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()
.
No comments:
Post a Comment