指向2-D阵列的指针 [英] POINTER TO 2-D ARRAY

问题描述

int a[3][4]={1,2,3,4,5,6,7,8,9,10,11,12};
int (*ptr)[4]=a;

ptr和* ptr在这种情况下的含义是什么？为什么ptr和* ptr指向相同的meory地址

what do ptr and *ptr mean in this context? Why ptr and *ptr point to the same meory address

推荐答案

ptr 是一个变量，在这种情况下它声明为指向4个整数数组的指针。



要读取声明，请从变量名称开始（ ptr 在这种情况下）并在可能的情况下正确行动：

ptr is a variable, in this case it is declared as a pointer to an array of 4 integers.

To read the declaration, start with the name of the variable (ptr in this case) and work out, going right when possible:

int (*ptr)[4]=a;
      ptr

"ptr is..."

我们不能正确，因为我们首先需要一个匹配的支架，所以我们往左走

We can''t go right because we need a matching bracket first, so we go left

int (*ptr)[4]=a;
    (*ptr)

"ptr is a pointer to..."

现在我们已经关闭了括号，所以我们可以再次右转：

Now we have closed the brackets so we can go right again:

int (*ptr)[4]=a;
    (*ptr)[4]

"ptr is a pointer to an array of four...

等于结束声明，所以我们必须去再次离开

The equals ends teh declaration, so we have to go left again

int (*ptr)[4]=a;
int (*ptr)[4]=a;

"ptr is a pointer to an array of integers"

并且其余部分指定了一个合适的值。



所以当你在你的代码中稍后使用 ptr ，你指的是由三个四个整数数组组成的数组的实例。

当你引用 * ptr 在你的代码中你指的是四个整数的一个数组，因为*取消引用指针。





这有意义吗？

And the rest of teh line assigns a suitable value.

So when you use ptr later in your code, you are referring to the instance of an array of three arrays of four integers.
When you refer to *ptr in your code you are referring to one of the arrays of four integers, because the "*" dereferences the pointer.


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