-----Original Message-----
From: linux-c-programming-owner@xxxxxxxxxxxxxxx [mailto:linux-c-
programming-owner@xxxxxxxxxxxxxxx] On Behalf Of RAM_LOCK
Sent: Wednesday, 29 July 2009 04:19 PM
To: linux-c-programming@xxxxxxxxxxxxxxx
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
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