This blog post presents an 8086 assembly program designed to count the number of 0s and 1s in a given 16-bit number. We’ll explore the bit manipulation techniques used and analyze the program’s execution flow.

DATA SEGMENT
NO DW 5648H
Z DW ?
O DW ?
DATA ENDS

CODE SEGMENT
ASSUME CS:CODE, DS:DATA
START:
MOV AX, DATA
MOV DS, AX
MOV AX, NO
MOV BX, 00H
MOV CX, 10H
MOV DX, 00H

UP:
ROL AX,1
JC ONE
INC BX
JMP NXT

ONE:
INC DX

NXT:
DEC CX
JNZ UP

MOV Z, BX
MOV O, DX

INT 3
CODE ENDS
END START

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Binary may seem like a bunch of 0s and 1s, but in the right hands, it dances to the rhythm of amazing programs! 💻🎶 Let’s count those bits! 0️⃣1️⃣

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Understanding the Code

This program utilizes bit manipulation to efficiently count 0s and 1s.

Data Segment:

• NO DW 5648H: Declares a 16-bit word variable NO (for “Number”), initialized to 5648H.
• Z DW ?: Reserves space for a 16-bit word variable Z to store the count of zeros.
• O DW ?: Reserves space for a 16-bit word variable O to store the count of ones.

Flowchart:

Code Segment:

• ASSUME CS:CODE, DS:DATA: Specifies segment registers.
• START: The program’s entry point.
• MOV AX, DATA; MOV DS, AX: Sets up the data segment.
• MOV AX, NO: Loads the input number (NO) into the AX register.
• MOV BX, 00H; MOV CX, 10H; MOV DX, 00H: Initializes counters: BX for zeros, CX for the loop (16 iterations for a 16-bit number), and DX for ones.
• UP: This is a loop label.
• ROL AX, 1: This instruction rotates the bits in AX to the left by one position. The least significant bit (LSB) is moved into the carry flag.
• JC ONE: This instruction (Jump if Carry) branches to the ONE label if the carry flag is set (meaning the LSB was a 1).
• INC BX: Increments the zero count (BX).
• JMP NXT: Jumps to the NXT label.
• ONE: This label is the target of the JC instruction.
• INC DX: Increments the one count (DX).
• NXT: This label is the target of the unconditional jump from UP.
• DEC CX: Decrements the loop counter.
• JNZ UP: Jumps back to the UP label if CX is not zero (loop continues).
• MOV Z, BX; MOV O, DX: Stores the final counts in Z and O.
• INT 3: Halts program execution for debugging.

High-Level Overview

1. Initialization: Input number is loaded; counters are initialized; loop counter is set.
2. Bit Iteration: The program iterates through each bit of the input number using ROL.
3. Counting 0s and 1s: Based on the carry flag after ROL, the appropriate counter is incremented.
4. Result Storage: The counts of 0s and 1s are stored in Z and O.
5. Program Termination: Execution halts.

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Output

 C:\TASM>masm czo.asm
Microsoft (R) Macro Assembler Version 5.00
Copyright (C) Microsoft Corp 1981-1985, 1987.  All rights reserved.

Object filename [czo.OBJ]:
Source listing  [NUL.LST]:
Cross-reference [NUL.CRF]:

  50422 + 450234 Bytes symbol space free

      0 Warning Errors
      0 Severe  Errors

C:\TASM>link czo.obj

Microsoft (R) Overlay Linker  Version 3.60
Copyright (C) Microsoft Corp 1983-1987.  All rights reserved.

Run File [CZO.EXE]:
List File [NUL.MAP]:
Libraries [.LIB]:
LINK : warning L4021: no stack segment

C:\TASM>debug CZO.EXE
-g

AX=5648  BX=000A  CX=0000  DX=0006  SP=0000  BP=0000  SI=0000  DI=0000
DS=0B97  ES=0B87  SS=0B97  CS=0B98  IP=0025   NV UP EI PL ZR NA PE NC
0B98:0025 CC            INT     3
-d 0b97:0000
0B97:0000  48 56 0A 00 06 00 00 00-00 00 00 00 00 00 00 00   HV..............
0B97:0010  B8 97 0B 8E D8 A1 00 00-BB 00 00 B9 10 00 BA 00   ................
0B97:0020  00 D1 C0 72 04 43 EB 02-90 42 49 75 F4 89 1E 02   ...r.C...BIu....
0B97:0030  00 89 16 04 00 CC E8 77-63 83 C4 06 FF 36 24 21   .......wc....6$!
0B97:0040  B8 0A 00 50 E8 47 5E 83-C4 04 5E 8B E5 5D C3 90   ...P.G^...^..]..
0B97:0050  55 8B EC 81 EC 84 00 C4-5E 04 26 80 7F 0A 00 74   U.......^.&....t
0B97:0060  3E 8B 46 08 8B 56 0A 89-46 FC 89 56 FE C4 5E FC   >.F..V..F..V..^.
0B97:0070  26 8A 47 0C 2A E4 40 50-8B C3 05 0C 00 52 50 E8   &.G.*[email protected].
-q

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Understanding the Memory Dump

This is the memory dump starting from address 0B97:0000, showing the contents of memory. Here is the breakdown:

0B97:0000 48 56 0A 00 06 00 00 00-00 00 00 00 00 00 00 00 HV…………..

In this memory dump:

• The first few bytes (48 56 0A 00 06 00 00 00) are the data segment where the program stores the values.
  • 48 56 0A 00: These represent the value 5648h (the number NO in the program).
  • 06 00: These represent the variable O (ones count), which is 6 (because there are 6 ones in 5648h).
  • 00 00: These represent the variable Z (zeroes count), which is 10 (because there are 10 zeroes in 5648h).
• The rest of the dump represents the program’s machine code and data.