在以下code运行时错误 [英] Runtime error in the following code

问题描述

以下code，据我要成功运行，但在runtime.I失败，没有得到的原因：

The following code,according to me should run successfully,but fails at runtime.I don't get the reason:

 void main()
 {
   int arr[5][3]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};
   int *m=arr[0];
   int **p=&m;
   p=p+1;
   printf("%d",**p);

  }

A.EXE已经停止在gcc编译运行工作，Windows 7的64位

a.exe has stopped working at runtime in gcc compiler,windows 7 64 bit

推荐答案

数组的数组和指针的指针是完全不同的，不能互换使用。

An array of arrays and a pointer to a pointer is quite different, and can't be used interchangeably.

例如，如果你看看你的阵列改编它看起来像这样在内存

For example, if you look at your array arr it looks like this in memory

+-----------+-----------+-----------+-----------+-----+-----------+
| arr[0][0] | arr[0][1] | arr[0][2] | arr[1][0] | ... | arr[4][2] |
+-----------+-----------+-----------+-----------+-----+-----------+

当你有指针到指针 P 节目真的不知道，它指向数组的数组，相反，它会被视为指针数组，它看起来像这样的记忆：

When you have the pointer-to-pointer p the program don't really knows that it points to an array of arrays, instead it's treated as an array of pointers, which looks like this in memory:

+------+------+------+-----+
| p[0] | p[1] | p[2] | ... |
+------+------+------+-----+
  |      |      |
  |      |      v
  |      |      something
  |      v
  |      something
  v
  something

所以，当你做 P + 1 你到 P [1] 这显然是不一样的改编[1] 。

So when you do p + 1 you get to p[1] which is clearly not the same as arr[1].

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