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ELEN [110]
3 years ago
11

The procedure call mystery(38) will yield which output? __________ a) 0 12 b) 12 0 c) 1 1 0 2 d) 1 1 1 1 e) 2 0 1 1 public void

mystery (int n) { if (n>2) mystery (n % 3); System.out.print( (n / 3) + " " ); }
Computers and Technology
1 answer:
Ivan3 years ago
3 0

Answer:

Option b is the correct answer for the above question.

Explanation:

  • The mystery function is a recursive function that calls for the two times when the user passes 38 on the argument.
  • The first value is 38 and the second value is 2 for which that function is called.
  • When the value 38 is passed, then again 2 is passed because of the "mystery (n % 3);" statement.
  • This statement holds by the if condition which gives the true when the argument value is greater than 2.
  • Hence for the 2 value the if condition will not true and the function is not called again.
  • Then the 38/3 and 2/3 are printed whose value is 12 0, but it will print 0 12 because of the recursive function.
  • Hence option b is the correct answer while the other is not because other options does not states the output of this program.
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What are the first and the last physical memory addressesaccessible using
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Answer

For First physical memory address,we add 00000 in segment values.

For Last physical memory address,we add 0FFFF in segment values.

<u>NOTE</u>-For addition of hexadecimal numbers ,you first have to convert it into binary then add them,after this convert back it in hexadecimal.

a)1000

For First physical memory address, we add 00000 in segment value

We add 0 at the least significant bit while calculating.

          1000<u>0</u> +00000 = 1000<u>0</u> (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           1000<u>0</u> +0FFFF =1FFFF    (from note)

b)0FFF

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

     0FFF<u>0</u> +00000=0FFF0 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           0FFF<u>0</u> +0FFFF =1FFEF (from note)

c)0001

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

     0001<u>0</u> +00000=00010 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           0001<u>0</u> +0FFFF =1000F (from note)

d) E000

For First physical memory address,we add 00000 in segment value

 We add 0 at the least significant bit while calculating.

    E000<u>0</u> +00000=E0000 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           E000<u>0</u> +0FFFF =EFFFF (from note)

e) 1002

For First physical memory address,we add 00000 in segment value

 We add 0 at the least significant bit while calculating.

    1002<u>0</u> +00000=10020  (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           1002<u>0</u> +0FFFF =2001F (from note)

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