Re: gdb shows source line multiple times before executing !!

[Nikola Kolev <nak26@drexel.edu>]

why don't you take the optimization flags (-ON) out, live the debug flag 
(-g) in, recompile, fire up the debugger, do your stuff, instead of 
wasting time, energy, bandwidth, etc. to explain something that is 
obvious to why it's happening...

--Nik

mohanlal jangir wrote:

>>Because GCC splits the code for the line into multiple parts.  Part of
>>the line has executed when you step over it, but it may not be
>>immediately obvious how.  For instance, given:
>>  a = b + c;
>>the first time you step over it may load b and c from memory; the
>>second time may add them; and only the third will store the result into
>>a.
>>    
>>
>hmm... but sometimes it behaves even worse. For example there are two lines
>a = b + c;
>d = e + f;
>then first gdb shows line a = b + c; 
>
that's when it loded b and c from memory, and possibly stored their sum 
in a temporary register

>then d = e + f and then again a = b +
>c. And while showing d = e + f first time if I try to print value of a, it
>will be some uninitialised (that indicates that a = b + c has not been
>executed). 
>
It is uninitialized, because it has not assigned the the sum to a yet, 
but skipped to load e and f because it determined it will be faster that 
way (here's where optimization comes to play)...

>Then on next "next" command I will see line a = b +c again and
>then valute of a will be correct one.
>  
>
Well, that's when it assigned a with what you're looking for...