In this program we will see how to reverse the digits of an 8-bit number using 8085.Problem StatementWrite 8085 Assembly language program to reverse an 8-bit number stored at location 8000H. Also, store the result at 8050H.DiscussionHere the task is too simple. There are some rotating instructions in 8085. The RRC, RLC are used to rotate the Accumulator content to the right and left respectively without carry. We can use either RRC or RLC to perform this task.InputAddressData……80004C……Flow Diagram ProgramAddressHEX CodesLabelsMnemonicsCommentsF0003A, 00, 80 LDA 8000HTake the number from memoryF0030F RRCRotate right without carry four timesF0040F RRC F0050F RRC F0060F RRC F00732, 50, 80 STA 8050HStore the result at memoryF00A76 HLTTerminate the ... Read More

In this program we will see how to find the minimum digit from a two-digit number.Problem StatementWrite 8085 Assembly language program to find the minimum digit from a two-digit number. The number is stored at location 8000H, store the result at 8050H.DiscussionHere we are performing this task by using masking operation. Each digit takes one nibbles. We are masking the upper nibble by ANDing with 0FH (0000 1111). Store the lower nibble into another register. After that, we are taking the upper nibble. To get it, we are shifting the number to the right four times to convert lower nibble ... Read More

In this program, we will see how to find the square of an 8-bit number.Problem StatementWrite 8085 Assembly language program to find the square of a number The number is stored at location 8000H, store the result at 8050H.DiscussionIn 8085, we cannot perform the multiplication operation directly. We are performing the multiplication by using repetitive addition. To get square of a number, we have to multiply the number with itself.InputAddressData……80000C……Flow Diagram ProgramAddressHEX CodesLabelsMnemonicsCommentsF00021, 00, 80 LXI H, 8000HLoad the number from 8000HF003AF XRA AClear accumulatorF00446 MOV B, MLoad data from memory to BF00586LOOPADD MAdd memory byte with AF00605 DCR BDecrease B by 1F007C2, 05, F0 JNZ ... Read More

In this program we will see how to find nth power of a number.Problem StatementWrite 8085 Assembly language program to find nth power of a number. The base is stored at location 8000H, and the exponent is stored at 8001H. Store the result at 8002H.DiscussionIn 8085, we cannot perform the multiplication operation directly. Here we are writing a subroutine to perform the multiplication by using repetitive addition. To perform nth power of a number, we have to multiply the number n times. The n value is used as a counter. If the base is 03H, exponent is 05H, then the ... Read More

In this program, we will see how to count the number of 1’s in an 8-bit number stored in register B.Problem StatementWrite 8085 Assembly language program to count the number of 1s in 8-bit number stored in register B.DiscussionIn this program, we are using the rotate operation to count the number of 1’s. As there are 8 different bits in 8-bit number, then we are rotating the number eight times. we can use RRC or RLC. Here we have used the RRC instruction. This instruction sends the LSb to MSb also to carry flag. So after each iteration, we can ... Read More

In this program we will see how to find the factorial of a number.Problem StatementWrite 8085 Assembly language program to find the factorial of an 8-bit number.DiscussionIn 8085, there is no direct instruction to perform multiplication. We need to perform repetitive addition to get the result of the multiplication. In each step we are decreasing the value of B and multiply with the previous value of B. We are repeating these steps until B reaches 1. and B – 1 to 0. Thus the factorial is generated.InputAddressData800005Flow Diagram ProgramAddressHEX CodesLabelsMnemonicsCommentsF00021, 00, 80 LXI H, 8000HLoad the numberF00346 MOV B, MTake number from memory ... Read More

In this program we will see how to multiply using logical operators.Problem StatementWrite 8085 Assembly language program to multiply two 8-bit numbers using logical operators.DiscussionWe are assuming the first number is in register B, and second number is in register C, and the result must not have any carry.Here we are multiplying with 04H. We can perform the multiplication by left rotate two times. Assign 06H into B, and 04H into C. Load B to A, then rotate the accumulator two times. Store the result into a specified memory.InputRegisterDataB06C04Flow Diagram ProgramAddressHEX CodesLabelsMnemonicsCommentsF00006, 06 MVI B, 06H F0020E, 04 MVI C, 04H F00478 MOV A, BLoad B ... Read More

In this program we will see how to add three 16-bit numbers stored in register pairs.Problem StatementWrite 8085 Assembly language program to add three 16-bit numbers stored in register pair BC, DE and HL. Store the result at DE register pair.DiscussionIn this program we are storing the 16-bit numbers into BC, DE and HL pair. We have DAD D instruction which helps to add HL and DE register pair, and store the result into HL pair. After that copy BC to DE, then again perform DAD D to add them. Finally using XCHG we are storing them into DE register ... Read More

The Program and data which are stored in the memory, are used externally to the microprocessor for executing the complete instruction cycle. Thus to execute a complete instruction of the program, the following steps should be performed by the 8085 microprocessor.Fetching the opcode from the memory;Decoding the opcode to identify the specific set of instructions;Fetching the remaining Bytes left for the instruction, if the instruction length is of 2 Bytes or 3 Bytes;Executing the complete instruction procedure.The given steps altogether constitute the complete instruction cycle. These above mentioned steps are described in detail later. The above instructions are assumed by ... Read More

The select unit of the register in the 8085 selects any one of the register pairs (BC, DE, HL, SP, PC, or WZ) for sending them to it to the latch unit specified for addressing. For example, the contents of the PC be C200H. If the selection unit be the register which selects the PC, and sends C200H from the PC to the address latch internally thereafter the latch holds the specified value and sends directly out to the pins of the address after buffering. The Most Significant Byte of the address i.e. the C2H, is sent out to the ... Read More