look at the program lines:

int a;
a=15+1.0;
printf("%d",a);
/*gives an output 16*/

NOW

printf("%d",15+1.0);

/*gives an output zero !!! why????*/

all the operations which contain any floating point number gives an output 0 .
could u please tell me why??????

The %d format specifier for printf expects an integer. You are passing a floating point value in the second example, so you are essentially asking for random output. In the first case, you assign the result back to a first, so it is automatically converted to an integer.

a=15+1.0
also doesn't work with _all_ compilers. Some will refuse to compile that... with some compilers you will need to do something like:
a=15+(int)1.0 or
a=(int)(15+1.0)
but the rest is as Smerdyakov exmaplied it. [addsig]