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
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### 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`
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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