Consider the following method: static int stackStuff (int n) { Stack stack = new Stack<>(); for (int count = 0; count <= n; count++) { if (count % 2 == 0) { stack.push(count); } } else { } return stack.pop(); } return 1; Give the result of: stackStuff (2) + stackStuff (0)
Consider the following method: static int stackStuff (int n) { Stack stack = new Stack<>(); for (int count = 0; count <= n; count++) { if (count % 2 == 0) { stack.push(count); } } else { } return stack.pop(); } return 1; Give the result of: stackStuff (2) + stackStuff (0)
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Transcribed Image Text:### Consider the following method:
```java
static int stackStuff(int n) {
Stack<Integer> stack = new Stack<>();
for (int count = 0; count <= n; count++) {
if (count % 2 == 0) {
stack.push(count);
} else {
return stack.pop();
}
}
return 1;
}
```
### Give the result of:
`stackStuff(2) + stackStuff(0)`
#### Explanation:
The method `stackStuff` takes an integer `n` and performs the following operations:
1. Initializes a new stack of integers.
2. Iterates from `0` to `n` inclusive:
- If the current counter `count` is even, it pushes `count` onto the stack.
- If the current counter `count` is odd, it pops the top value from the stack and returns it.
3. If none of the numbers are odd (or after processing all numbers), it returns `1`.
#### Calculating `stackStuff(2)`:
- `count = 0`: Pushes `0` onto the stack.
- `count = 1`: Pops `0` from the stack and returns it.
Result: `0`
#### Calculating `stackStuff(0)`:
- `count = 0`: Pushes `0` onto the stack.
- Exits loop (since no odd numbers occur) and returns `1`.
Result: `1`
### Therefore, the answer to `stackStuff(2) + stackStuff(0)` is:
`0 + 1 = 1`
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