What is value of printf("%d",12.5);

Dear friends,
I am facing a problem in my small code. I was just wondering that what is the output of the code and why did it happen.

The code is

#include<stdio.h>
main()
{
float i=12.5;
printf("%d",i);
getchar();
}


It gives me 0 as output.
But i am confused why is it so.

Please help me in finding out the reason behind it.

Its giving as 0 that is due to the change in formats specified in both declaration and printf statements...
It is to be noted if u declare an integer variable and in printf if u give %f then also a garbagge value is output.

but can I know what exactly happened in the storage of floats? how are floats stored in memory?

actually when u give
float i=12.5;
4 byte of memory is allocated for the float variable.But when u give
printf("%d",i); it accesses the first 2 bytes of the float variable (as it is now asked to o/p [SIZE="3"][U][B]the integer i[/B][/U][/SIZE] .Hence you get to see a 0 not 12.