8085 program to count number of elements which are less than 0A

In this section we will count elements which are lesser than 0AH using 8085.

problem Statement

There is an array of some elements. Write 8085 Assembly language program to count number of elements that are lesser than 0AH.

Discussion

The array is placed at location F051H onwards. The F050 is holding the size of the array. The logic is simple. At first we will take the array size into the B register. The C register will count number of elements less than 0AH. We will take numbers one by one from memory, then compare it with 0A. if the CY flag is set it indicates that the Accumulator is holding smaller value, so increase C by one, otherwise ignore it.

Input

Address	Data
F050	FE
F051	02
F052	07
F053	A5
F054	48
F055	08

Flow Diagram

program

Address	HEX Codes	Labels	Mnemonics	Comments
8000	21, 50, F0		LXI H,F050	Point the memory location F050
8003	46		MOV B,M	Load size of the array to B
8004	0E, 00		MVI C,00H	Clear C register to count numbers
8006	23	LOOP	INX H	Point to the first element of array
8007	7E		MOV A,M	Load memory element to Acc
8008	FE, 0A		CPI 0AH	Compare Acc and 0AH
800A	D2, 0E, 80		JNC SKIP	If number is greater, skip it
800D	0C		INR C	Increase C by 1
800E	05	SKIP	DCR B	decrease B by 1
800F	C2, 06, 80		JNZ LOOP	If array is not completed, jump to Loop
8012	79		MOV A,C	Take the number from C to A
8013	32, 50, 51		STA F150	Store result at F150H
8016	76		HLT	Terminate the program.