main:
li $s1, i
# i
li $s2, j
# j
li $s3, x
li $s0, A
$L8:
addiu $s2,$s2,1
#x+A[j];
$L11:
movz $s3,$s3,$0
#x=0
slt $s0,$s2,100
#j<100
bne
$s0,$0,$L11 # branch to L11 if
$j=0
addiu $s2,$s2,1
#j = j+1
addiu $s1,$s1,1
#i++
slt
$s2,$s1,100 # i<100
beq
$s2,$0,$L3 # branch to L3 if
$j=0
nop
b $L8
move $s2,$s1
$L3:
j $s3
move $s2,$0
Compile the following C while loop into MIPS assembly code assuming the following register-variable mapping shown...
Translate the following C code to MIPS assembly code. Use a minimum number of instructions. Register allocations - i $s0 - j $s1 - base of A[] $s2 - base of B[] $s3 2) A[3] = B[i] + B[j]; 3) i = 0; while (j != A[i]) { i++; }
Translate the following C code to MIPS assembly. Assume that the values of a, b, i, and j are in registers $s0, $s1, $t0, and $t1, respectively. Also assume that $s2 holds the base address of the array D. for (i = 0: i < a: i++) for (j = 0: j < b: j++) D[2 * j] = i + j;
IN MIPS AND MUST RUN IN QTSPIM
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 stored in registers Ss0, Ss1, St0 and Stl, respectively. Also assume that register Ss2 holds the base address of the array D. for (i=0; i<a; itt) for (i-0j<b:jt+)
B2. Convert the C code to MIPS assembly with only 2 efficient instructions: Register assignment: timer-v0 int timer = 0x0AC8 0001; B3. Write MIPS assembly code segment for the following C code snippet for (i - 0, i < 100; i++) -array Register assignment: i-) $ao Base of array -> $s0 array [ i+1] [i] / 2;
B2. Convert the C code to MIPS assembly with only 2 efficient instructions: Register assignment: timer-v0 int timer = 0x0AC8 0001; B3. Write...
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++)
Using beq only, not bge! 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 $s0, $s1, $t0, and $t1, respectively. Also, assume that register $s2 holds the base address of the integer array D. Comments are required. for(i=1; i<a; i++) for(j=1; j<b; j++) D[2*j] = i + j;
Translate the following C code to MIPS assembly code. Assume that the value of i is in register $t0, and $s0 holds the base address of the integer MemArray if (i > 10) MemArray[i] = 0; else MemArray[i] = -MemArray[i];
C to MIPS Conversion C variable h i j k x int a[] or &a[0] MIPS register replacement $s0 $s1 $s2 $s3 $s4 $a0 Translate to MIPS. DO NOT USE pseudo MIPS instructions (e.g., BGE). Answer MUST use true 32-bit MIPS instructions: Note that all variables (h,i,j,x,a[]) are 32-bit signed integers. while ( h < 3 ) { a[j++]= 0; x = i >> 3; }
2. The table below holds MIPS assembly code fragments with different branch instructions LOOP addi $s2. $s2. 2 subi $t1. st1. 1 bne t1. 0. LOOP DONE: LOOP: it st2. $0. stl beq t2. 0. DONE addi $s2. Ss2. 2 LOOP DONE: For the loops written in MIPS assembly in the above table, assume that the register Şt1 is initialized to the value of 10. What is the value in register $s2 assuming that $s2 initially has a value of...
For the following C statement, what is the corresponding MIPS assembly code? Assume that the variables i, and j are assigned to registers $s0, $s1, respectively. Assume that the base address of the arrays A and B are in registers $s2 and $s3, respectively. B[i] = A[i] - 10