Anitha Boyapati wrote: > Usually, (well I thought so) the argument expected in f() is a pointer > variable. The above gives the result as 2 for the first call to f()! > Why is t giving out 2 without being dereferenced in case 1 ? You explicity cast an integer into a pointer to an integer. The whole point of a cast is to tell the compiler to consider one thing as another. In this case you're telling it that an integer's value is actually a pointer -- a completely invalid pointer, but a pointer nonetheless. There is no dereference because you didn't ask for a dereference. Nothing is automatic in C, if you want the address of i you must write &i. Brian
