Question
Consider the following code: long y = 9223372036854775807L; y = y + 1; System.out.println(y); What number does the program print when compiled and run? Hint: think about the max value you just looked up in your previous answer. Group of answer choices This program crashes and raises a longOverflow exception. -9223372036854775808 -1 9223372036854775808 0
Answers
Answered by
GPT-4o mini
The code snippet you've provided declares a long variable `y` initialized to the maximum value of a long in Java, which is `9223372036854775807L` (this is \(2^{63} - 1\)). When you add 1 to this value, it causes an overflow.
In Java, when a long integer overflows, it wraps around to the minimum value of a long, which is `-9223372036854775808L` (this is \(-2^{63}\)). Therefore, when the line `y = y + 1;` is executed, `y` becomes `-9223372036854775808`.
So, when you print `y`, the output will be:
```
-9223372036854775808
```
To summarize, the program prints:
**-9223372036854775808**
In Java, when a long integer overflows, it wraps around to the minimum value of a long, which is `-9223372036854775808L` (this is \(-2^{63}\)). Therefore, when the line `y = y + 1;` is executed, `y` becomes `-9223372036854775808`.
So, when you print `y`, the output will be:
```
-9223372036854775808
```
To summarize, the program prints:
**-9223372036854775808**