RE: What compiler is doing when we pass unnecessary parameters in scanf

Thanks Trevor for your details explaination...
It is clear to me now... thanks again for clearing this concept :)



-----Original Message-----
From: Trevor Woollacott [mailto:[email protected]] 
Sent: Wednesday, July 29, 2009 9:32 PM
To: [email protected]; Sarkar, Kaushik
Subject: RE: What compiler is doing when we pass unnecessary parameters
in scanf

> -----Original Message-----
> From: [email protected] [mailto:linux-c-
> [email protected]] On Behalf Of RAM_LOCK
> Sent: Wednesday, 29 July 2009 04:19 PM
> To: [email protected]
> Subject: What compiler is doing when we pass unnecessary parameters in
> scanf
>
>
> Hi,
> In the second scenario what value is it printing when i have given
> extra
> parameter in scanf?
> Does it vary from compiler to compiler?
>
> Scenario : I
> -------------
> root@kaushik_Fedora11 ~/C/LET_US_C/ch-1> cat simple-interest.c
> #include <stdio.h>
>
> void main ()
> {
>         int p;
>         float i=0;
>         printf ("enter the principal amount\n");
>         scanf ("%d",&p);
>         i = (p*5*5)/100;
>         printf ("Interterest is : %f\n",i);
> }
> root@kaushik_Fedora11 ~/C/LET_US_C/ch-1> ./a.out
> enter the principal amount
> 100
> Interterest is : 25.000000
>
>
> Scenario : II
> -------------
> > cat simple-interest.c
> #include <stdio.h>
>
> void main ()
> {
>         int p;
>         float i=0;
>         printf ("enter the principal amount\n");
>         scanf ("p:%d",&p);
>         i = (p*5*5)/100;
>         printf ("Interterest is : %f\n",i);
> }
> root@kaushik_Fedora11 ~/C/LET_US_C/ch-1> ./a.out
> enter the principal amount
> 100
> Interterest is : -9321198.000000
>
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Hi

With scenario II your input of 100 is not matching, so you end up with 
whatever value was stored at &p before the scanf.
i.e.

(gdb) r
Starting program: ./a.out

Breakpoint 1, main () at test.c:6
6             float i=0;
(gdb) p p
$4 = -37284792
(gdb) s
7             printf ("enter the principal amount\n");
(gdb) s
enter the principal amount
8             scanf ("p:%d",&p);
(gdb) s
100
9             i = (p*5*5)/100;
(gdb) p p
$6 = -37284792
(gdb) s
10            printf ("Interterest is  : %f\n",i); }
(gdb) s
Interterest is : -9321198.000000

Here you can see p = -37284792 before and after your scanf
Then i = (-37284792*5*5)/100 = -9321198.000000

If you check the value returned by scanf, you can determine how many
input 
values were successfully matched. In the case where you entered 100 in 
scenario I, scanf would have returned 1, and in scenario II it would
have 
returned 0.

Regards,
Trevor