Just curious, but would someone be able to explain to me this info frame output.
Here's a toy program:
#include <stdio.h>
int foo(int argc) {
printf("Hello world! %d\n", argc);
}
int main(int argc, char *argv[]) {
foo(argc);
return 0;
}
and here's the gdb info frame output when breaking at foo:
(gdb) info frame
Stack level 0, frame at 0x28abf0:
eip = 0x401196 in foo (a.c:4); saved eip 0x4011c4
called by frame at 0x28ac10
source language c.
Arglist at 0x28abe8, args: argc=1
Locals at 0x28abe8, Previous frame's sp is 0x28abf0
Saved registers:
ebp at 0x28abe8, eip at 0x28abec
(gdb) p &argc
$1 = (int *) 0x28abf0
(gdb)
Why are locals and arglist at the same location? As far as I understand,
Locals at should denote the current ebp value as the upper-bound address
for the locals of the current frame (which it does). But why does arglist
point to the same location? Based on printing &argc, the value for
Arglist at definitely doesn't seem to point to the arguments.
As an aside, I do understand the x86 calling convention and the structure
of the stack frames. It just seemed odd to me that all the online
info frames I could find had identical Arglist at and Locals at values,
but only this one lonely post about this inconsistency:
http://forums.devshed.com/programming-42/gdb-info-frames-arglist-locals-address-782598.html
At any rate, wondering to some degree whether there's a known reason for this, and/or how one actually goes about posting a gdb bug. Thanks!
