没有一个二维数组衰变的指针的指针 [英] Doesn't a 2D array decay to pointer to pointer

问题描述

到现在为止我是pretty多少确定

Up till now I was pretty much sure that

int arr[4][5];

然后改编将衰减的指针的指针。

Then arr will decay to pointer to pointer.

但此链接证明我是错的。

我不知道我怎么得到关于改编是指针的指针，但它似乎pretty明显，我。因为改编[I] 将是一个指针，因此改编应该是一个指针的指针。

I am not sure how did I get about arr being pointer to pointer but it seemed pretty obvious to me. Because arr[i] would be a pointer, and hence arr should be a pointer to pointer.

难道我错过了一些东西。

Am I missing out on something.

推荐答案

是的，你是在很多错过了：）

Yep you are missing out on a lot :)

要避免文本的另一面墙上，我会链接到<一个href=\"http://stackoverflow.com/questions/22983734/multiarray-and-multiarray0-and-multiarray0-same/22984784#22984784\">an回答我今天早些时候写道解释多维数组。

To avoid another wall of text I'll link to an answer I wrote earlier today explaining multi-dimensional arrays.

考虑到这一点，改编是4个元素，每一个都是5 INT 秒。
当一个前$ P $以外＆放pssion使用，改编或 sizeof的改编，这个衰减到＆放大器;常用3 [0] 。但是，什么是＆放大器;常用3 [0] ？它是一个指针，并且重要的是，一个右值

With that in mind, arr is a 1-D array with 4 elements, each of which is an array of 5 ints. When used in an expression other than &arr or sizeof arr, this decays to &arr[0]. But what is &arr[0]? It is a pointer, and importantly, an rvalue.

由于＆放大器; ARR [0] 是一个指针，它不能进一步衰减。 （数组衰变，指针不会）。此外，它是一个右值。即使它可能衰变成一个指针，哪来的指针呢？你不能在一个右值点。 （什么是＆放大器;？（X + Y））

Since &arr[0] is a pointer, it can't decay further. (Arrays decay, pointers don't). Furthermore, it's an rvalue. Even if it could decay into a pointer, where would that pointer point? You can't point at an rvalue. (What is &(x+y) ? )

看它的另一种方式是记住 INT ARR [4] [5]; 20的连续集团 INT S，分为4个标段的5编译器的脑海中，但在运行时内存中没有特殊标记。

Another way of looking at it is to remember that int arr[4][5]; is a contiguous bloc of 20 ints, grouped into 4 lots of 5 within the compiler's mind, but with no special marking in memory at runtime.

如果有双重衰退，那么会是什么 INT ** 点？它必须指向一个为int * 定义。但在内存中是为int * ？当然也有不是一堆指针徘徊在内存刚刚在这种情况发生时的情况下。

If there were "double decay" then what would the int ** point to? It must point to an int * by definition. But where in memory is that int * ? There are certainly not a bunch of pointers hanging around in memory just in case this situation occurs.

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