引用变量如何在内存中表示? [英] How is a reference variable represented in memory?

问题描述

int num = 0;
int *ptrNum = #
int &refNum = num;

在存储表1或表2中，引用变量的正确表示是哪一个?
如果是表2，那么为什么指针是对象而引用不是?
如果两种表示都不正确，请提供正确的表示，并解释为什么引用不是对象.

Which one is the correct representation of reference variable in the memory-table 1 or table 2?
If table 2,then why a pointer is an object and a reference isn't?
If both representations are incorrect then please provide a correct representation and an explanation for why a reference isn't an object.

推荐答案

在存储表1或表2中，引用变量的正确表示是哪一个?

Which one is the correct representation of reference variable in the memory-table 1 or table 2?

可以是，也可以不是.

在这种情况下，不需要存储对象的地址，因此表1就足够了.在另一个示例中，可能需要引用变量的地址.例如，当引用是非内联函数的参数时.

In this case, there is no need for storing the address of the object, so the table 1 would be enough. In another example, an address of the referenced variable could be needed. Such as when the reference is an argument of a non-inlined function.

请提供正确的表示形式

please provide a correct representation

C ++标准[dcl.ref]/4的草稿N4140:

draft N4140 of the C++ standard [dcl.ref]/4:

尚不确定引用是否需要存储

It is unspecified whether or not a reference requires storage

完全没有指定

如何进行存储.

How it could be stored is not specified at all.

为什么引用不是对象的说明.

an explanation for why a reference isn't an object.

同一文档将对象定义为

[对象简介]/1

...对象是存储区域...

... An object is a region of storage ...

如果未将引用定义为具有存储，则该引用也未定义为对象.

If a reference is not defined to have storage, then it's not defined to be an object.

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