Still cannot print variable

Still cannot print variable

[JS]

I am still have problems printing a variable. I have this code in a file
called test.:

int main()
{

        int d = 119;

        return(0);

}

I then compile it like this:

        gcc -g -Wall -o test test.c

(I have also tried with option -O0 but that changes nothing).

I run "gdb test" followed by "print d" and then I get this error:

No symbol "d" in current context.

Why is it not possible to see the contents of "d" when debugging?

Re: Still cannot print variable

[Konstantin Karganov]

Hello JS,

> I run "gdb test" followed by "print d" and then I get this error:

> No symbol "d" in current context.

> Why is it not possible to see the contents of "d" when debugging?

Because you are not yet debugging!
As described above, you started the debugger but not started the
program being debugged.
Try this:
 gdb test
 break main            // to stop your program run
 run
 print d

-- 
Best regards,
 Konstantin 

Re: Still cannot print variable

[JS]

Konstantin Karganov wrote:

> Hello JS,
> 
>> I run "gdb test" followed by "print d" and then I get this error:
> 
>> No symbol "d" in current context.
> 
>> Why is it not possible to see the contents of "d" when debugging?
> 
> Because you are not yet debugging!
> As described above, you started the debugger but not started the
> program being debugged.
> Try this:
>  gdb test
>  break main            // to stop your program run
>  run
>  print d
> 


Ok now it work but when I write "print d" I get:

$1 = 134513410

But I declared d to be 119.

I have tried the different print arguments:

x
d
u
o
etc. but nothig seems to print 119.

Re[2]: Still cannot print variable

[Konstantin Karganov]

Hello JS,

>> Try this:
>>  gdb test
>>  break main            // to stop your program run
>>  run
>>  print d
>> 
> Ok now it work but when I write "print d" I get:

> $1 = 134513410

> But I declared d to be 119.
Sure! Because you stand before the initialization.
Try to do "step" or "next" and after that you'll see your variable
initialized.

-- 
Best regards,
 Konstantin 

Re: Re[2]: Still cannot print variable

[JS]

Konstantin Karganov wrote:

> Hello JS,
> 
>>> Try this:
>>>  gdb test
>>>  break main            // to stop your program run
>>>  run
>>>  print d
>>> 
>> Ok now it work but when I write "print d" I get:
> 
>> $1 = 134513410
> 
>> But I declared d to be 119.
> Sure! Because you stand before the initialization.
> Try to do "step" or "next" and after that you'll see your variable
> initialized.
> 

Yes I just read Elis reply...should have done that before posting :-)

Re: Still cannot print variable

[Joakim Hove]

JS <fedevaps@yahoo.dk> writes:


> I run "gdb test" followed by "print d" and then I get this error:

You must explicitly start test with the run command from gdb:

bash% gdb test
<...>
(gdb) run


However, that is also not enough because the current program will
immediately complete, and hence when you get back to the (gdb) prompt
the context of the (no longer) running program is void. Hence you must
set a breakpoint to halt the execution:


bash% gdb test
<...>
(gdb) break test.c:5    # Break at line 5 in file test.c
(gdb) run
(gdb) print d



HTH - Joakim

-- 
Joakim Hove
hove AT ift uib no                 /    
Tlf: +47 (55 5)8 27 90            /     Stabburveien 18		 
Fax: +47 (55 5)8 94 40           /      N-5231 Paradis		 
http://www.ift.uib.no/~hove/    /      	55 91 28 18 / 92 68 57 04

Re: Still cannot print variable

[Eli Zaretskii]

> From: JS <fedevaps@yahoo.dk>
> Date: Fri, 15 Apr 2005 11:09:09 +0200
> 
> I am still have problems printing a variable. I have this code in a file
> called test.:
> 
> int main()
> {
> 
>         int d = 119;
> 
>         return(0);
> 
> }
> 
> I then compile it like this:
> 
>         gcc -g -Wall -o test test.c
> 
> (I have also tried with option -O0 but that changes nothing).
> 
> I run "gdb test" followed by "print d" and then I get this error:
> 
> No symbol "d" in current context.

You need to get to the main function, since "d" is a local variable.
After you do get to main, step one line to get past the assignment of
119 to "d", and then print its value.

When I do this, I get the value of "d" printed as I'd expect.

Re: Still cannot print variable

[Bob Rossi]

On Fri, Apr 15, 2005 at 11:09:09AM +0200, JS wrote:
> I am still have problems printing a variable. I have this code in a file
> called test.:
> 
> int main()
> {
> 
>         int d = 119;
> 
>         return(0);
> 
> }
> 
> I then compile it like this:
> 
>         gcc -g -Wall -o test test.c
> 
> (I have also tried with option -O0 but that changes nothing).
> 
> I run "gdb test" followed by "print d" and then I get this error:

You know, I just realized it's kind of odd that GDB doesn't look on the
users PATH to figure out which program to debug. For instance, it must
first look in the . directory, which is why this works in the first
place. I'd expect GDB to start debugging /usr/bin/test, as if he'd run
'test' on the command line.

Bob Rossi

RE: Still cannot print variable

[Dave Korn]

----Original Message----
>From: Bob Rossi
>Sent: 15 April 2005 12:55


> You know, I just realized it's kind of odd that GDB doesn't look on the
> users PATH to figure out which program to debug. For instance, it must
> first look in the . directory, which is why this works in the first
> place. I'd expect GDB to start debugging /usr/bin/test, as if he'd run
> 'test' on the command line.


  Perhaps gdb *does* look in $PATH, and JS has the current directory ('.')
in his $PATH settings?

  I just tried 'gdb yes' and it WFM....

    cheers,
      DaveK
-- 
Can't think of a witty .sigline today....

Re: Still cannot print variable

[Daniel Jacobowitz]

On Fri, Apr 15, 2005 at 07:54:54AM -0400, Bob Rossi wrote:
> You know, I just realized it's kind of odd that GDB doesn't look on the
> users PATH to figure out which program to debug. For instance, it must
> first look in the . directory, which is why this works in the first
> place. I'd expect GDB to start debugging /usr/bin/test, as if he'd run
> 'test' on the command line.

It does search $PATH.  However, it searches the current directory
first, because it's more like a program taking a file argument than it
is like a command.

-- 
Daniel Jacobowitz
CodeSourcery, LLC