当已经声明** p时，* p是什么意思

问题描述

代码

Code

short **p = (short **)malloc(sizeof(short *));
*p = malloc(sizeof(short));
**p = 10;
printf("**p = %d", **p);

输出

在此代码中，声明了多指针**p，并且使用了*p而没有任何声明(可能是**p).

In this code, a multiple pointer **p is declared and *p is used without any declaration(maybe it's by **p).

在我的情况下，*p是什么意思?对不起，一个非常简单的问题.

What does *p mean in my case? Sorry for very simple question.

我看到了C标准和堆栈溢出，但是我找不到东西.

I saw C standard and stack overflow, but I couldn't find out something.

推荐答案

对于任何数组或指针p和索引i，表达式p[i]完全等于*(p + i)(其中*是一元解引用运算符，它在指针上的结果就是指针指向的值.

For any array or pointer p and index i, the expression p[i] is exactly equal to *(p + i) (where * is the unary dereference operator, the result of it on a pointer is the value that the pointer is pointing to).

因此，如果我们有p[0]，则它精确等于*(p + 0)，等于*(p)，等于*p.从此倒退，*p等于p[0].

So if we have p[0] that's then exactly equal to *(p + 0), which is equal to *(p) which is equal to *p. Going backwards from that, *p is equal to p[0].

所以

*p = malloc(sizeof(short));

等于

p[0] = malloc(sizeof(short));

还有

**p = 10;

等于

p[0][0] = 10;

(**p等于*(*(p + 0) + 0)等于*(p[0] + 0)等于p[0][0])

请务必注意，星号*在不同的上下文中可能表示不同的意思.

It's important to note that the asterisk * can mean different things in different contexts.

可以在声明变量时使用，然后表示声明为指针":

It can be used when declaring a variable, and then it means "declare as pointer":

int *p;  // Declare p as a pointer to an int value

它可以用来取消引用一个指针，以获取该指针指向的值:

It can be used to dereference a pointer, to get the value the pointer is pointing to:

*p = 0;  // Equal to p[0] = 0

它可以用作乘法运算符:

And it can be used as the multiplication operator:

r = a * b;  // Multiply the values in a and b, store the resulting value in r

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