I am trying to write a program which can determine if a number is prime or not. It accepts inputs from 2-9 though the result is NOT PRIME in case of 2 and 3, for all other number it works. Where is my mistake? I am using GUI Turbo Assembler for assembly.
.MODEL SMALL
.stack 100H
.DATA ; the data section contains anything that we want to be automatically
; initialized by the system before it calls the entry point of the program.
VAL1 DB ? ; Define Byte (the size is 1 byte), VAL1 is a variable, "?" means that the values are not initialized.
NL1 DB 0AH,0DH, 'ENTER NO: ','$' ; prints the text, $ means the current address according to the assembler.
NL2 DB 0AH,0DH, 'IT IS NOT PRIME','$' ; prints the text, $ means the current address according to the assembler.
NL3 DB 0AH,0DH, 'IT IS PRIME','$' ; prints the text, $ means the current address according to the assembler.
.CODE ; start coding.
MAIN:
MOV AX,@DATA ; DATA is the name of the Data Segment, we declare our variables in a segment, move DATA to AX.
MOV DS,AX ; move AX to DS, AX is called Initialization of Data Segment and it makes all the variables in Data Segment accessible.
LEA DX,NL1 ; move NL1 label's data into DX.
MOV AH,09H ; move the string into AH.
INT 21H ; standard input and output interrupts are found in the DOS Interrupt.
MOV AH,01H ; read a character and move into AH.
INT 21H ; standard input and output interrupts are found in the DOS Interrupt.
SUB AL,30H ; subtract 30H from the input value.
MOV VAL1,AL ; save the result into VAL1 label.
MOV AH,00 ; clear AH.
MOV CL,2 ; CL usually used to for loop counters.
DIV CL ; divides the value by CL and keeps the result's quotient in CL and remainder in CH.
MOV CL,AL ; move AL to CL.
LBL1:
MOV AH,00 ; clear AH.
MOV AL,VAL1 ; move the value in VAL1 to AL.
DIV CL ; divides the loop.
CMP AH,00 ; check if AH is 0.
JZ LBL2 ; (Jump if Zero) if CMP returns false, jump to LBL2, else continue.
DEC CL ; decrea8se the value by 1.
CMP CL,1 ; compare if the CL value is 1.
JNE LBL1 ; Jump if Not Equal to 0.
JMP LBL3 ; jump to LBL3.
LBL2:
MOV AH,09H ; move the string to AH.
LEA DX,NL2 ; equals to - MOV DX, offset NL2.
INT 21H ; standard input and output interrupts are found in the DOS Interrupt.
JMP EXIT ; jump to EXIT label.
LBL3:
MOV AH,09H ; move the string to AH.
LEA DX,NL3 ; equals to - MOV DX, offset NL3.
INT 21H ; standard input and output interrupts are found in the DOS Interrupt.
EXIT:
MOV AH,4CH ; exit from the program.
INT 21H ; standard input and output interrupts are found in the DOS Interrupt.
END MAIN ; the ending point of the code written in code segment.
According to your your algorithm: 3 DIV 2 = 1 and 3 DIV 1 = 3 Remainder 0, so it is treated as not-prime.
If you move the CL check to the beginning of the loop block you can intercept these cases:
...
MOV CL,2 ; CL usually used to for loop counters.
DIV CL ; divides the value by CL and keeps the result's quotient in CL and remainder in CH.
MOV CL,AL ; move AL to CL.
LBL1:
CMP CL,1 ; compare if the CL value is 1.
JBE LBL3 ; Jump if CL=1 or CL=0.
MOV AH,00 ; clear AH.
MOV AL,VAL1 ; move the value in VAL1 to AL.
DIV CL ; divides the loop.
CMP AH,00 ; check if AH is 0.
JZ LBL2 ; (Jump if Zero) if CMP returns false, jump to LBL2, else continue.
DEC CL ; decrea8se the value by 1.
JMP LBL1 ; jump to LBL1.
...