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Print the Table of a Given Number in 8086 Microprocessor

In this tutorial, we will learn how to print the table of a given number in 8086 Microprocessor? By Ayush Sharma Last updated : May 22, 2023

Problem Statement

Write an assembly language program in 8086 to print the table of a given integer.

Assumptions: Suppose the inputted number is at memory location 500 and the table will be printed from starting location 600 till 609 in hexadecimal.

Algorithm

1. Load input number address in SI and also load the address where we want output in DI .
2. Store 00 in CH register.
3. Increment value of CH by 1 and move the content of [SI] into AH register.
4. 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.
5. Compare the value of CH and 0A, if not equal then go to step number 3 otherwise halt the program.

8086 program to print the table of input integer

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

1. MOV SI, 500: load 500 in SI.
2. MOV DI, 600: load 600 in DI.
3. MOV CH, 00: load 00 data in CH register.
4. INC CH: increment the value inside CH register by 1.
5. MOV AL, SI: move the content of SI into AL register.
6. MUL CH: multiply the contents of AL and CH register and store in AX register.
7. MOV [DI], AL: move the contents of AL register into [DI].
8. INC DI: increment the value of DI by 1.
9. CMP CH, 0A: subtract data inside CH register and 0A.
10. JNZ 408: jump to address 408 if zero flag is 0.
11. HLT: terminate the program.

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