C++ 如何从引用中获取指针？
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How to get a pointer from a reference?
提问by Oliver
There seems to be many relavent questions talking about pointer vs. reference, but I couldn't find what I want to know. Basically, an object is passed in by a reference:
似乎有很多关于指针与引用的相关问题，但我找不到我想知道的。基本上，对象是通过引用传入的：
funcA(MyObject &objRef) { ... }
Within the function, can I get a pointer to that object instead of the reference? If I treat the reference objRefas an alias to the MyObject, would &objRefactually give me a pointer to the MyObject? It doesn't seem likely. I am confused.
在函数中，我可以获得指向该对象的指针而不是引用吗？如果我将引用objRef视为 的别名MyObject，&objRef实际上会给我一个指向 MyObject 的指针吗？似乎不太可能。我很迷惑。
Edit: Upon closer examination, objRefdoes give me back the pointer to object that I need - Most of you gave me correct info/answer, many thanks. I went along the answer that seems to be most illustrative in this case.
编辑：经过仔细检查，objRef确实给了我指向我需要的对象的指针 - 你们中的大多数人给了我正确的信息/答案，非常感谢。在这种情况下，我采用了似乎最能说明问题的答案。
回答by Praetorian
Yes, applying the address-of operator to the reference is the same as taking the address of the original object.
是的，将 address-of 运算符应用于引用与获取原始对象的地址相同。
#include <iostream>
struct foo {};
void bar( const foo& obj )
{
std::cout << &obj << std::endl;
}
int main()
{
foo obj;
std::cout << &obj << std::endl;
bar( obj );
return 0;
}
Result:
结果：
0x22ff1f
0x22ff1f
回答by Joseph Mansfield
Any operator applied to a reference will actually apply to the object it refers to (§5/5 [expr]); the reference can be thought of as another name for the same object. Taking the address of a reference will therefore give you the address of the object that it refers to.
应用于引用的任何运算符实际上都将应用于它所引用的对象（§5/5 [expr]）；可以将引用视为同一对象的另一个名称。因此，获取引用的地址将为您提供它所引用的对象的地址。
It as actually unspecified whether or not a reference requires storage (§8.3.2/4 [dcl.ref]) and so it wouldn't make sense to take the address of the reference itself.
实际上未指定引用是否需要存储（§8.3.2/4 [dcl.ref]），因此获取引用本身的地址是没有意义的。
As an example:
举个例子：
int x = 5;
int& y = x;
int* xp = &x;
int* yp = &y;
In the above example, xpand ypare equal - that is, the expression xp == ypevaluates to truebecause they both point to the same object.
在上面的例子中，xpand ypare equal - 也就是说，表达式的xp == yp计算结果为，true因为它们都指向同一个对象。
回答by Kerrek SB
The general solution is to use std::addressof, as in:
一般的解决方案是使用std::addressof，如：
#include <type_traits>
void foo(T & x)
{
T * p = std::addressof(x);
}
This works no matter whether Toverloads operator&or not.
无论是否T过载，这都有效operator&。
回答by Ben Voigt
Use the address-of (&) operator on the reference.
&在引用上使用 address-of ( ) 运算符。
&objRef
Like any other operator used on a reference, this actually affects the referred-to object.
与在引用上使用的任何其他运算符一样，这实际上会影响被引用的对象。
As @Kerrek points out, since the operator affects the referred-to object, if that object has an overloaded operator&function, this will call it instead and std::address_ofis needed to get the true address.
正如@Kerrek 指出的那样，由于操作符会影响被引用的对象，如果该对象具有重载operator&函数，则会调用它，std::address_of并且需要获取真实地址。
回答by Drew Dormann
Use the address operator on the reference.
在引用上使用地址运算符。
MyObject *ptr = &objRef;
回答by damson
In C++, a reference is a restricted type of pointer. It can only be assigned once and can never have a NULL value. References are most useful when used to indicate that a parameter to a function is being Passed by Reference where the address of the variable is passed in. Without a Reference, Pass By Value is used instead.
在 C++ 中，引用是一种受限制的指针类型。它只能分配一次，并且永远不能有 NULL 值。当用于指示函数的参数通过引用传递，其中传递变量的地址时，引用最有用。如果没有引用，则使用传递值。
