1. Consider the following Java code in which most of the instructions have been commented out. java /* A */ for ( /* B */ ; x <= 0 ; /* D */) { /* E */ } /* F */ * Convert the above Java code into MIPS assemble code, including the comments. mips
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Consider the following Java code in which most of the instructions have been commented out.
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- please use java and keep the solution and code very basic Q2)solve thisIndicate the exact machine-user interaction sequence for the single-line input sequence of 26 4: System.out.print("Enter two values: "); int1 = keyboard.nextint); int2 = keyboard.nextint); keyboard.nextLine; result = int1 % int2; System.out.printf("%d rem %d = %d\n", int1, int2, result); Consider the code from the previous question, reproduced below:System.out.print("Enter two values: ");int1 = keyboard.nextint);int2 = keyboard.nextint);keyboard.nextLine();result = int1 % int2;System.out.printf("%d rem %d = %d\n", int1, int2, result);The above requires some declaration statements to run. You do not need to indicate where to place each declaration, but write the statements needed for the above code to work. (Don't forget the keyboard!) Again, we reproduce the code of this Question System.out.print("Enter two values: ");int1 = keyboard.nextint);int2 = keyboard.nextint();keyboard.nextLine();result = int1 %…java please, Complete main() to read dates from input, one date per line. Each date's format must be as follows: March 1, 1990. Any date not following that format is incorrect and should be ignored. Use the substring() method to parse the string and extract the date. The input ends with -1 on a line alone. Output each correct date as: 3/1/1990. Ex: If the input is: March 1, 1990 April 2 1995 7/15/20 December 13, 2003 -1 then the output is: 3/1/1990 12/13/2003Implement a subprogram which takes 4 numbers in the argument registers $a0...$a3, andreturns the largest value and the average in $v0 and $v1 to the calling program. Theprogram must be structured as follows:Subprogram largestAndAverage($a1, $a2, $a3, $a4){ int var0 = $a0, var1 = $a1, var2 = $a2, var3 = $a3; $s0 = getLarger($a1, $a2); $s0 = getLarger($s0, $a3); $v0 = getLarager(s0, $a4); // Largest is in $v0 $v1 = (var0 + var1 + var2 + var3)/ 4; // Aversge is in $v1 return;}Subprogram getLarger($a0, $a1) { $v0 = $a0 if ($a1 > $a0) $v0 = $a1 return;}Note the use of the variables var0...var3. Because the values of $a0 and $a1 (at least) arechanged on the call to getLarger, they will not be available when they are needed to calculatethe average, and must be stored on the stack. To do this problem correctly, you mustcalculate the maximum value using the getLarger subprogram shown here, and it must becalled before the average is calculated. This implies that at a minimum $a0 and $a1 must…
- Implement a subprogram which takes 4 numbers in the argument registers $a0...$a3, andreturns the largest value and the average in $v0 and $v1 to the calling program. Theprogram must be structured as follows:Subprogram largestAndAverage($a1, $a2, $a3, $a4){ int var0 = $a0, var1 = $a1, var2 = $a2, var3 = $a3; $s0 = getLarger($a1, $a2); $s0 = getLarger($s0, $a3); $v0 = getLarager(s0, $a4); // Largest is in $v0 $v1 = (var0 + var1 + var2 + var3)/ 4; // Aversge is in $v1 return;}Subprogram getLarger($a0, $a1) { $v0 = $a0 if ($a1 > $a0) $v0 = $a1 return;}Take note of how var0...var3 are used. The values of $a0 and $a1 (at least) must be placed on the stack because they are not immediately available when needed to calculate the average because they are modified during the call to getLarger. You must use the getLarger subprogram displayed above to calculate the greatest value for this problem, and it must be called before the average calculation. This indicates that at the very least $a0…INSTRUCTION: Answer Question 3 by refering to the funtion (2) below. #include <iostream>using namespace std; int sortSwapCount(int data[], int size, double &proportion){ int swapCount = 0; for (int i = 0; i < size - 1; i++) { for (int j = 0; j < size - i - 1; j++) { if (data[j] > data[j + 1]) { int temp = data[i]; data[i] = data[i+1]; data[i+1] = temp; swapCount++; } } } double expectedSwapCount = (double)size*(size-1); double swapCount1 = (double)swapCount; proportion = (swapCount1/expectedSwapCount)*100; return swapCount;}Question 9 Na I need help with this homework question please. Analyze the following pseudocode and give a tight (Θ) bound on the runningtime as a function of n. You can assume that all individual instructions are elementary(i.e. take time in Θ(1)). Show your work
- bitcount: Given x, set y to the number of bits in x that are 1. Write the code in Java. Provided input(s): x Permitted: 40 operations (may use !, ~, +, -, <<, >>, &, ^, |) Hint: The obvious solution would be something like ans=0; for(int i=0; i<32; i+=1) { a += x&1; x >>=1; } We don’t allow for loops, but even if you replace it with 32 copies that’s still 96 operations, and we only allow 40 for this task. The trick is to do things in parallel, treating a number like a vector of smaller numbers. Suppose I wanted to count the bits of an 8-bit number with bits abcdefgh. With a little shifting and masking I could make three numbers 0b00e00h 0a00d00g 0000c00f and add them to get xx0yy0zz where xx = a+b, yy = c+d+e, and zz = f+g+h. Extending this trick to several rounds on 32 bits will solve this problem.Knowing that (x + 1)2 = x2 + 2x + 1,and that multiplication is more time consuming than addition, write an efficientprogram that displays the first ten natural numbers and their squares. Knowingthat (x + 1)3 = x3 + 3x2 + 3x + 1, the same process can be followed to displaythe cubes of these numbers.Here is a sample run: 0^2 = 01^2 = 12^2 = 4...8^2 = 649^2 = 8110^2 = 100Computer Science Is it possible to convert this to java? #include <stdio.h> void printBinary(int n, int i) { // Prints the binary representation // of a number n up to i-bits. int k; for (k = i - 1; k >= 0; k--) { if ((n >> k) & 1) printf("1"); else printf("0"); } } typedef union { float f; struct { // Order is important. // Here the members of the union data structure // use the same memory (32 bits). // The ordering is taken // from the LSB to the MSB. unsigned int mantissa : 23; unsigned int exponent : 8; unsigned int sign : 1; } raw; } myfloat; // Function to convert real value // to IEEE floating point representation void printIEEE(myfloat var) { // Prints the IEEE 754 representation // of a float value (32 bits) printf("%d | ", var.raw.sign);…
- Interestingly some numbers can be perfectly represented in Base 10, but not in Base 2. An example is one-tenth (in Base 10, this is 0.1; in Base 2 this is a repeating decimal. Suppose we set both f and g to one-tenth using the statements below. f = 0.1; g = 0.1; Answer the following questions. For each, mark the best answer. Which will be closer to the actual value of one-tenth? f g Which can hold higher values? f g Which takes more memory? f g When using real values I should always use type double, to be safe. True Falseconvert the following java code program into (mips assembly language) program with the same output with the same loops i've been used no additional thing please *write the code as a text* class java{ public static void main(String[] args){ int count=0; System.out.println("All numbers"); while (true) { count++; System.out.print(count+" "); if (count==5) break; } System.out.println("\nOdd numbers"); for (count=0; count <= 5; count++) { odd(count); } System.out.println("\nEven numbers"); for (count=1; count <= 5; count++) { even(count); } System.out.println(" "); } public static void odd(int count){ int result =0; if (count%2==1){ result = count; System.out.print(result+" "); } } public static void even(int count){ int result =0; if (count%2==0){ result = count; System.out.print(result+" "); } } }Write a Java program called Div that takes 2 (two) double command-line arguments as inputs, dividend, and divisor (in that order) and performs a division operation. Your program either print the quotient or an error if the divisor is zero. The divisor is the number you divide the dividend by.