Knowee
Questions
Features
Study Tools

The following segment of code contains at least one error - mark ALL of the errors that occur in this segment of code:int val = 0;while ( val < 10 ) { String msg = ""; int num = val; while ( num < 10 ) { if (num % 2 == 0 ) { msg = msg + num; num = num + 1; } }}System.out.println(msg);Group of answer choicesThe variable msg is used outside of its scopeThe variable num is used outside of its scopeThe inner loop will loop foreverThe outer loop will loop foreverYou cannot add together a String and an int value

Question

The following segment of code contains at least one error - mark ALL of the errors that occur in this segment of code:int val = 0;while ( val < 10 ) { String msg = ""; int num = val; while ( num < 10 ) { if (num % 2 == 0 ) { msg = msg + num; num = num + 1; } }}System.out.println(msg);Group of answer choicesThe variable msg is used outside of its scopeThe variable num is used outside of its scopeThe inner loop will loop foreverThe outer loop will loop foreverYou cannot add together a String and an int value

...expand
🧐 Not the exact question you are looking for?Go ask a question

Solution

The errors in the code are:

  1. The variable msg is used outside of its scope: The variable msg is declared inside the while loop and is being used outside of it in the System.out.println statement. This will cause a compile-time error because the scope of msg is limited to the while loop where it is declared.

  2. The inner loop will loop forever: The increment of the variable num is inside an if condition that checks if num is even. If num is odd, it will not be incremented and the inner loop will run indefinitely.

  3. The outer loop will loop forever: The variable val is never incremented within the outer loop, so the condition val < 10 will always be true, causing an infinite loop.

The other two options are not errors:

  • The variable num is not used outside of its scope: The variable num is declared and used only within the while loop, so there is no scope issue with num.

  • You can add together a String and an int value in Java: In Java, you can use the + operator to concatenate a String and an int. The int is automatically converted to a String before the concatenation.

This problem has been solved

Similar Questions

The following segment of code contains at least one error - mark ALL of the errors that occur in this segment of code:int val = 0;while ( val < 10 ) { String msg = ""; int num = val; while ( num < 10 ) { if (num % 2 == 0 ) { msg = msg + num; num = num + 1; } }}System.out.println(msg);

Select the correct answerWhat will be the output of the following Java program? class scopevariable { public static void main(String args[]) { int c; c = 15; { int d = 36; System.out.print(c + " " + d); } System.out.println(c + " " + d); } }Options5 6 55 6 5 6Runtime errorCompilation error

Select the correct answerWhat is the output of the following code snippet?for (int i = 0; i < 5; i++) {  System.out.print(i + " ");}Options0 1 2 3 40 1 2 3Compilation error1 2 3 4 5

Select the correct answerWhat is the output of the following program?public class Test{ public static void main(String []args){ int i = 0; for(i = 0; i < 10; i++){ continue; } System.out.println(i); }}OptionsCompilation error0109

Given1. public class Program{2. public static void main(String[] args) {3.            4. for(int j = 0,k=5; j < k; k--) ;5. for(int j = 0; j++ < 3;) ;6. for(int i = 0; i < 5; i++, System.out.print(i + ".go ")) ;7. 8. }9. }What will be the result?*Compilation fails due to multiple errors.Compilation fails due to an error on line 6.0.go 1.go 2.go 3.go 4.go 5.goAn exception is thrown at runtime.1.go 2.go 3.go 4.go 5.go

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.