引用的引用地址是否与指针的地址相同？ [英] Is the address of a reference to a dereferenced pointer the same as the address of the pointer?

问题描述

在C ++中，引用的引用地址是否保证与指针的地址相同？

In C++, is the address of a reference to a dereferenced pointer guaranteed to be the same as the address of the pointer?

或者，下面的断言保证总是保持真的？

Or, written in code, is the following assertion guaranteed to always hold true?

SomeType *ptr = someAddress;
SomeType &ref = *ptr;
assert(&ref == ptr);

推荐答案

是的，当然， someAddress 不是空指针，否则不允许被取消引用。在这种情况下，行为是不确定的，虽然你的实现可能表现得好像是平等的，特别是在低优化级别。

Yes, provided of course that someAddress is not a null pointer, or otherwise not allowed to be dereferenced. In that case, behavior is undefined, although your implementation might well behave as though they are equal, especially with low optimization levels.

如果你想要精确， code>& ref 实际上不是引用的地址，它是引用的引用的地址。由于 ref 被绑定到 * ptr ，这意味着 ref 和 ptr 的referand（或pointee如果你喜欢）是同一个对象，因此两个地址& ref 和 ptr 是相等的。

If you want to be precise, then &ref isn't really the "address of a reference", it's the "address of the referand of a reference". Since ref was bound to *ptr, that means the referand of ref and the referand (or pointee if you prefer) of ptr are the same object, and hence the two addresses &ref and ptr are equal.

正如Bo指出的， c $ c>& ref with是指针的值或存储在指针中的地址，而不是指针的地址。

And as Bo points out, what you're comparing &ref with is the "value of the pointer", or the "address stored in the pointer", rather than "the address of the pointer".

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