8086 program to reverse 8 bit number using 8 bit operation
Problem – Write an assembly language program in 8086 microprocessor to reverse 8 bit number using 8 bit operation.
Example – Assume 8 bit number is stored at memory location 2050
Algorithm –
- Load contents of memory location 2050 in register AL
- Assign 0004 to CX Register Pair
- Rotate the contents of AL by executing ROL instruction using CX
- Store the content of AL in memory location 2050
Program –
Memory Address |
Mnemonics |
Comments |
400 |
MOV AL, [2050] |
AL<-[2050] |
404 |
MOV CX, 0004 |
CX <- 0004 |
407 |
ROL AL, CX |
Rotate AL content left by 4 bits(value of CX) |
409 |
MOV [2050], AL |
[2050]<-AL |
40D |
HLT |
Stop Execution |
Explanation –
- MOV AL, [2050] loads contents of memory location 2050 in AL
- MOV CX, 0004 assign 0004 to CX register pair
- ROL AL, CX rotate the content of AL register left by 4 bits i.e. value of CX register pair
- MOV [2050], AL stores the content of AL in 2050 memory address
- HLT stops executing the program
Last Updated :
14 Jun, 2018
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