Explanation:
An application interface or user interface ,is the set of features an application provides so that user may supply input to and recieve output from,the program.
Answer:
Explanation:
Since there are no line numbers in this question I will start counting from public class SelectionSort{ as line 1 and so on, as well as provide the code on that line.
The index of the smallest value is returned on line 8 where it says return min which shouldn't have any spaces and should be return minPosition;
The input array is given as an argument at the beginning of the function on line 2 where it says private static int positionMin (int] vals, int startPosition) {, as the variable vals.
This input array is also used as an argument on line 10 where it says Position; private static void swap(int] vals, int firstPosition, int secondPosition) and line 15 where it says vals[secondPosition] temp return public static void selSort(int| vals) {
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
