8086 program to print the table of input integer
Last Updated :
29 Oct, 2021
Problem – Write an assembly language program in 8086 to print the table of input integer.
Assumption – Suppose the inputted number is at memory location 500 and the table will be printed from starting location 600 till 609 in hexadecimal.
Example –
Algorithm –
- Load input number address in SI and also load the address where we want output in DI .
- Store 00 in CH register.
- Increment value of CH by 1 and move the content of [SI] into AH register.
- Multiply content of AL and CH and store it in AX and then move content of AL into [DI], then increment value of DI by 1.
- Compare the value of CH and 0A, if not equal then go to step number 3 otherwise halt the program.
Program –
ADDRESS |
MNEMONICS |
COMMENTS |
400 |
MOV SI, 500 |
SI<-500 |
403 |
MOV DI, 600 |
DI<-600 |
406 |
MOV CH, 00 |
CH<-00 |
408 |
INC CH |
CH<-CH+1 |
409 |
MOV AL, [SI] |
AL<-[SI] |
40B |
MUL CH |
AX<-AL*CH |
40D |
MOV [DI], AL |
[DI]<-AL |
40F |
INC DI |
DI<-DI+1 |
410 |
CMP CH, 0A |
CH-0A |
413 |
JNZ 408 |
jump to address 408 if zero flag is 0 |
415 |
HLT |
Terminates the program |
Explanation –
- MOV SI, 500: load 500 in SI.
- MOV DI, 600: load 600 in DI.
- MOV CH, 00: load 00 data in CH register.
- INC CH: increment the value inside CH register by 1.
- MOV AL, SI: move the content of SI into AL register.
- MUL CH: multiply the contents of AL and CH register and store in AX register.
- MOV [DI], AL: move the contents of AL register into [DI].
- INC DI: increment the value of DI by 1.
- CMP CH, 0A: subtract data inside CH register and 0A.
- JNZ 408: jump to address 408 if zero flag is 0.
- HLT: terminate the program.
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