# Assuming that y is stored in $s0 and base address of A array is stored in $s1. lw $t0, 800($s1) # $t0 = A[200] add $t0, $s0, $t0 # $t0 = y + A[200] sw $t0, 2000($s1) # A[500] = y + A[200]
7. The value for variable y. A[500]=y+ A[200]; Compile the given array in C to MIPS...
Compile the following C while loop into MIPS assembly code assuming the following register-variable mapping shown below. Also assume the array A holds integers.//$s0: A, $s1: I, $s2: j, $s3: x for(i=0; i<100; i++) {x=0; for(j = i + l; j < 100; j++) {x = x + A [j];} A[i] = x;}
MIPS assembly language
Implement the following code in MIPS int array [ ] {2, 3, 4, 5, 6); int main) int num, position; scanf("%d",&num) ; position search(array, printf("The position is: num, 5); %d\n",positio int search(int array, int num, int size int position =-1; for(int i-0;i<size; i++) if(array [i]=num) { position-i; break; return position; Register map $s1: position $a0: array address $a1: num . $a2: size . $VO: return value
int a = 200; // $1 int b = 500; // $2 int xMax = 1000; // $3 int yMax = 2000; // $4 if (xMax < a){ a = xMax; } if (yMax < b){ b = yMax; } How would I translate the above C code into MIPS assembly language?
MIPS assembly language question
If $s0 has the base address of array A, and A has the following elements: 9, 6, 4, 24, 33, 0, 1; What is the value of $t1 after the instructions are executed? addi $t1, $zero, 20 add $t0, $t1, $S0 lw $t1, (Sto)
Name B. (7 pts) MIPS short answer 1. (3pt) For the following MIPS assembly language program: loop: addi Sto, $to,-1 bne $to, $zero, loop Translate the second instruction into MIPS machine language and write it in hex. 2. (2 pt) Which best describes the reason that we maintain the stack pointer in a register? (circle one) i. The hardware forces use of a stack pointer. ii. We need a local pointer because we are often limited to relative addressing. ili....
1. Convert the following C statement to MIPS code. Assume x, y and z are assigned to registers $s0, $s1 and $s2 respectively and the starting address of array A is $s6. x = y + z + A[3] 2. 2.Convert the following C statement to MIPS code. Assume the base address of array A is $s3 and h is assigned to register $s2 . A[7] = h + A[7]
7. Translate the following C code to MIPS assembly code. Use a minimum number of instructions. Assume that the values of a,b, i and j are in registers Ss0, Ss1, St0, and St1, respectively. Also, assume that register SS2 holds the base address of the array D. for(i-0; i<a; i++) for(j=0 ; j<b; j++)
==> Please In C Language <== Insert a value in a sorted array: Given an integer array a[9] = { 2, 3, 5, 7, 11, 13, 17, 19 }, read in an integer k from the screen and insert k into the array so the new array remains sorted. Note that array a has size 9, but there are only 8 initial values. This ensures that no value will be lost after the insertion. For example, when k=12 , the...
Write a MIPS assembly instruction that loads the value of a signed char (C++ char datatype uses 1 byte) variable in register $s1. The address of this variable is stored in register $s4.
For Computer Organization and Design: Suppose we have 3 arrays A, B, and C, array A’s base address is in $S3, B’s base address is in $S4, C’s base address is in $S5. You can only use two temporal registers $t0 and $t1. Convert the following code into MIPS assembly language code. Suppose every array element occupies 1 word in MIPS (32 bits). A[20] = B[0] + C[6] - 1