Final answer:
The MIPS machine code for the given code is provided, along with an explanation of each instruction and its format. The code is then converted to C.
Step-by-step explanation:
The MIPS machine code for the given code is as follows:
addi $s5, $zero, 0
Binary: 001000 10000 10001 0000 0000 0000 0000 0000
Decimal: 848977408
- sll $t1, $s3, 2
Binary: 000000 00000 01100 01001 00000 00000 00001 00000
Decimal: 30976 - add $t1, $t1, $s6
Binary: 000000 01000 01100 10001 00000 100000
Decimal: 67177344 - lw $t0, 0($t1)
Binary: 100011 01100 01000 0000000000000000
Decimal: 12582912 - bne $t0, $s5, Exit
Binary: 000101 01000 10001 0000 0000 0000 0000 0000
Decimal: 262983168 - sll $s4, $s3, 1
Binary: 000000 00000 01100 10000 00000 000010
Decimal: 2048 - sub $s4, $s4, $10
Binary: 000000 10000 10000 10011 00000 100010
Decimal: 16912976 - j Loop
Binary: 000010 00000000000000000000001111
Decimal: 671088592 - Exit
Binary: 000000 00000 00000 00000 00000 101000
Decimal: 40
The given MIPS code consists of various instructions including addi (I format), sll (R format), add (R format), lw (I format), bne (I format), sub (R format), and j (J format).
The given code in C would be:
int save[];
int i, j, k;
k = 0;
while(true) {
j = i << 2;
j = j + save[j];
if(save[j] != k) {
break;
}
k = i << 1;
k = k - 10;
}
Exit: