I know that gdb allows for an already declared variable to be set using the set command.
Is it possible for gdb to dynamically declare a new variable inside the scope of a given function?
3 Answers
You can dynamically allocate some space and use it to store a new variable. Depending on what you mean by "scope of the current function" it may not be what you want.
But here is how it looks like, when you have function func() that takes a pointer to an output parameter:
set $foo = malloc(sizeof(struct funcOutStruct))
call func($foo)
p *$foo
call (void) free($foo)
6 Comments
nullromo
In my case, calling
func($foo) from gdb did not work (Undefined command: "func".), but calling set $garbage = func($foo) worked. In addition, I had to use malloc instead of alloc.
Illia Bobyr
@nullromo Thanks - corrected the example, added
call - it executes an expression and discards the result. And fixed malloc.
aitzkora
It seems now you have to cast to void when a function such as
free is use like in ``` (gdb) call free(0) 'free' has unknown return type; cast the call to its declared return type (gdb) call (void) free(0) ```Illia Bobyr
@aitzkora Thank you, updated answer. What about the
call func($foo) part? It would be strange for gdb to require an explicit type to be specified there. But, at the same time, it is unclear to me, why it would need it for the free call.aitzkora
You right, I have also to add a cast (void) to call func($foo) if that function has no return type. I do not know if it is due to my gdb implementation, but it is also needed
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For C (and probably C++) code, that would be very hard, since doing so in most implementations would involve shifting the stack pointer, which would make the function's exit code fail due to it no longer matching the size of the stack frame. Also all the code in the function that accesses local variables would suddenly risk hitting the wrong location, which is also bad.
So, I don't think so, no.
1 Comment
r_2
Strictly speaking this is correct, it would be a pain to get the variables on the stack. But if you're ok with static variables, Ilya's solution works.
that's how I used to print variables
(gdb) set $path=((ngx_path_t **)ngx_cycle->paths.elts)[2]
(gdb) print *$path
$16 = {
name = {
len = 29,
data = 0x80ed15c "/usr/local/nginx/fastcgi_temp"
},
len = 5,
level = {1, 2, 0},
manager = 0,
loader = 0,
data = 0x0,
conf_file = 0x0,
line = 0
}
set $foo = ...and later reference$foo. Obviously such variables are in no way visible to the running code, however.