QUESTION 8 What is the value of x after the call to m1 in line 8 of the following code? public static void main(String[] args) { int x = 5 ; m1(x); 50 6 8. 9 10 } public static void m1 (int x) { 110 12 13 if ( x > 0 || x < 3) x = x++ + (x + 2); else 14 15 16 17 x--; 18 19 m2(x); 20 21 } 22 public static void m2 (int x) { while (x > 0) { 230 24 25 26 x -- 3; } } 27 28 29 O Compiler error none of the above
QUESTION 8 What is the value of x after the call to m1 in line 8 of the following code? public static void main(String[] args) { int x = 5 ; m1(x); 50 6 8. 9 10 } public static void m1 (int x) { 110 12 13 if ( x > 0 || x < 3) x = x++ + (x + 2); else 14 15 16 17 x--; 18 19 m2(x); 20 21 } 22 public static void m2 (int x) { while (x > 0) { 230 24 25 26 x -- 3; } } 27 28 29 O Compiler error none of the above
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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java
![### Question 8
**Question:**
What is the value of `x` after the call to `m1` in line 8 of the following code?
```java
public static void main(String[] args) {
int x = 5;
m1(x);
}
public static void m1(int x) {
if (x > 0 || x < 3)
x = x++ + (x + 2);
else
x--;
m2(x);
}
public static void m2(int x) {
while (x > 0) {
x -= 3;
}
}
```
**Multiple Choice Options**
- [ ] 5
- [ ] -1
- [ ] 0
- [ ] 1
- [ ] Compiler error
- [ ] None of the above
### Explanation:
1. **Initialization:**
In the `main` method, `x` is initialized to 5.
2. **Calling m1:**
The method `m1` is called with `x` as 5.
3. **Within m1 Method:**
- The condition `(x > 0 || x < 3)` is checked. Since `x` (which is 5) is greater than 0, the condition is true.
- `x = x++ + (x + 2);` is evaluated.
- `x++` uses the current value of `x` (5) and then `x` is incremented to 6.
- The expression inside the parentheses `(x + 2)` uses the updated value of `x` (6), thus it becomes `6 + 2 = 8`.
- Therefore, `x = 5 + 8` which results in `x` being set to 13.
4. **Calling m2:**
The method `m2` is called with `x` as 13.
5. **Within m2 Method:**
- The `while` loop `(x > 0)` is executed:
- 1st iteration: `x` becomes `13 - 3 = 10`
- 2nd iteration: `x` becomes `10 - 3 = 7`
- 3rd iteration: `x` becomes `7 - 3 = 4`
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0e6f6223-c696-4938-9015-1aa854831435%2F12442091-21d5-4200-b1d0-da5743d66286%2Fyc71hwk_processed.png&w=3840&q=75)
Transcribed Image Text:### Question 8
**Question:**
What is the value of `x` after the call to `m1` in line 8 of the following code?
```java
public static void main(String[] args) {
int x = 5;
m1(x);
}
public static void m1(int x) {
if (x > 0 || x < 3)
x = x++ + (x + 2);
else
x--;
m2(x);
}
public static void m2(int x) {
while (x > 0) {
x -= 3;
}
}
```
**Multiple Choice Options**
- [ ] 5
- [ ] -1
- [ ] 0
- [ ] 1
- [ ] Compiler error
- [ ] None of the above
### Explanation:
1. **Initialization:**
In the `main` method, `x` is initialized to 5.
2. **Calling m1:**
The method `m1` is called with `x` as 5.
3. **Within m1 Method:**
- The condition `(x > 0 || x < 3)` is checked. Since `x` (which is 5) is greater than 0, the condition is true.
- `x = x++ + (x + 2);` is evaluated.
- `x++` uses the current value of `x` (5) and then `x` is incremented to 6.
- The expression inside the parentheses `(x + 2)` uses the updated value of `x` (6), thus it becomes `6 + 2 = 8`.
- Therefore, `x = 5 + 8` which results in `x` being set to 13.
4. **Calling m2:**
The method `m2` is called with `x` as 13.
5. **Within m2 Method:**
- The `while` loop `(x > 0)` is executed:
- 1st iteration: `x` becomes `13 - 3 = 10`
- 2nd iteration: `x` becomes `10 - 3 = 7`
- 3rd iteration: `x` becomes `7 - 3 = 4`
-
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