运算符的地址是否返回变量引用的值的地址 [英] Does address-of operator return address of value referenced by the variable

问题描述

我正在阅读本章有关指针的内容，其中指出:

I'm reading this chapter on pointers and it states the following:

变量的地址可以通过在变量名前面加上 带有&"号的变量，称为地址运算符.为了 例如:

The address of a variable can be obtained by preceding the name of a variable with an ampersand sign (&), known as address-of operator. For example:

int myvar = 23...;
foo = &myvar;

我的理解是否正确，它是对象new MyStruct()的地址，而不是变量本身?我对变量(而不是它们引用的对象)的存储方式还没有很好的了解，很可能在编译程序时根本不使用它们.

Is my understanding correct that it's the address of an object new MyStruct(), not the variable itself? I don't yet have good understanding of how variables (not objects they reference) are stored, and it's very likely that they are not used at all when program is compiled.

推荐答案

没有对象23...，它是一个值.有一个名为myvar的对象"，它有一个地址.如果您为变量分配新值，则该地址不会更改，这是变量本身的属性.

There is no object 23..., that is a value. There is an "object" named myvar, and it has an address. That address doesn't change if you assign a new value to the variable, it's a property of the variable itself.

int var = 23;
int *foo = &var;
var = 35; // foo isn't changed here, it holds the same address.

*foo = 42; // again, foo itself isn't modified

assert(var == 42); // But we modified var through the address in foo

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作为旁注，因为您还标记了此C ++.我只想提一下，由于C ++允许您为自己的类重载运算符(包括&)，因此整个混合过程中增加了一定的复杂性.但是对于原始变量，使用&将始终产生该变量的地址.

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As a side note, because you also tagged this C++. I'll just mention that there is certain complexity added to the whole mix, due to the fact C++ allows you to overload operators (among them &) for your own classes. But for primitive variables, using & will always yield the address of said variable.

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