Translate the following C code into MIPS assembly code, using the minimum number of instructions. Assume that the values of a, b. 1, and are stored in registers $se $51 ste, and $t1, respectively. Also, assume that register $52) holds the base address of the array D. for(i = 0; i < a; i++) for(j = 0; j < b; j++) D[4*j] = i + j; move $10, $zero loop_i: blt $10, $50, loop_jj end_loop_i loop_j: move $11, $zero blt $t1, $s1, update_dj end_loop_j update_d: sll $t2, $11, 2 add $13, $10, $t1 sw $13,0($s2) addi $11, $t1, 1 j loop_j end_loop_j: addi $to, $t0, 1 j loop_iend_loop_i: move $10, $zero outer loop: bge $10, $50, end outer move $11, $zero inner_loop: bge $t1, $s1, end_inner sll $12, $11, 2 add $13, $to, $t1 sw $13, 0($s2) addi $t1, $t1, 1j inner_loop end_inner: addi $to, $t0, 1 jouter loop end outer: move $to, $zero outer loop: blt $to, $s0, continue outer j end outer continue outer: move $t1, $zero inner_loop: blt $t1, $s1, continue_inner j end_inner continue_inner: sll $t2, $11, 2 add $12, $s2, $12 add $13, $to, $t1 sw $13, 0($12) addi $t1, $t1, 1 j inner_loop end_inner: addi $to, $t0, 1j outer_loop end_outer: move $10, $zero outer loop: bge $to, $s0, end outer move $t1, $zero inner_loop: bge $t1, $s1, end_inner sll $12, $11, 2 add $13, $to, $t1 add $12, $s2, $t2 sw $t3, 0($12) addi $t1, $t1, 1 j inner_loop end_inner: addi $10, $10, 1 j outer_loop end outer:
Translate the following C code into MIPS assembly code, using the minimum number of instructions. Assume that the values of a, b. 1, and are stored in registers $se $51 ste, and $t1, respectively. Also, assume that register $52) holds the base address of the array D. for(i = 0; i < a; i++) for(j = 0; j < b; j++) D[4*j] = i + j; move $10, $zero loop_i: blt $10, $50, loop_jj end_loop_i loop_j: move $11, $zero blt $t1, $s1, update_dj end_loop_j update_d: sll $t2, $11, 2 add $13, $10, $t1 sw $13,0($s2) addi $11, $t1, 1 j loop_j end_loop_j: addi $to, $t0, 1 j loop_iend_loop_i: move $10, $zero outer loop: bge $10, $50, end outer move $11, $zero inner_loop: bge $t1, $s1, end_inner sll $12, $11, 2 add $13, $to, $t1 sw $13, 0($s2) addi $t1, $t1, 1j inner_loop end_inner: addi $to, $t0, 1 jouter loop end outer: move $to, $zero outer loop: blt $to, $s0, continue outer j end outer continue outer: move $t1, $zero inner_loop: blt $t1, $s1, continue_inner j end_inner continue_inner: sll $t2, $11, 2 add $12, $s2, $12 add $13, $to, $t1 sw $13, 0($12) addi $t1, $t1, 1 j inner_loop end_inner: addi $to, $t0, 1j outer_loop end_outer: move $10, $zero outer loop: bge $to, $s0, end outer move $t1, $zero inner_loop: bge $t1, $s1, end_inner sll $12, $11, 2 add $13, $to, $t1 add $12, $s2, $t2 sw $t3, 0($12) addi $t1, $t1, 1 j inner_loop end_inner: addi $10, $10, 1 j outer_loop end outer:
Computer Networking: A Top-Down Approach (7th Edition)
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Chapter1: Computer Networks And The Internet
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Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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![Translate the following C code into MIPS assembly code, using the minimum number of instructions. Assume that the values of a, b. 1, and are stored in registers $se $51 ste, and $t1, respectively. Also, assume that
register $52) holds the base address of the array D.
for(i = 0; i < a; i++)
for(j = 0; j < b; j++)
D[4*j] = i + j;
move $10, $zero loop_i: blt $10, $50, loop_jj end_loop_i loop_j: move $11, $zero blt $t1, $s1, update_dj end_loop_j update_d: sll $t2, $11, 2 add $13, $10, $t1 sw $13,0($s2) addi $11, $t1, 1 j loop_j end_loop_j: addi $to, $t0, 1 j loop_iend_loop_i:
move $10, $zero outer loop: bge $10, $50, end outer move $11, $zero inner_loop: bge $t1, $s1, end_inner sll $12, $11, 2 add $13, $to, $t1 sw $13, 0($s2) addi $t1, $t1, 1j inner_loop end_inner: addi $to, $t0, 1 jouter loop end outer:
move $to, $zero outer loop: blt $to, $s0, continue outer j end outer continue outer: move $t1, $zero inner_loop: blt $t1, $s1, continue_inner j end_inner continue_inner: sll $t2, $11, 2 add $12, $s2, $12 add $13, $to, $t1 sw $13, 0($12) addi $t1, $t1, 1 j
inner_loop end_inner: addi $to, $t0, 1j outer_loop end_outer:
move $10, $zero outer loop: bge $to, $s0, end outer move $t1, $zero inner_loop: bge $t1, $s1, end_inner sll $12, $11, 2 add $13, $to, $t1 add $12, $s2, $t2 sw $t3, 0($12) addi $t1, $t1, 1 j inner_loop end_inner: addi $10, $10, 1 j outer_loop end outer:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdaad9e7e-1f14-44d5-98ac-51623b651f9b%2F3b8a6146-768d-490f-b2c2-c0b2bb6a32fc%2F7g3jw7u_processed.png&w=3840&q=75)
Transcribed Image Text:Translate the following C code into MIPS assembly code, using the minimum number of instructions. Assume that the values of a, b. 1, and are stored in registers $se $51 ste, and $t1, respectively. Also, assume that
register $52) holds the base address of the array D.
for(i = 0; i < a; i++)
for(j = 0; j < b; j++)
D[4*j] = i + j;
move $10, $zero loop_i: blt $10, $50, loop_jj end_loop_i loop_j: move $11, $zero blt $t1, $s1, update_dj end_loop_j update_d: sll $t2, $11, 2 add $13, $10, $t1 sw $13,0($s2) addi $11, $t1, 1 j loop_j end_loop_j: addi $to, $t0, 1 j loop_iend_loop_i:
move $10, $zero outer loop: bge $10, $50, end outer move $11, $zero inner_loop: bge $t1, $s1, end_inner sll $12, $11, 2 add $13, $to, $t1 sw $13, 0($s2) addi $t1, $t1, 1j inner_loop end_inner: addi $to, $t0, 1 jouter loop end outer:
move $to, $zero outer loop: blt $to, $s0, continue outer j end outer continue outer: move $t1, $zero inner_loop: blt $t1, $s1, continue_inner j end_inner continue_inner: sll $t2, $11, 2 add $12, $s2, $12 add $13, $to, $t1 sw $13, 0($12) addi $t1, $t1, 1 j
inner_loop end_inner: addi $to, $t0, 1j outer_loop end_outer:
move $10, $zero outer loop: bge $to, $s0, end outer move $t1, $zero inner_loop: bge $t1, $s1, end_inner sll $12, $11, 2 add $13, $to, $t1 add $12, $s2, $t2 sw $t3, 0($12) addi $t1, $t1, 1 j inner_loop end_inner: addi $10, $10, 1 j outer_loop end outer:
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