Can anyone explain this behavior of gdb?
900 memset(&new_ckpt_info,'\0',sizeof(CKPT_INFO));
(gdb)
**903 prev_offset = cp_node->offset;**
(gdb)
**905 m_CPND_CKPTINFO_READ(ckpt_info,(char *)cb->shm_addr.ckpt_addr+sizeof(CKPT_** HDR),i_offset);
(gdb)
**903 prev_offset = cp_node->offset;**
(gdb)
**905 m_CPND_CKPTINFO_READ(ckpt_info,(char *)cb->shm_addr.ckpt_addr+sizeof(CKPT_ HDR),i_offset);**
(gdb)
**908 bitmap_offset = client_hdl/32;**
(gdb)
**910 bitmap_value = cpnd_client_bitmap_set(client_hdl%32);**
(gdb)
**908 bitmap_offset = client_hdl/32;**
(gdb)
**910 bitmap_value = cpnd_client_bitmap_set(client_hdl%32);**
(gdb)
**908 bitmap_offset = client_hdl/32;**
(gdb)
**910 bitmap_value = cpnd_client_bitmap_set(client_hdl%32);**
(gdb)
913 found = cpnd_find_exact_ckptinfo(cb , &ckpt_info , bitmap_offset , &offset , &prev_offset);
(gdb)
916 if(!found)
(gdb) p found
$1 = <value optimized out>
(gdb) set found=0
Left operand of assignment is not an lvalue.
Why after executing line 903 it again executes the same for 905 908 910?
Another things is found is a bool-type variable, so why it is showing value optimized out?
I am not able to set the value of found as well.
This seems to be a compiler optimization (in this case its -O2); how can I still set the value of found?
To debug optimized code, learn assembly/machine language.
Use the GDB TUI mode. My copy of GDB enables it when I type the minus and Enter. Then type C-x 2 (that is hold down Control and press X, release both and then press 2). That will put it into split source and disassembly display. Then use stepi and nexti to move one machine instruction at a time. Use C-x o to switch between the TUI windows.
Download a PDF about your CPU's machine language and the function calling conventions. You will quickly learn to recognize what is being done with function arguments and return values.
You can display the value of a register by using a GDB command like p $eax