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Study the given program and answer the questions given below. \[ \text { STC } \] CMC LXI B , 1234H MOV A, B RAR MOV H, A MOV A, C RAR MOV L, A HLT (i) Write the purpose of the program.

Ask by Bob Fuentes. in India
Jan 26,2025

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Answer

The program initializes registers B and C with the value `1234H`, performs a right bitwise rotation on each register through the carry flag, and stores the rotated values into registers H and L, respectively.

Solution

### (i) **Purpose of the Program** The given assembly program performs the following operations step-by-step: 1. **Initialize the Carry Flag:** - `STC` (Set Carry): Sets the Carry flag to **1**. - `CMC` (Complement Carry): Inverts the current Carry flag. Since `STC` set it to **1**, `CMC` changes it to **0**. 2. **Load Immediate Data into Registers B and C:** - `LXI B, 1234H`: Loads the hexadecimal value **1234H** into the register pair **B** and **C**. - **B** = `12H` (0001 0010 in binary) - **C** = `34H` (0011 0100 in binary) 3. **Rotate Register B Right Through Carry:** - `MOV A, B`: Moves the content of register **B** (`12H`) into the **Accumulator (A)**. - `RAR` (Rotate Accumulator Right): - **Before Rotation:** - **A** = `0001 0010` (12H) - **Carry Flag** = 0 - **After Rotation:** - **A** becomes `0000 1001` (09H) - **Carry Flag** remains **0** - `MOV H, A`: Stores the rotated value (`09H`) into register **H**. 4. **Rotate Register C Right Through Carry:** - `MOV A, C`: Moves the content of register **C** (`34H`) into the **Accumulator (A)**. - `RAR` (Rotate Accumulator Right): - **Before Rotation:** - **A** = `0011 0100` (34H) - **Carry Flag** = 0 - **After Rotation:** - **A** becomes `0001 1010` (1AH) - **Carry Flag** remains **0** - `MOV L, A`: Stores the rotated value (`1AH`) into register **L**. 5. **Terminate the Program:** - `HLT`: Halts the execution of the program. **Summary of Operations:** - **Initialization:** - Registers **B** and **C** are loaded with `12H` and `34H` respectively. - Carry flag is initially set to **1** and then complemented to **0**. - **Rotation:** - **Register B (`12H`)** is rotated right through the carry flag resulting in `09H` stored in **H**. - **Register C (`34H`)** is rotated right through the carry flag resulting in `1AH` stored in **L**. **Overall Purpose:** The program **initializes the register pair B and C with the value `1234H`**, **performs a right bitwise rotation** on each of these registers **through the carry flag**, and **stores the resulting rotated values into registers H and L respectively**. Finally, it halts execution. #### **Detailed Flow:** | Instruction | Operation | B | C | A | H | L | Carry Flag | |-------------|-----------------------------------|-----|-----|-----|-----|-----|------------| | STC | Set Carry | 12H | 34H | | | | 1 | | CMC | Complement Carry | 12H | 34H | | | | 0 | | LXI B,1234H | Load B=12H, C=34H | 12H | 34H | | | | 0 | | MOV A,B | A = B | 12H | 34H | 12H | | | 0 | | RAR | Rotate A right through Carry (12H→09H) | 12H | 34H | 09H | | | 0 | | MOV H,A | H = A | 12H | 34H | 09H | 09H | | 0 | | MOV A,C | A = C | 12H | 34H | 34H | 09H | | 0 | | RAR | Rotate A right through Carry (34H→1AH) | 12H | 34H | 1AH | 09H | | 0 | | MOV L,A | L = A | 12H | 34H | 1AH | 09H | 1AH | 0 | | HLT | Halt | 12H | 34H | 1AH | 09H | 1AH | 0 | ### **Final Register States:** - **H** = `09H` - **L** = `1AH` - **Carry Flag** = **0**

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This program is designed to manipulate binary data stored in registers and shift bits to the right. Specifically, it takes the content of the B and C registers, performs a right arithmetic shift (RAR) on each, and stores the results in the H and L registers, respectively. The final output consists of the higher byte in register H and the lower byte in register L, effectively creating a new 16-bit value from the original 8-bit values in registers B and C. By executing this sequence, you can achieve bitwise operations that are often critical for efficient data processing or communication protocols, showcasing the power of low-level programming in microcontrollers.

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