Consider the following code with your current instruction pointer (the line that will be executed next, indicated by ->) at the f(x) line in g(), having been called by the g(2) line in main():
public class testprog {
static void f (int x) {
System.out.println ("num is " + (x+0)); // <- STEP INTO
}
static void g (int x) {
-> f(x); //
f(1); // <----------------------------------- STEP OVER
}
public static void main (String args[]) {
g(2);
g(3); // <----------------------------------- STEP OUT OF
}
}
If you were to step into at that point, you will move to the println() line in f(), stepping into the function call.
If you were to step over at that point, you will move to the f(1) line in g(), stepping over the function call.
Another useful feature of debuggers is the step out of or step return. In that case, a step return will basically run you through the current function until you go back up one level. In other words, it will step through f(x) and f(1), then back out to the calling function to end up at g(3) in main().