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SSSSS [86.1K]
3 years ago
11

Analyze the following code. II: public class Test { public static void main(String[] args) { System.out.println("Welcome to Java

!"); } } Both I and II can compile and run and display Welcome to Java, but the code in II has a better style than I. Only the code in I can compile and run and display Welcome to Java. Only the code in II can compile and run and display Welcome to Java. Both I and II can compile and run and display Welcome to Java, but the code in I has a better style than II.
Computers and Technology
1 answer:
Grace [21]3 years ago
8 0

COMPLETE QUESTION

I. public class Test {

public static void main(String[] args){

System.out.println("Welcome to Java!");

}

}

II. public class Test { public static void main(String[] args) {System.out.println("Welcome to Java!");}}

Answer:

Both codes will compile and run and display Welcome to Java, but the code in II has a better style than I

Explanation:

When written codes, paying attention to proper coding styles and efficient memory management enables us to create programs that are highly efficient, coding styles refer to proper indentions and avoiding too lenghty lines of code (as is in code I), adding approprite comments etc.

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#include <iostream>

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   char address[2];

   int tag, firstBit, secondBit, setNumber;

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           firstBit = 0;

        } else if (address[0] == '1'){

           tag = 0;

           firstBit = 1;

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           tag =1;

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}

Explanation:

The C++ source code prompts the user for an input for the address variable, then the nested if statement is used to assign the value of the firstBit value given the value in the first index in the address character array. Another if statement is used to assign the value for the secondBit and then the setNumber is calculated.

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