what is printf doin....?

The behavior of printf is undefined if there is insufficient no. of arguments for the format string
when i am initializing a variable it print variable value, consider this:
main()
{
int a=10;
printf("%d");
} ANSWER IS 10. This was asked by one of my friends. he says," it has to do something with segmentation",so the question is of couse logicaly valid..

The correct answer to the question is undefined behaviour.But if you are wondering why the result prints 10, here's a possible explanation (which seems to work for your platform):
1. Local variables are stored on the stack (int a)
2. The arguments you pass to a function are pushed to the same stack
3. Printf sees a "%d", and assumes that an integer argument is avaliable on the stack.
4. It reads an integer from the stack, and picks up the variable 'a', and prints itNote that the above explanation seems to work on your (and perhaps many other) platform. There is no guarantee that this will be the case with some other compiler on some other OS. That's precisely why it is called 'undefined behaviour'.