为什么一个指针的指针数组不兼容的指针？ [英] Why is a pointer to a pointer incompatible with a pointer to an array?

问题描述

好吧，我无法理解指针的指针VS指向数组的指针。
请看下面的code：

OK, I'm having trouble understanding pointers to pointers vs pointers to arrays. Consider the following code:

char s[] = "Hello, World";
char (*p1)[] = &s;
char **p2 = &s;
printf("%c\n", **p1); /* Works */
printf("%c\n", **p2); /* Segmentation fault */

为什么第一个printf的工作，而第二个不？

Why does the first printf work, while the second one doesn't?

据我了解，S是一个指向数组的第一个元素（即H）。
因此宣告P2为char **意味着它是一个指针指向一个字符。使得它指向的应该是合法的，因为S是一个指向一个char。因而取消引用它（即** P2）应该给H。但事实并非如此！

From what I understand, 's' is a pointer to the first element of the array (that is, 'H'). So declaring p2 as char** means that it is a pointer to a pointer to a char. Making it point to 's' should be legal, since 's' is a pointer to a char. And thus dereferencing it (i.e. **p2) should give 'H'. But it doesn't!

推荐答案

您误会出在什么取值是。它的不的指针：它是一个数组

Your misunderstand lies in what s is. It is not a pointer: it is an array.

现在在大多数情况下，取值计算为一个指向数组的第一个元素：相当于＆放大器; S [0] ，一个指向'H'。虽然这里最重要的事情是，你得到的指针值评估时，取值是一个临时的，短暂的价值 - 就像＆放大器; S [0] 。

Now in most contexts, s evaluates to a pointer to the first element of the array: equivalent to &s[0], a pointer to that 'H'. The important thing here though is that that pointer value you get when evaluating s is a temporary, ephemeral value - just like &s[0].

由于该指针是不是一个永久性的对象（它实际上不是什么存储在取值），你不能在这做一个指针到指针点。使用指针到指针，你必须有一个指针对象指向 - 例如，下面是确定：

Because that pointer isn't a permanent object (it's not actually what's stored in s), you can't make a pointer-to-pointer point at it. To use a pointer-to-pointer, you must have a real pointer object to point to - for example, the following is OK:

char *p = s;
char **p2 = &p;

如果您评估 * P2 ，你告诉编译器来加载东西 P2 点和把它当作一个指针到字符。这很好，当 P2 在没有实际指向一个指针到字符;但是当你做的char ** P2 =安培; S; 的东西 P2 点是不是一个指针在所有 - 它是一个数组（在这种情况下，它是13 字符 s的块）。

If you evaluate *p2, you're telling the compiler to load the thing that p2 points to and treat it as a pointer-to-char. That's fine when p2 does actually point at a pointer-to-char; but when you do char **p2 = &s;, the thing that p2 points to isn't a pointer at all - it's an array (in this case, it's a block of 13 chars).

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