// Example 8.27a void Func(int a[], int & r) { int i; for (i = 0; i < 100; i++) { a[i] = r + i/2; } } Write the assembly code that The Intel compiler generates from example 8.27a (32 bit mode):
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// Example 8.27a
void Func(int a[], int & r) {
int i;
for (i = 0; i < 100; i++) {
a[i] = r + i/2;
}
}
Write the assembly code that The Intel compiler generates from example 8.27a (32 bit mode):
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- 8. Consider the following C code snippet. // c code void setArray(int num) { int i=5; compare (4, i); int compare (int a, int b) { if (sub (a, b) >= 0) return 1; else return 0; int sub (int a, int b) { return a - b; Implement the C code snippet in ARM assembly language. Use R4 to hold the variable i.MIPS Programming Assignment Part I Rewrite C statements int i = 11; int j = 10; int A[] = { 0x11, 0x22, 0x33 }; int B[] = { 0x0, 0x1, 0x2, 0x3, 0x4, 0x5, 0x6, 0x7, 0x8 }; B[ 8 ] = A[ i - j ]; in MIPS assembly. Assume that addresses of variables i, j, A, and B are loaded into the registers $s3, $s4, $s6, and $s7, respectively: .data i: .word 11 j: .word 10 A: .word 0x11, 0x22, 0x33 B: .word 0x0, 0x1, 0x2, 0x3, 0x4, 0x5, 0x6, 0x7, 0x8 .text la $s3, i # load address of i la $s4, j # load address of j la $s6, A # load address of A la $s7, B # load address of B When you finished programming, add a comment at the end of your code which specifies what value is assigned to the location of B[8] after the program executes. Part IIWrite MIPS Assembly program that allows the user to enter a string of text. Determine whether the entered string is a palindrome (a word or a phrase that reads the same backward as forward, like "kayak" and "level") and print the result on the screen.Systems Programming Purpose: The purpose of this assignment is to practice loop instructions and array iterations in M6800 assembly language programming. Write an assembly language program which implements the following C code that is a find minimum problem solution implementation: int A[10] = {8, 14, 6, 16, 5, 5, 10, 9, 4, 11}; int i = 0; int min = 255; // This is the largest number 8-bit accumulators can hold. while (i < 10) { if (A[i] < min) min = A[i]; } Your solution should be able to handle all possible array elements and orderings for any value an accumulator can hold. You should treat variable į as XR in assembler (index register) and store variable min in the address 60H. 0010: 00 00 00 00 00 00 00 00 0O 00 00 00 00 00 00 00 8628: 00 060 00 00 00 600 00 00 00 08 08 00 6 00 00 00 0030: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0040: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0050: 00 G0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 006 0: 00 00 00 00 00 00 00 00 00 08 00…
- Implement the following expression in assembly language: EDX = (val3 + val4) - (val2 - val1) - (5/3)*7 Assume that val1, val2, val3 and val4 are 16-bit integer variables You need to implement the expression the way it is provided, you cannot do any reduction on the expression while implementing it. Initialize val1 with 120 (hexadecimal), val2 with 39 (hexadecimal), val3 with 20 (hexadecimal) and val4 with 27 (hexadecimal) You are NOT allowed to update the values of any variables. Use ONLY mov, add, sub, movzx, movsx, or neg instructions whenever needed. Use the debugger to verify your answer.Compile the following C code snippet with different processor architecture. int square(int num) { return num * num; } 1) compile using armv7-a clang 11.0, and explain the program. 2) compile using x86-64 gcc 11.2, and explain the program.2) Translate the function origami into a MIPS assembly code fragment: int origami (int Aŋ, int BØ, int a2) { int to = 0; int vo = 0; int t1; do { t1 = A[t0]; if (t1 >= 0) { B[v0] = A[t0]; vo++; to++; } while (t0Write a RISC-V code for the following C code: #define size 1024 int main () { int reps, int step, option, i; reps=10; step=2; option=1; char a[size]; for (i=0; reps;i++) { for (j=0;jImplement the following expression in assembly language: ECX = –(val3 - val1) + (-val4 + val2) + 3 Assume that val1 is 16-bit variable, val2 is 32-bit variable, val3 is 8-bit variable, and val4 is 8-bit variable. You need to implement the expression the way it is provided, you cannot do any reduction on the expression while implementing it. Initialize val1 with 12 (decimal), val2 with 9 (decimal), val3 with 2 (decimal), val4 with 20 (decimal), You are NOT allowed to update the values stored in val1, val2, val3 and val4 Use mov, add, sub, movsx, movzx, or neg instructions whenever needed. Use the debugger to verify your answer. Submit the following: Save your source code using your last name, Lastname2.asm and upload the Lastname2.asm Screenshot (showing the code and register window) of ECX register contains the correct result. Give the Whole Code and screen shot of the Registerint total; int i; total = 0; for (i = 10; i > 0; i--) { total + i; } The following translates this into ARM assembler: /* -- sum-to-ten.s */ .text .global start _start: mov rº, #0 mov r1, #10 again: end: add r0, r0, r1 subs r1, r1, #1 bne again mov r7, #1 swi 0 @r0 := 0 @r1 = 10 @r0 := r0 + r1 @ r1 := r1 - 1 @loop again if we haven't hit zero @setup exit @exit In the above assembly language program, we first use the MOV instruction to initialize RO at O and R1 at 10. The ADD instruction computes the sum of RO and R1 (the second and third arguments) and places the result into RO (the first argument); this corresponds to the total += i; line of the equivalent C program. Note the use of the label at the beginning of the ADD instruction. The label is simply the relative memory address of where that instruction is when the assembler loads it into memory. The subsequent SUBS instruction decreases R1 by 1. To understand the next instruction, we need to understand that in addition to the…Convert the C function below to MIPS assembly language. Make sure you follow the MIPS calling conventions. unsigned int sum(unsigned int n) { if (n == 0) return %; else return n+ sum(n-1); %3D%3DQ: CONVERT FOLLOWING C CODE INTO ARM ASSEMBLY CODE #include<stdio.h>int main() {int i, space, rows = 8, star=0; for(i = 0; i < rows-1; i++) {for(space = 1; space < rows-i; space++) {printf(" ");}for (star = 0; star <= 2*i; star++) {if(star==0 || star==2*i)printf("*");elseprintf(" ");}printf("\n");}for(i=0; i<2*rows-1; i++){printf("*");}return 0;}Create a x86 assembly function for this c code: /*float calculateAvg ( float blocks[ ], int SIZE ) { float AVG = 0.0; for ( int i = 0; i < SIZE; i++ ) AVG += fabsf ( blocks[ i ] ); AVG = AVG / ( float ) SIZE; return AVG;} */SEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. 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