Write an assembly language program which implements the following C code that is a find minimum problem solution implementation:

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Systems Programming
Purpose: The purpose of this assignment is to practice loop instructions and
array iterations in M6800 assembly language programming.
Write an assembly language program which implements the following C code that
is a find minimum problem solution implementation:
int A[10] = {8, 14, 6, 16, 5, 5, 10, 9, 4, 11};
int i = 0;
int min = 255; // This is the largest number 8-bit accumulators can hold.
while (i < 10) {
if (A[i] < min)
min = A[i];
}
Your solution should be able to handle all possible array elements and orderings
for any value an accumulator can hold.
You should treat variable į as XR in assembler (index register) and store variable
min in the address 60H.
0010: 00 00 00 00 00 00 00 00 0O 00 00 00 00 00 00 00
8628: 00 060 00 00 00 600 00 00 00 08 08 00 6 00 00 00
0030: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0040: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0050: 00 G0 00 00 00 00 00 00 00 00 00 00 00 00 00 00
006 0: 00 00 00 00 00 00 00 00 00 08 00 00 00 00 00 00
0070: 00 00 00 00 0O 00 00 00 00 00 00 00 00 0O 00 00
008 0: 05 G0 00 00 00 00 00 00 00 00
009 0: 00 60 00 00 00 00 00 00 00 00
BBA 8: 00 B0 00 00 00 00 00 00 00 00 00 00 00 00 00 00
OOB 0: 00 60 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0OC 0: 00 00 00 00 00 00 00 00 0O 00 00 00 00 00 00 00
OOD 8: 00 00 00 00 00 00 00 00 60 00 00 00 00 00 00 00
OGE 0: 00 G0 00 00 00 00 00 00 00 00 00 00 00 00 00 00
AAEA- AA AA AA AG 00 00 00 00 00 00 00 00 00 0O 00 00
8100: 86 05 97 0 00 00 00 00 6O 00
00 00 00 00 00
00 00 00 00 00
0120: 00 00 00 00 00 00 00 00 60 00
6130: 00 00 00 00 00 00 00 00 08 00
Transcribed Image Text:Systems Programming Purpose: The purpose of this assignment is to practice loop instructions and array iterations in M6800 assembly language programming. Write an assembly language program which implements the following C code that is a find minimum problem solution implementation: int A[10] = {8, 14, 6, 16, 5, 5, 10, 9, 4, 11}; int i = 0; int min = 255; // This is the largest number 8-bit accumulators can hold. while (i < 10) { if (A[i] < min) min = A[i]; } Your solution should be able to handle all possible array elements and orderings for any value an accumulator can hold. You should treat variable į as XR in assembler (index register) and store variable min in the address 60H. 0010: 00 00 00 00 00 00 00 00 0O 00 00 00 00 00 00 00 8628: 00 060 00 00 00 600 00 00 00 08 08 00 6 00 00 00 0030: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0040: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0050: 00 G0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 006 0: 00 00 00 00 00 00 00 00 00 08 00 00 00 00 00 00 0070: 00 00 00 00 0O 00 00 00 00 00 00 00 00 0O 00 00 008 0: 05 G0 00 00 00 00 00 00 00 00 009 0: 00 60 00 00 00 00 00 00 00 00 BBA 8: 00 B0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 OOB 0: 00 60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0OC 0: 00 00 00 00 00 00 00 00 0O 00 00 00 00 00 00 00 OOD 8: 00 00 00 00 00 00 00 00 60 00 00 00 00 00 00 00 OGE 0: 00 G0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 AAEA- AA AA AA AG 00 00 00 00 00 00 00 00 00 0O 00 00 8100: 86 05 97 0 00 00 00 00 6O 00 00 00 00 00 00 00 00 00 00 00 0120: 00 00 00 00 00 00 00 00 60 00 6130: 00 00 00 00 00 00 00 00 08 00
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