Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
Answer:
Explanation:
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Answer:
A.)
arr[0] = 10;
arr[1] = 10;
Explanation:
Given the array:
arr = {1,2,3,4,5}
To set the first two elements of array arr to 10.
Kindly note that ; index numbering if array elements starts from 0
First element of the array has an index of 0
2nd element of the array has an index of 1 and so on.
Array elements can be called one at a time using the array name followed by the index number of the array enclosed in square brackets.
arr[0] = 10 (this assigns a value of 10 to the index value, which replace 1
arr[1] = 10 (assigns a value of 10 to the 2nd value in arr, which replaces 2
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