blocks|key|1055747|text|因为如果不是通过引用，它就是通过值。为此，您需要创建一个副本，为此，您需要调用复制构造函数。但要做到这一点，我们需要创建一个新值，因此我们调用复制构造函数，依此类推……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1055748|(你会有无限的递归，因为“为了复制，你需要复制一份”。)|1055749|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Because if it's not by reference, it's by value. To do that you make a copy, and to do that you call the copy constructor. But to do that, we need to make a new value, so we call the copy constructor, and so on...

(You would have infinite recursion because "to make a copy, you need to make a copy".)

blocks|key|1270594|text|因为按值传递将调用复制构造函数:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1270595|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Because pass-by-value would invoke the copy constructor :)

blocks|key|1055804|text|如果通过值传递，它将是无限递归的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1055805|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It would be infinitely recursive if you passed it in by value

blocks|key|1270687|text|将对象作为引用传递是非常重要的。如果一个对象作为值传递给复制构造函数，那么它的复制构造函数将调用自身，将实际参数复制到形参。因此，将启动对复制构造函数的无休止的调用链。这个过程会一直持续到系统内存耗尽。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1270688|因此，在复制构造函数中，参数应始终作为引用传递。|1270689|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

It is very essential to pass objects as reference. If an object is passed as value to the Copy Constructor then its copy constructor would call itself, to copy the actual parameter to the formal parameter.
Thus an endless chain of call to the copy constructor will be initiated. This process would go on untill the system run out of memory.

Hence, in a copy constructor, the parameter should always be passed as reference.

blocks|key|1056010|text|如果它不是通过引用传递的，那么它将通过值传递。如果参数是通过值传递的，则其复制构造函数将调用自身将实际参数复制到形参。这个过程会一直持续到系统内存耗尽。因此，我们应该通过引用传递它，这样复制构造函数就不会被调用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1056011|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

If its not passed by reference then it would pass by value. If the argument is passed by value, its copy constructor would call itself to copy the actual parameter to formal parameter. This process would go on until the system runs out of memory.
So, we should pass it by reference , so that copy constructor does not get invoked.

blocks|key|1056063|text|有必要将对象作为引用传递，而不是通过值传递，因为如果您通过值传递它，它的副本是使用复制constructor.This构造的，这意味着复制构造函数将调用自己，使copy.This过程将继续进行，直到编译器耗尽内存。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1056064|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It is necessary to pass object as reference and not by value because if you pass it by value its copy is constructed using the copy constructor.This means the copy constructor would call itself to make copy.This process will go on until the compiler runs out of memory.

blocks|key|1056111|text|复制构造函数定义了复制的含义，因此如果我们只传递一个对象(我们将传递该对象的副本)，但为了创建副本，我们将需要一个复制构造函数，因此它会导致无限递归。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1056112|因此，复制构造函数必须有一个引用作为参数。|1056113|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

A copy constructor defines what copying means,So if we pass an object only (we will be passing the copy of that object) but to create the copy we will need a copy constructor, Hence it leads to infinite recursion.
So, A copy constructor must have a reference as an argument.

Why must a copy constructor's parameter be passed by reference?

Why should the copy constructor accept its parameter by reference in C++?

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为什么复制构造函数的参数必须通过引用传递？

问为什么复制构造函数应该在C++中通过引用接受它的参数？
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