C++ “对类型的非常量左值引用无法绑定”错误与引用（类型 &）但不是指针（类型 *）
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"non-const lvalue reference to type cannot bind" error with reference (Type &) but not with pointer (Type *)
提问by Alexis Cofion
I am getting this error "Non-const lvalue to type 'Cell' cannot bind to a temporary of type 'Cell *' with this code :
我收到此错误“输入'Cell'的非const左值无法使用以下代码绑定到'Cell *'类型的临时文件：
class RegionHolder
{
public:
RegionHolder(Region& Region1):m_RegionCellNOO(&(Region1.m_NOO))
~RegionHolder();
protected:
Cell & m_RegionCellNOO; // difference is here
};
but not with this one :
但不是这个：
class RegionHolder
{
public:
RegionHolder(Region& Region1):m_RegionCellNOO(&(Region1.m_NOO))
~RegionHolder();
protected:
Cell * m_RegionCellNOO; // difference is here
};
I don't understand the problem and would really like to use references and not pointers.
我不明白这个问题，真的很想使用引用而不是指针。
Thanks
谢谢
回答by Mike Seymour
You forgot to show us the definition, but presumably Region1.m_NOOis an object of type Cell. Your first example is taking the address of it, and trying to use the resulting pointer to initialise a reference. References aren't initialised from pointers, but from the object itself:
你忘了向我们展示定义，但大概Region1.m_NOO是一个类型的对象Cell。您的第一个示例是获取它的地址，并尝试使用结果指针来初始化引用。引用不是从指针初始化，而是从对象本身初始化：
RegionHolder(Region& Region1):m_RegionCellNOO(Region1.m_NOO) {}
// ^ no & ^^ don't forget that
There is one caveat with using references rather than pointers: they are not assignable, and so neither is your class. Often that's not a problem; but if you do need your class to be assignable, then you'll need to use a pointer instead.
使用引用而不是指针有一个警告：它们是不可赋值的，所以你的类也不是。通常这不是问题；但是如果你确实需要你的类是可分配的，那么你需要使用一个指针来代替。
回答by Hyman Aidley
The unary &operator gets a pointer to a variable. So, in C++ (as in C) a = &bgets a pointer to band stores this value in a, so if bis of type int, aneeds to be of type int*. Assignment to references, on the other hand, is implicit, so if a is of type int&you simply need to write a=bto make aa reference to b. So in your case, you need to rewrite your constructor as
一元运算&符获取指向变量的指针。因此，在 C++ 中（如在 C 中）a = &b获取指向b并将此值存储在 中的指针a，因此如果b是 type int，则a需要是 type int*。另一方面，对引用的赋值是隐式的，因此如果 a 是类型，int&您只需要编写a=b以a引用b. 因此，在您的情况下，您需要将构造函数重写为
RegionHolder(Region& Region1):m_RegionCellNOO(Region1.m_NOO) {}
However, I think you're better off using pointers than references here anywayand trying to use C++ without getting comfortable with pointers is a very bad idea. So I suggest you take the time to make yourself comfortable with using pointers instead of trying to avoid them.
但是，我认为无论如何在这里使用指针比使用引用更好，并且尝试使用 C++ 而不熟悉指针是一个非常糟糕的主意。所以我建议你花点时间让自己适应使用指针而不是试图避免它们。
