Why GDB jumps unpredictably between lines and prin

Question

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?

Answer 1

Declare found as "volatile". This should tell the compiler to NOT optimize it out.

volatile int found = 0;

Answer 2

Im using QtCreator with gdb.

Adding

QMAKE_CXXFLAGS += -O0
QMAKE_CXXFLAGS -= -O1
QMAKE_CXXFLAGS -= -O2
QMAKE_CXXFLAGS -= -O3

Works well for me

Answer 3

Typically, boolean values that are used in branches immediately after they're calculated like this are never actually stored in variables. Instead, the compiler just branches directly off the condition codes that were set from the preceding comparison. For example,

int a = SomeFunction();
bool result = --a >= 0; // use subtraction as example computation
if ( result ) 
{
   foo(); 
}
else
{
   bar();
}
return;

Usually compiles to something like:

call .SomeFunction  ; calls to SomeFunction(), which stores its return value in eax
sub eax, 1 ; subtract 1 from eax and store in eax, set S (sign) flag if result is negative
jl ELSEBLOCK ; GOTO label "ELSEBLOCK" if S flag is set
call .foo ; this is the "if" black, call foo()
j FINISH ; GOTO FINISH; skip over the "else" block
ELSEBLOCK: ; label this location to the assembler
call .bar
FINISH: ; both paths end up here
ret ; return

Notice how the "bool" is never actually stored anywhere.

Answer 4

Recompile without optimizations (-O0 on gcc).

Answer 5

When debugging optimized programs (which may be necessary if the bug doesn't show up in debug builds), you often have to understand assembly compiler generated.

In your particular case, return value of cpnd_find_exact_ckptinfo will be stored in the register which is used on your platform for return values. On ix86, that would be %eax. On x86_64: %rax, etc. You may need to google for '[your processor] procedure calling convention' if it's none of the above.

You can examine that register in GDB and you can set it. E.g. on ix86:

(gdb) p $eax
(gdb) set $eax = 0 

Answer 6

The compiler will start doing very clever things with optimisations turned on. The debugger will show the code jumping forward and backwards alot due to the optimized way variables are stored in registers. This is probably the reason why you can't set your variable (or in some cases see its value) as it has been cleverly distributed between registers for speed, rather than having a direct memory location that the debugger can access.

Compile without optimisations?