Difference between pointer and array -- Sun interview

If you can understand this interview question, you can basically understand the difference between the pointer and the array name.
In addition, it's okay to repost an explanation. I am too lazy to write it myself. If you want to know the details, read "C expert programming".
======================
/* File1.c */
Char buff [128];
 
/* File2.c */
Extern char * Buff;/* notice: extern char buff [] */
 
Int main ()
{
* Buff = 'a ';
Return 0;
}
 
The result is as follows:
$ GCC file1.c-C
$ GCC file2.c-C
$ GCC file1.o file2.o-o ff
$./FF
Segmentation fault
 
Q: Why is this result?
 
I tried it. Replace it with extern char buff [] In file2,
But I don't know why ..
 
 
An explanation on the Forum:
---------------------
Arrays and pointers are completely different.
The value of the array element is placed in the array, and the pointer is an address.
The compiler often treats arrays as pointers for processing convenience.
For example:
Int Buf [5] = {0 };
Int * P = Buf;
At this time, the compiler knows that Buf is an array.
Therefore, the array address instead of its content will be passed to the P pointer.
However, if the array is in another compilation unit (the compiler processes each compilation unit independently)
Pointer P. It does not know that it points to an array.
No Buf is an array information during the link.
So at the end of the link, P links the Buf content rather than its address.
 
My experiment results in Linux
---------------------------------
In file1.c
Char Buf [128] = {0x10, 0x20, 0x30, 0x40 };
 
In file2.c
Extern int Buf // note that the char Buf in file1.c is forcibly converted to an integer.
Int main ()
{
Printf ("Buf = % x/N", Buf );
}
The output result is
Buf = 40302010