10-Convert the following C code to MIPS. Assume the address of base array is associated with $s0, n is associated with $s1, position is associated with $t0, c is associated with $t1, d is associated with $t2, and swap is associated with $t3 for (c = 0; c < (n - 1); c++) { position = c; for (d = c + 1; d < n; d++) if (array[position] > array[d]) position = d; if (position != c) { swap = array[c]; array[c] = array[position]; array[position] = swap;

Database System Concepts
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ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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10-Convert the following C code to MIPS. Assume the address of base array is
associated with $s0, n is associated with $s1, position is associated with $t0, c is
associated with $t1, d is associated with $t2, and swap is associated with $t3
for (c = 0; c < (n - 1); c++)
{
position = c;
for (d = c + 1; d < n; d++)
if (array[position] > array[d])
position = d;
if (position != c)
swap = array[c];
array[c] = array[position];
array[position] = swap;
}
Transcribed Image Text:10-Convert the following C code to MIPS. Assume the address of base array is associated with $s0, n is associated with $s1, position is associated with $t0, c is associated with $t1, d is associated with $t2, and swap is associated with $t3 for (c = 0; c < (n - 1); c++) { position = c; for (d = c + 1; d < n; d++) if (array[position] > array[d]) position = d; if (position != c) swap = array[c]; array[c] = array[position]; array[position] = swap; }
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