Printf function problem

[satyanagendra]

#include<stdio.h>
#include<conio.h>
main()
{
double a,b;
a=2e+4;
b=3e+5;
printf("%d %f",a,b);
}


in the above ex i expected result as

0 3e+5
but i got 0,0.00000000 if anybody know the reason then please mail me

[pradeep kaltari]

#include<stdio.h>
#include<conio.h>
main()
{
double a,b;
a=2e+4;
b=3e+5;
printf("%d %f",a,b);
}


in the above ex i expected result as

0 3e+5
but i got 0,0.00000000 if anybody know the reason then please mail me

Hi,
Try this:

#include<stdio.h>
#include<conio.h>
main()
{
double a,b;
a=2e+4;
b=3e+5;
printf("%d %e",(int)a,b);
}
 

This will give you your expected output.

As you were using the control string %d to display a (of type double), you need to cast it as int and as expected it returns 0.
Also, to display double in the format you expect you need to use "%e". If you use %f to print the value of b, the output would be 300000.000000.
I hope this was helpful.

Regards,
Pradeep