1. ZONG Wen Department of Computer Science and Engineering The Chinese University of Hong Kong wzong@cse.cuhk.edu.hk

2. 1. Write an assembly program with given instruction 2. Two’s complement operation 3. Ordering of bytes in data

3. Basic component of a computer

4. There are only 6 machine instructions on a single- accumulator processor, namely: Load [Memory]{to ACC}e.g. Load [2872] Store [Memory]{from ACC}e.g. Store [1536] Add Constant{to ACC}e.g. Add -95 Add [Memory]{to ACC}e.g. Add [2132] Multiply Constant{to ACC}e.g. Multiply 23 Multiply [Memory]{to ACC}e.g. Multiply [298]

5. Variable x and y are stored in address 1000 and 1004, try to implement x 2 + y 2, store the result in 1008. Load [1000]; load data to register Multiply [1000]; perform memory + register operation Store [1008]; store temporary result to memory Load[1004]; Multiply[1004]; Add[1008]; Store[1008]; store result

6. Example: (–6) + 8 * 3 – 1 Step 1  11111010 + (00001000 * 00000011) – 00000001 Step 2  (11111010 + 00011000) – 00000001 Step 3  00010010 – 00000001 Step 4  00010001 Step 5  17 (decimal answer without overflow)

7. Example: (–6) + 8 * 3 – 1 Step 1  11111010 + (00001000 * 00000011) – 00000001 Step 2  (11111010 + 00011000) – 00000001 Step 3  00010010 – 00000001 Step 4  00010001 Step 5  17 (decimal answer without overflow)

8. In big endian, you store the most significant byte in the smallest address In little endian, you store the least significant byte in the smallest address

9. For a data of 4 bytes: 90AB12CD Big endian:

10. For a data of 4 bytes: 90AB12CD Little endian: