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JAVA Programming :: Basic Concepts

Home > JAVA Programming > Basic Concepts > General Questions

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Answer: Option A

Explanation:

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62. What gets written on the screen when the following program is compiled and run. Select the one right answer.
public class test {
public static void main(String args[]) {
inti;
float f = 2.3f;
double d = 2.7;
i = ((int)Math.ceil(f)) * ((int)Math.round(d));
System.out.println(i);
}
}

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Answer: Option E

Explanation:

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63. What is the Output o the below code:

class Ideone
{
public static void main (String[] args) throws javlang.Exception
{
VariablePass vp = new VariablePass ();
vp.method(5);

}
}

class VariablePass {
int x=20;
public void method (int x) {
x+=x;
System.out.println(x);
}
}

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Answer: Option B

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64. You have two class files name Test.class and Test1.class inside
javaproject directory.
Test.java source code is :

public class Test{
public static void main (String[] args){
System.out.println("Hello Test");
}
}
Test1.java source code is :
public class Test1{
public static void main (String[] args){
System.out.println("Hello Test1");
}
}
you have issued below commands from command prompt.
cd javaproject
java Test Test1

What is the output ?

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Answer: Option A

Explanation:

You must specify exactly one class file to execute. If more than one then first one will be executed.

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65. You have a java file name Test.java .
Test.java needs access to a class contained in app.jar in "exam"
directory.

Which of the follwing command set classpath to compile clean?

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Answer: Option A

Explanation:

javac -classpath exam/app.jar Test.java is the correct command to set exam/app.jar in classpath.

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