Example: Determine the 4 bytes of the interrupt vector for Type 12 interrupt. Solution: Address of the first memory location: Segment --- 0000, offset --- 12*4=48=30H, 20-bit address --- 00030H Based on the following copy of the debug screenshot, the 4 bytes are: 6E 00 91 03. 6E 00 are two bytes of the offset, and 91 03 are the two bytes of the segment. Don't forget byte swapping. Then the 20-bit address of the Type 12 ISR is : 0391:006E. Memory contents of the first 32 interrupt vectors: 0000:0000 68 10 A7 00 8B 01 70 00-16 00 91 03 8B 01 70 00 h............. 0000:0010 8B 01 70 00 B9 06 OC 02-40 07 OC 02 FF 03 OC 02.p...@.. 0000:0020 46 07 OC 02 OA 04 OC 02-3A 00 91 03 54 00 91 03 F........T... 0000:0030 6E 00 91 03 88 00 91 03-A2 00 91 03 FF 03 OC 02 n....... 0000:0040 A9 08 OC 02 A4 09 OC 02-AA 09 OC 02 5D 04 OC 02 0000:0050 BO 09 OC 02 OD 02 DB 02-C4 09 OC 02 8B 05 OC 02 0000:0060 OE OC OC 02 14 0C OC 02-1F OC OC 02 AD 06 OC 02 0000:0070 AD 06 OC 02 A4 FO 00 F0-37 05 OC 02 3C 26 00 CO ...........