这个问题有很多很好的答案,包括前向声明、多态性等重要用例,但我觉得你的问题的“灵魂”部分没有得到回答,即Java和C++中不同的语法意味着什么。
让我们来比较一下这两种语言的情况:
Java语言:
Object object1 = new Object(); //A new object is allocated by Java
Object object2 = new Object(); //Another new object is allocated by Java
object1 = object2;
//object1 now points to the object originally allocated for object2
//The object originally allocated for object1 is now "dead" - nothing points to it, so it
//will be reclaimed by the Garbage Collector.
//If either object1 or object2 is changed, the change will be reflected to the other
与此最接近的等效值为:
C++:
Object * object1 = new Object(); //A new object is allocated on the heap
Object * object2 = new Object(); //Another new object is allocated on the heap
delete object1;
//Since C++ does not have a garbage collector, if we don't do that, the next line would
//cause a "memory leak", i.e. a piece of claimed memory that the app cannot use
//and that we have no way to reclaim...
object1 = object2; //Same as Java, object1 points to object2.
让我们看看另一种C++方式:
Object object1; //A new object is allocated on the STACK
Object object2; //Another new object is allocated on the STACK
object1 = object2;//!!!! This is different! The CONTENTS of object2 are COPIED onto object1,
//using the "copy assignment operator", the definition of operator =.
//But, the two objects are still different. Change one, the other remains unchanged.
//Also, the objects get automatically destroyed once the function returns...
最好的方法是——或多或少——Java(隐式)处理指向对象的指针,而C++可以处理指向对象或对象本身的指针。这是有例外的——例如,如果您声明Java“原始”类型,它们是复制的实际值,而不是指针。所以
Java语言:
int object1; //An integer is allocated on the stack.
int object2; //Another integer is allocated on the stack.
object1 = object2; //The value of object2 is copied to object1.
也就是说,使用指针不一定是正确或错误的处理方式;然而,其他答案已经令人满意地涵盖了这一点。不过,总的想法是,在C++中,您可以对对象的生存期以及它们将生存的位置进行更多的控制。
重点是——Object*Object=newObject()构造实际上最接近典型的Java(或C#)语义。