gdb: how does it know the variable type and size?
Question
I'm trying to figure this out as I'm trying to do the same thing (hopefully) with a home grown script:
Example C code:
typedef struct _B
{
A aa;
double b;
char c[LEN];
int d;
char *a_ptr[10];
} B;
B this_b;
If I compile this with gcc -g
and gdb a.out
afterwards, gdb
knows exactly what and where a_ptr
is:
(gdb) p &(this_b.a_ptr)
$1 = (char *(*)[10]) 0x804a084
how does it do that? And can I do the same thing (knowing it's address and type) through other utilities?
Solution
When you build with the -g
flag, GCC (and most other compilers) stores additional "debugging info" in your binary (a.out
).
You can examine that info with tools other than GDB. For example, readelf -w a.out
(assuming you are on Linux or another ELF
platform).
You can also compare the size of a.out
when built with and without -g
. It is not uncommon for the debug binary to be 5 to 10 times larger.
OTHER TIPS
This info is known on compile time. Gcc gathers it and stores it to be used later by different tools (gdb in this case).