Runtime error in the following code

问题

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

回答1:

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

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] |
+-----------+-----------+-----------+-----------+-----+-----------+

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

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

回答2:

With the line

int **p=&m

you create a pointer to a pointer to an integer.

Then, you add one to the address - one memory address, that is, not one times the number of bytes to point to the next integer.

Then you deference it twice:

both dereferences will return unspecified values, so the second dereference may break memory boundaries for the OS you are using,
both times it will be off boundary alignmemnt, which may cause issues in some OSes.

回答3:

int **p=&m;

p points to address, where m is placed:

... |  m  | sth | ... |  p  | ...
       ^                 V
       |_________________|

Now, increment it:

... |  m  | sth | ... |  p  | ...
             ^           V
             |___________|

So, now p points to sth. What is sth? Nobody knows. But you're trying to get access to the address sth contains. This is undefined behavior.

回答4:

Here int **p=&m;. p points to m. Then when p = p + 1; p will point to the address next to m (integer). That address may not be accessible.

来源：https://stackoverflow.com/questions/17808535/runtime-error-in-the-following-code

标签

runtime-error