I'm struggling to grasp how the methods `foo` and `bar` work in this Java program. My study guide for the midterm includes this, and I'm really hoping someone can shed some light on it. Here's the code snippet:
```java
class Driver {
public static void main(String[] args) {
int a = bar(2);
int b = foo(a);
System.out.print(b);
}
static int foo(int a) {
a = bar(a) - 2;
return a;
}
static int bar(int a) {
System.out.print(a);
return a + 1;
}
}
```
What output can I expect from this program?
1 Answer
When you call `bar(2)`, it prints `2` and returns `3` because of `return a + 1;`. Then in `foo`, you call `bar(a)`, which means `bar(3)`, printing `3` and returning `4`. Finally, `foo` returns `4 - 2`, which equals `2`, so the output of the program is just `2`. Don't get too hung up on the names `foo` and `bar`, they are often used as generic placeholders in programming examples.
Ah, thank you! That really clarifies things for me.