.file 1 "" .section .mdebug.abi32 .previous .nan legacy .module fp=32 .module nooddspreg .abicalls .globl array .data .align 2 .type array, @object .size array, 20 array: .word 2 .word 3 .word 4 .word 5 .word 6 .rdata .align 2 $LC0: .ascii "%d\000" .align 2 $LC1: .ascii "The Position is: %d\012\000" .text .align 2 .globl main .set nomips16 .set nomicromips .ent main .type main, @function main: .frame $fp,40,$31 # vars= 8, regs= 2/0, args= 16, gp= 8 .mask 0xc0000000,-4 .fmask 0x00000000,0 .set noreorder .cpload $25 .set nomacro addiu $sp,$sp,-40 sw $31,36($sp) sw $fp,32($sp) move $fp,$sp .cprestore 16 movz $31,$31,$0 addiu $2,$fp,28 move $5,$2 lw $2,%got($LC0)($28) nop addiu $4,$2,%lo($LC0) lw $2,%call16(__isoc99_scanf)($28) nop move $25,$2 .reloc 1f,R_MIPS_JALR,__isoc99_scanf 1: jalr $25 nop lw $28,16($fp) lw $2,28($fp) li $6,5 # 0x5 move $5,$2 lw $4,%got(array)($28) lw $2,%got(search)($28) nop move $25,$2 .reloc 1f,R_MIPS_JALR,search 1: jalr $25 nop lw $28,16($fp) sw $2,24($fp) lw $5,24($fp) lw $2,%got($LC1)($28) nop addiu $4,$2,%lo($LC1) lw $2,%call16(printf)($28) nop move $25,$2 .reloc 1f,R_MIPS_JALR,printf 1: jalr $25 nop lw $28,16($fp) move $2,$0 move $sp,$fp lw $31,36($sp) lw $fp,32($sp) addiu $sp,$sp,40 j $31 nop .set macro .set reorder .end main .size main, .-main .align 2 .globl search .set nomips16 .set nomicromips .ent search .type search, @function search: .frame $fp,24,$31 # vars= 8, regs= 1/0, args= 0, gp= 8 .mask 0x40000000,-4 .fmask 0x00000000,0 .set noreorder .set nomacro addiu $sp,$sp,-24 sw $fp,20($sp) move $fp,$sp sw $4,24($fp) sw $5,28($fp) movz $31,$31,$0 sw $6,32($fp) li $2,-1 # 0xffffffffffffffff sw $2,8($fp) sw $0,12($fp) b $L4 nop $L7: lw $2,12($fp) nop sll $2,$2,2 lw $3,24($fp) nop addu $2,$3,$2 lw $3,0($2) lw $2,28($fp) nop bne $3,$2,$L5 nop li $2,1 # 0x1 sw $2,8($fp) b $L6 nop $L5: lw $2,12($fp) nop addiu $2,$2,1 sw $2,12($fp) $L4: lw $3,12($fp) lw $2,32($fp) nop slt $2,$3,$2 bne $2,$0,$L7 nop $L6: lw $2,8($fp) move $sp,$fp lw $fp,20($sp) addiu $sp,$sp,24 j $31 nop .set macro .set reorder .end search .size search, .-search .ident "GCC: (Ubuntu 5.4.0-6ubuntu1~16.04.9) 5.4.0 20160609"
MIPS assembly language Implement the following code in MIPS int array [ ] {2, 3, 4,...
MIPS assembly language Covert this code to MIPS: #include <stdio.h> int function (int a) int main)i int x=5 ; int y: y function(x); printf "yd",y); return 0; int function (int a) return 3*a+5; Assumptions: . Place arguments in $a0-$a3 . Place return values in $vO-$v1 Return address saved automatically in $ra . lgnore the stack for this example. (Thus, the function will destroy registers used by calling function
Convert the following C fragment to equivalent MIPS assembly language. Assume that the variables a, b, c, d, i and x are assigened to registers $t1, $t2, $t3, $t4, $s0 and $s1 respectively. Assume that the base address of the array A and B is in register $a0 and $a1 respectively. if (a > 0) b = a + 10; else b = a - 10;
How can I convert the following C code to MIPS Assembly? +++++++++++++++++++++++++++++++++ MIPS main program ++++++++++++++++++++++++++++++++ .data # Defines variable section of an assembly routine. array: .word x, x, x, x, x, x, x, x, x, x # Define a variable named array as a word (integer) array # with 10 unsorted integer numbers of your own. # After your program has run, the integers in this array # should be sorted. .text # Defines the start of the code...
5. Write the MIPS minimal sequence of instructions for the following C procedure code: int array_sum (a[], b[]) { int i; i =1; for (i=1; i<100; i=i+1) { b[i] = D + a[i-1] + a[i] + a[i+1] ;} } Assume that: a and b are arrays of words and the base address of “a” is in $a0 and the base address of “b” is in $a1, Register $S1 is associated with...
WRITE THE FOLLOWING CODE IN FLOATING POINT NUMBERS IN ASSEMBLY LANGUAGE USING MIPS IN MARS .data prompt: .asciiz "\nMaximum number is : " prompt1: .asciiz "\nMinimum number is : " prompt2: .asciiz "\nRange of the array is : " size: .word 10 #load array array: .word 23, -12, 45, -32, 52, -72, 8, 13,22,876 .text #load address of array and size la $s4,array #load address of A lw $t0,size #load i to t0 jal getArrayRange li $v0, 4 la...
Implement the functionality of "main" function for the following C code in MIPS assembly. Comment your code including which registers represent the variables in the main function. Include the .data and the .text sections. Do not write the bit_count function. int main() { int count = 149; int num = 432; if (count + num > 534){ count = bit_count(num); printf("%d bits to store the number", count); }else{ printf("cannot count bit in %d",) } }
Can anyone help to solve this MIPS assembly language problem? Please help. Translate the fowlowwing C code to MIPS assembly language. void main() { int i, sum, begin, end; for(i=0; i < 5; i++){ scanf(“%d, %d”, &begin, &end); sum = accum_range(begin, end); printf(“sum[%d: %d] = %d\n”, begin, end, sum); } int accum_range(int a, int b){ return accum (b) - accum(a); } int accum(int final) { int sum = 0; for (int I = 1; I <= final; I = I...
Need to implement an IterativeMax function in assembly language using the MIPS calling convention. IterativeMax take two arguments: The address of an array of word-long (four byte) signed integers, held in $a0. The length of the array as an unsigned integer, held in $a1. The main purpose of the function is to determine which element of the provided array is the largest (i.e., the maximum element in the array). This is achieved by going through the array one element at...
Turn the Following c-code into MIPS assembly code. You are given the main procedure which calls multiply. You are also given the argument registers to be used. /* C-program */ int multiply (int number, int times) { int f; f = number * times return f; } # MIPS PROGRAM # assumes we have called the program leaf_example # $a0=number, $a1=times, $s0=f main: $addi $a0,$zero,3 $addi $a1,$zero,2 jal multiply # place the address into $ra j EXIT2 EXIT2: j OS...
Write an assembly language program that corresponds to the following C program: int width; int length; int perim; int main () { scanf ("%d%d", &width, &length); perim = (width + length) * 2; printf ("width = %d\n", width); printf ("length = %d\n\n", width); printf ("perim = %d\n", perim); return 0; }