C ++引用如何看，内存方面？ [英] How does a C++ reference look, memory-wise?

问题描述

给定：

int i = 42;
int j = 43;
int k = 44;

通过查看变量地址，我们知道每个变量占用4个字节（在大多数平台上）。

By looking at the variables addresses we know that each one takes up 4 bytes (on most platforms).

但是，考虑到：

int i = 42;
int& j = i;
int k = 44;

我们将看到变量 i 4字节，但 j 采用无和 k 再次在堆栈4个字节。

We will see that variable i indeed takes 4 bytes, but j takes none and k takes again 4 bytes on the stack.

这里发生了什么？看起来 j 在运行时是不存在的。而作为函数参数，我接收的引用呢？ 必须在堆栈上占用一些空间...

What is happening here? It looks like j is simply non-existent in runtime. And what about a reference I receive as a function argument? That must take some space on the stack...

当我们在它 - 为什么我不能定义一个数组或参考？

And while we're at it - why can't I define an array or references?

int&[] arr = new int&[SIZE]; // compiler error! array of references is illegal

推荐答案

j 时，会替换为我的地址。因此，基本上，引用内容地址在编译时被解析，并且不需要像在运行时指针那样去引用它。

everywhere the reference j is encountered, it is replaced with the address of i. So basically the reference content address is resolved at compile time, and there is not need to dereference it like a pointer at run time.

意味着地址i：

Just to clarify what I mean by the address of i :

void function(int& x)
{
    x = 10;
}

int main()
{
    int i = 5;
    int& j = i;

    function(j);
}

在上述代码中， j < 的 参数将在其堆栈上占用一个位置。也就是说，以 j 作为参数调用 function 时，函数强>。编译器可以且不应该为 j 在主栈上保留空间。

In the above code, j should not take space on the main stack, but the reference x of function will take a place on its stack. That means when calling function with j as an argument, the address of i that will be pushed on the stack of function. The compiler can and should not reserve space on the main stack for j.

say ::

C ++标准8.3.2 / 4：

没有对引用的引用，没有引用数组，
和没有指向引用的指针。

There shall be no references to references, no arrays of references, and no pointers to references.

为什么引用数组是非法的？

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