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Motorola Coldfire 68000 microprocessor 1.[5 points] Use the Programming Card to calculate the memory locations of instructions in the Main Routine, 2.[18 points] Show the

Motorola Coldfire 68000 microprocessorimage text in transcribedimage text in transcribed

1.[5 points] Use the Programming Card to calculate the memory locations of instructions in the Main Routine, 2.[18 points] Show the contents of the memory map from Ox4FEO to 0x5000 after the entire program is executed. If the contents are unknown, put 0xFF to that location. The main routine passes two long word numbers store at locations 0x2000 and 0x3000 to the SUB1 subroutine. The subroutine multiplies the two numbers and stores the long word result at the location of 0x4FEC. 3.[3 points] Show the contents of D2, D3 and A7 after the program is finished. Initial Contents: D0-0xFFCCAA88, D1-0xEEDDOOBB, A0-0xFFFF0000 At memory location 0x2000-2003: OxFF FF FF FE At memory location 0x3000-3003: 0x00 03 00 02 Main Routine: Memory Location .ORG Ox1000 0x1000 LEA 0x5000,%A7 /* Points the stack to addr 0x5000 LEA 0x1500,%A1 * Points A1 to addr 0x1500 PEA Ox2000 PEA 0x3000 BSR SUB1 TRAP #15 /Push addr 0x2000 to stack /*Push addr 0x3000 to stack /Branch to a subroutine ORG 0x4000 %A1, #-4 #12,%A7 /*Set the Frame Pointer /*Save the Register values SUB1 LINK SUBA.I 0x18 ( %A7 ) , %A3 0x1C (%A7) , %A4 (8A3) ,&D2 ( %A4 ) , % 93 &D2,8D3/*multiply MOVEA.L MOVEA.L MOVE.IL MOVE. L MULU. W MOVE. L MOVEM.L ADDA. L UNLK RTS the values /*Restore (%A7) , %D0/9D1/9A0 #12 , %A7 %A1 the Registers /Release the Frame Pointer Show the memory contents in the table below (Note: not all locations are modified) 1.[5 points] Use the Programming Card to calculate the memory locations of instructions in the Main Routine, 2.[18 points] Show the contents of the memory map from Ox4FEO to 0x5000 after the entire program is executed. If the contents are unknown, put 0xFF to that location. The main routine passes two long word numbers store at locations 0x2000 and 0x3000 to the SUB1 subroutine. The subroutine multiplies the two numbers and stores the long word result at the location of 0x4FEC. 3.[3 points] Show the contents of D2, D3 and A7 after the program is finished. Initial Contents: D0-0xFFCCAA88, D1-0xEEDDOOBB, A0-0xFFFF0000 At memory location 0x2000-2003: OxFF FF FF FE At memory location 0x3000-3003: 0x00 03 00 02 Main Routine: Memory Location .ORG Ox1000 0x1000 LEA 0x5000,%A7 /* Points the stack to addr 0x5000 LEA 0x1500,%A1 * Points A1 to addr 0x1500 PEA Ox2000 PEA 0x3000 BSR SUB1 TRAP #15 /Push addr 0x2000 to stack /*Push addr 0x3000 to stack /Branch to a subroutine ORG 0x4000 %A1, #-4 #12,%A7 /*Set the Frame Pointer /*Save the Register values SUB1 LINK SUBA.I 0x18 ( %A7 ) , %A3 0x1C (%A7) , %A4 (8A3) ,&D2 ( %A4 ) , % 93 &D2,8D3/*multiply MOVEA.L MOVEA.L MOVE.IL MOVE. L MULU. W MOVE. L MOVEM.L ADDA. L UNLK RTS the values /*Restore (%A7) , %D0/9D1/9A0 #12 , %A7 %A1 the Registers /Release the Frame Pointer Show the memory contents in the table below (Note: not all locations are modified)

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