64 位环境中的引用大小 [英] Size of references in 64bit environments

问题描述

在浏览关于 SO 的另一个问题的回复时遇到了这个问题(引用 Vs 变量获取).我的问题是，对于所有 64 位环境，是否保证对变量的引用是 64 位，即使原始大小较小?就像在 64 位环境中的 char 引用一样，>sizeof(char)?标准中是否有任何部分明确规定了这一点?

Came across this one while browsing the response to another question on SO (References Vs Variable Gets). My question is that for all 64bit environments is it guaranteed that a reference to a variable will be of 64 bits even if the original had a lesser size? As in char references in 64bit environment would be >sizeof(char)? Is there any section in the standard which specifies this explicitly?

为了更清楚——字符 c1 = 'a';字符&c2 = c1;我的问题是 64 位机器中的 sizeof(c2) > sizeof(c1)?

For more clarity -- char c1 = 'a'; char& c2 = c1; My question is sizeof(c2) > sizeof(c1) in 64bit machines?

推荐答案

The Standard (ISO C++-03) 说到以下关于引用的事情

The Standard (ISO C++-03) says the following thing about references

未指定引用是否需要存储(3.7).

如果我错了或者我没有正确理解他的问题，请有人纠正我.

Please someone correct me if I am wrong or if I have not understood his question correctly.

编辑:

我的问题是 64 位机器中的 sizeof(c2) > sizeof(c1)?

My question is sizeof(c2) > sizeof(c1) in 64bit machines?

不，因为 @Chubsdad 注意到 sizeof(c2) = sizeof (c1)，该标准的相关引用是

No, as @Chubsdad noticed sizeof(c2) = sizeof (c1), the relevant quote from the Standard is

应用于引用或引用类型时，结果是引用类型的大小.(ISO C++ $5.3.3/2)

When applied to a reference or a reference type, the result is the size of the referenced type. (ISO C++ $5.3.3/2)

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