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Svetach [21]
3 years ago
14

There are 55 registers, and total 60 instructions available in a general-purpose computer. The computer allows only 2-address in

structions, where one operand can be a register, and another can be a memory location. The memory is byte addressable with 64KB (Kilo bytes) in size. The minimum number of bits to encode the instruction will be…………?
Computers and Technology
1 answer:
Andrei [34K]3 years ago
7 0

Answer:

31 bits.

Explanation:

Given, total number of registers = 55

Total instructions = 60

Size of memory = 16 KB

Now, no of registers are 55. We find the next greater or equal power of 2 which is 64 = 26. Hence, 6 bits are required to represent a register operand.

Number of instructions = 60. We find the next greater or equal power of 2 which is 64 = 26. Hence, 6 bits are required to represent a instruction.

Size of memory = 64 KB = 26 * 210 * 23 bits = 219 bits. Hence, 19 bits are required to represent a memory location.

Now, an instruction has 2 parts, opcode and operand. As given there are only two address instructions which are memory operand and register operand.

Hence, total bits would be: 6 bits (opcode) + 6 bits (register operand) + 19 (memory operand) = 31 bits.

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import java.util.ArrayList;

import java.util.List;

public class Vowels {

   public static void main(String[] args) {

   String word = "I am David from Nigeria";

       System.out.println(which_vowels_present(word));

   }

   public static List which_vowels_present(String enteredWord){

       List<Character>list = new ArrayList<>();

       String word = enteredWord.toLowerCase();

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

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           }

       }

       return list;

   }

}

Explanation:

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  • return the list
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