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assume nums = $s0, i = $s1, j = $s2, nums[I] = $t0, nums[k] != (not equal) $t1 $t2, size = $s5 int k; for (j = 0; j < size-1; j++) { i = j; for (k = j+1; k < size; k++) { if (nums[i] >= nums[k]) { i = k; } int tmp = nums[i]; nums[i] = nums[j]; sw $s3, 0($t1) <---- This two lines like so nums[j] = tmp;   sw $t2, 0($t0) <------ }

assume nums = $s0, i = $s1, j = $s2, nums[I] = $t0, nums[k] != (not equal) $t1 $t2, size = $s5 int k; for (j = 0; j < size-1; j++) { i = j; for (k = j+1; k < size; k++) { if (nums[i] >= nums[k]) { i = k; } int tmp = nums[i]; nums[i] = nums[j]; sw $s3, 0($t1) <---- This two lines like so nums[j] = tmp;   sw $t2, 0($t0) <------ }

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Computer Science

Convert to Mips Assembly Language Please!

assume nums = $s0, i = $s1, j = $s2, nums[I] = $t0, nums[k] != (not equal) $t1 $t2, size = $s5

int k;

for (j = 0; j < size-1; j++)
{
i = j;
for (k = j+1; k < size; k++) {
if (nums[i] >= nums[k]) {
i = k;
}

int tmp = nums[i];
nums[i] = nums[j]; sw $s3, 0($t1) <---- This two lines like so
nums[j] = tmp;   sw $t2, 0($t0) <------
}

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