GDB debugger : char array type examine and print command

Question

In C program i have declared a buffer of characters: char buffer_in[500] ; When i run this program step by step on GDB i test the buffer reference with this commands:

(gdb) ptype buffer_in
type = char [500]
(gdb) ptype &buffer_in
type = char (*)[500]
(gdb) p &buffer_in
$9 = (char (*)[500]) 0x7fffffffdb60
(gdb) x buffer_in
0x7fffffffdb60: 0x2e
(gdb) x &buffer_in
0x7fffffffdb60: 0x2e

In C if I declared and array of characters the object is referenced like a pointer. I &buffer_in it is the address of first element of the array why the output of command x buffer_in is the same than x &buffer_in ?. I think that x buffer_in must trie to examine 0x2e address and so it is wrong referenced.

Thanks

Answer 1

So, gdb's x command expects a memory address - the command's purpose is to dump some memory in hex. If you give it an array variable, it will assume you mean you want it to dump starting at the address the array is stored at. If you give it a pointer to an array variable, it will assume you mean you want it to dump starting at that pointer. Those two are the same - this is much like the way C is actually compiled.

To put a finer point on it,

 printf("0x%8.8lX 0x%8.8lX\n", (unsigned long)buffer_in, (unsigned long)&buffer_in);

prints the same number twice. So you'd expect gdb to dump the same byte from memory when asked to dump each address expression.