Analyze the following code: public class Test { public static void main(String argv[]) { System.out.println("argv.length is " + argv.length); } } Group of answer choicesThe program has a compile error because String argv[] is wrong and it should be replaced by String args[].If you run this program without passing any arguments, the program will have a runtime error because argv is null.If you run this program without passing any arguments, the program will display argv.length is 0.
Question
Analyze the following code: public class Test { public static void main(String argv[]) { System.out.println("argv.length is " + argv.length); } } Group of answer choicesThe program has a compile error because String argv[] is wrong and it should be replaced by String args[].If you run this program without passing any arguments, the program will have a runtime error because argv is null.If you run this program without passing any arguments, the program will display argv.length is 0.
Solution 1
The correct answer is: "If you run this program without passing any arguments, the program will display argv.length is 0."
Explanation:
-
The program does not have a compile error because "String argv[]" is not wrong. In Java, you can declare an array in multiple ways and "String argv[]" is one of them. It's just an array of Strings, which is what the main method expects.
-
If you run this program without passing any arguments, the program will not have a runtime error because argv is not null. In Java, if you don't pass any arguments to the main method, it receives an empty array (not null).
-
If you run this program without passing any arguments, the program will display "argv.length is 0". This is because the length of the argv array is 0 when no arguments are passed to the main method. The System.out.println statement is printing the length of the argv array, which would be 0 if no arguments are passed.
Solution 2
The correct answer is: "If you run this program without passing any arguments, the program will display argv.length is 0."
Explanation:
-
The program does not have a compile error because "String argv[]" is not wrong. In Java, you can declare an array in multiple ways and "String argv[]" is one of them. So, it doesn't need to be replaced by "String args[]". Both are valid.
-
If you run this program without passing any arguments, the program will not have a runtime error because argv is not null. In Java, if no arguments are passed to the main method, the JVM initializes the argument array as an empty array (not null). So, argv.length will return 0, not a null pointer exception.
-
If you run this program without passing any arguments, the program will indeed display "argv.length is 0". This is because the length of the argv array is 0 when no arguments are passed to the main method.
Similar Questions
Analyze the following code carefully.public class Test { int x; public Test(String t) { System.out.println("Test"); } public static void main(String[] args) { Test test = new Test(); System.out.println(test.x); }}Group of answer choicesThe program has a compile error because you cannot create an object from the class that defines the object.The program has a compile error because x has not been initialized.The program has a compile error because System.out.println method cannot be invoked from the constructor.The program has a compile error because Test does not have a default construc
elect the correct answerPredict the output ?class Test { public static void main(String args[]) { int arr[] = new int[2]; System.out.println(arr[0]); System.out.println(arr[1]); }}OptionsCompiler Errorgarbage valuegarbage value00Exception
Analyze the following code. Please select all that apply. public class Test { public static void main(String[] args) { Object a1 = new A(); Object a2 = new Object(); System.out.println(a1); System.out.println(a2); } } class A { int x; @Override public String toString() { return "A's x is " + x; } } Group of answer choicesThe program cannot be compiled, because System.out.println(a1) is wrong and it should be replaced by System.out.println(a1.toString());When executing System.out.println(a1), the toString() method in the Object class is invoked.When executing System.out.println(a2), the toString() method in the Object class is invoked.When executing System.out.println(a1), the toString() method in the A class is invoked.
public class Test { public static void main(String[] args) { for (int i = 0; i < args.length; i++) { System.out.print(args[i] + " "); } } } What is the output, if you run the program using java Test 1 2 3?Group of answer choices311 2 31 2 Flag question: Question 15Question 151 ptsWhich correctly creates an array of five empty Strings? Group of answer choicesString[] a = new String [5];String[] a = {"", "", "", "", ""};String[5] a;String[ ] a = new String [5]; for (int i = 0; i < 5; a[i++] = null);
Analyze the following code:public class Test { public static void main(String[] args) { int[] x = new int[5]; int i; for (i = 0; i < x.length; i++) x[i] = i; System.out.println(x[i]); }}Group of answer choicesThe program displays 4.The program displays 0 1 2 3 4.The program has a compile error because i is not defined in the last statement in the main method.The program has a runtime error because the last statement in the main method causes ArrayIndexOutOfBoundsException.
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.