Assignment is by default really by value.
如果希望变量只是一个引用，则可以使用引用或指针类型，具体取决于要使用的变量.这些类型声明为T*和T&.
If you want a variable to be just a reference, you use either a reference or a pointer type, depending what you want to with it. These types are declared T* and T&.
为了进一步说明这一点:
To illustrate this a little more:
在Java中，当您说MyClass obj时，会创建一个对象，但是引用/指针存储在变量obj中.
In Java, when you say MyClass obj, an object is created, but a reference/pointer is stored in the variable obj.
在C ++中，MyClass obj创建对象并将其存储在obj中.如果要使用引用/指针，则需要将变量明确声明为MyClass* objPointer或MyClass& objReference.
In C++, MyClass obj creates the object and will stored it in obj. If you want to work with references/pointers, you need to declare variables explicity as MyClass* objPointer or MyClass& objReference.
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