复制构造函数相关问题（本机 c++）
我知道复制构造函数必须有一个引用作为参数，以避免对其自身的“无限次调用”。我的问题是 - 为什么会发生这种情况，其背后的逻辑是什么？
CExample(const CExample& temp)
{
length = temp.length;
}
Possible Duplicate:
Why should the copy constructor accept its parameter by reference in C++?
i know that a copy constructor must have a reference as a parameter, to avoid an 'infinite number of calls' to itself. my question is - why exactly that happens, what is the logic behind it?
CExample(const CExample& temp)
{
length = temp.length;
}
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假设你的参数复制 C'tor 是按值传递的，C'tor 要做的第一件事就是复制参数 [这就是每个函数，包括构造函数对按值参数所做的事情]。为了做到这一点，它必须再次调用 C'tor，从原始变量到局部变量......[一遍又一遍......]这最终将导致无限循环。
assume your argument to the copy C'tor was passed by value, the first thing the C'tor would have done, was copying the argument [that's what every function, including constructors do with by-value arguments]. in order to do so, it would have to invoke the C'tor again, from the original to the local variable... [and over and over again...] which will eventually cause an infinite loop.
C++ 中有时会调用复制构造函数。其中之一是当您有一个像这样的函数时
，在这种情况下，当您调用
CExample::CExample时，会调用CExample::CExample来从x构造obj>。如果您具有以下签名
（请注意，
obj现在是通过引用传递的），则复制构造函数CExample::CExample不会 被调用。如果您的构造函数接受按值复制的对象（与第一个示例中的函数
f一样），编译器必须按顺序调用复制构造函数first创建一个副本（再次与函数f一样），但是...哎呀，我们必须调用复制构造函数才能调用复制构造函数。这听起来很糟糕，不是吗？Copy constructors are called on some occasions in C++. One of them is when you have a function like
In this case, when you make a call
CExample::CExamplegets called to constructobjfromx.In case you have the following signature
(note that
objis now passed by reference), the copy constructorCExample::CExampledoes not get called.In the case your constructor accepts the object to be copied by value (as with the function
fin the first example), compiler will have to call the copy constructor first in order to create a copy (as with the functionf, again), but... oops, we have to call the copy constructor in order to call the copy constructor. This sounds bad, doesn't it?参见此处
“这是一个参考，因为值参数需要制作一个副本，这将调用复制构造函数，这将复制其参数，这将调用复制构造函数，这......”
或者以不同的方式，构造函数的值参数必须调用构造函数复制参数中的值。这个新的构造函数需要做同样的事情 - 导致无限递归！
See here
"It is a reference because a value parameter would require making a copy, which would invoke the copy constructor, which would make a copy of its parameter, which would invoke the copy constructor, which ... "
Or put a different way a value parameter to the constructor would have to call the constructor to copy the value in the parameter. This new constructor would need to do the same - leading to infinite recursion!