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Harman [31]
4 years ago
8

Which line(s) immediately flush the output buffer?

Computers and Technology
1 answer:
Murljashka [212]4 years ago
3 0

Answer:

lines 2,3 and 5 ( c )

Explanation:

The lines that immediately flush the output buffer are :  System.out.print("Two for the show\n"); .   System.out.println("Three to get ready"); and   System.out.flush();

making use of the new line character or printIn() statement the buffered output gets flushed therefore making autoflush true

but System.out.print() doesn't  use any flush technique so we need to flush the buffered output done by the print() statement hence we choose lines 2,3,5

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Answer:

C

Explanation:

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3 years ago
Consider an interface p ublic interface NumberFormatter { String format (in n); } Provide four classes that implement this inter
sesenic [268]

Answer and Explanation:

Here is the code for the question with output.

public interface NumberFormatter {

  public String format(int n);

}

class DefaultFormatter implements NumberFormatter

{

  public String format(int n)

  {

      return String.valueOf(n);

  }

}

class DecimalFormatter implements NumberFormatter

{

  public String format(int n)

  {

      return String.format("%,d",n); //formats the number by putting comma for 3 digits

  }

}

class AccountingFormatter implements NumberFormatter

{

  public String format(int n)

  {

      return String.format("(%d)",n);

  }

}

class BaseFormatter implements NumberFormatter

{

  int base;

 

  public BaseFormatter(int b)

  {

         

      base = b;

      if(base < 2 || base > 36) //if values out of range 2-36 inclusive are given, set default to 2

          base = 2;

  }

  public String format(int n)

  {

      return Integer.toString(n, base);      

  }

  int getBase()

  {

      return base;

  }

}

public class TestNumberFormatter {

  private static void print(int n, NumberFormatter formatter)

  {

      String s = String.format("%15s",formatter.format(n));

      System.out.print(s);

  }

  public static void main(String[] args) {

      int numbers[]= {100,55555,1000000,35,5};

      DefaultFormatter defFmt = new DefaultFormatter();

      DecimalFormatter deciFmt = new DecimalFormatter();

      AccountingFormatter accFmt = new AccountingFormatter();

      BaseFormatter baseFmt=new BaseFormatter(16) ; //base 16

      String s;

     

      System.out.println("\tDefault \t Decimal \tAccounting \t Base("+baseFmt.getBase()+")");

      for(int i = 0; i < numbers.length; i++)

      {

          print(numbers[i], defFmt);

          print(numbers[i], deciFmt);

          print(numbers[i], accFmt);

          print(numbers[i], baseFmt);

          System.out.println();//new line

      }

     

  }

}

output

  Default    Decimal    Accounting    Base(16)

100 100 (100) 64

55555 55,555 (55555) d903

1000000 1,000,000 (1000000) f4240

  35 35 (35) 23

5 5 (5) 5

3 0
3 years ago
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AveGali [126]

Answer:

void printC()  

{  

   int i, j;  

   for (i = 0; i < 4; i++) //i indicate row number. Here we have 5 rows

       {  

         printf("C"); //print C for every row  

         for (j = 0; j < 6; j++) //j indicate column number. Here we have 7 Rows

         {  

           if (i == 0 || i == 4) //For first and last row  

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          else if (i = 1|| i= 3) //for Second forth row  

                printf("C        +      +"); //print 'C    +    +'

          else if (i = 2) For second row  

                printf("C       +++++"); //print 'C +++++'

           else

               continue; //to jump to next iteration

         }  

         printf("\n"); // print in next line

}  

}

4 0
3 years ago
Given a sorted list of integers, output the middle integer. assume the number of integers is always odd.
Svetllana [295]

integer userInput

integer i

integer mid

integer array(20) number

userInput = 1

for i = 0; userInput >= 0; i = i + 1

  if number[i] > -1

     userInput = Get next input

     number[i] = userInput

i = i - 1

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if i > 9

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elseif i % 2 == 0

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else

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Allushta [10]

Even numbers are numbers whigh are divisible by 2. Therefore, the first even number is 2. A pseudocode which adds the first 100 even numbers can be written thus :

counter = 0

sum = 0

interval = 2

while counter < 100 :

sum = sum + interval

interval += 2

counter +=1

print(sum)

  • A counter takes count of the number of values summed

  • Initializes a variable which holds the sum of even values

  • Since even numbers are divisible by 2; every factor ; increase every added value by 2

  • The program ends once counter is 100

Learn more : brainly.com/question/25327166

6 0
2 years ago
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