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I was trying to find the nth Fibonacci number e.x n=3, output = 1
so my logic was this
a = 0
b = 1
input n
si 0
n>2
loop
temp = b
b = a+b
a = b
loop if si/=cx
print b
This is my pseudo code logic. When I tried to implement this I am stuck in an infinite loop
.MODEL SMALL
.STACK 100h
.DATA
STRING0 DB 'Enter INDEX $'
.CODE
MAIN PROC
MOV AX,#DATA
MOV DS,AX
LEA DX, STRING0
MOV AH,9
INT 21H
MOV AH, 2
MOV DL,0AH ;NEW LINE
INT 21H
MOV DL,0DH
INT 21H
MOV AH,1
INT 21H
SUB CX,CX
MOV CL,AL
MOV SI,0
MOV AX,0
MOV BX,1
LOOP1:
PUSH BX
ADD BX,AX
POP AX
INC SI
LOOP LOOP1
MOV DX,BX
MOV AH,9
INT 21H
MAIN ENDP
END MAIN
I use EMU 4.08. The code us stuck at an infinite loop. I have no idea why
I did SUB cx,cx to move the AL value to CL and use CL as counter otherwise it gives me error that the code failed to send 8bit data to 16bit
I was trying to find the nth Fibonacci number e.x n=3, output = 1
From your example I understand that you consider the Fibonacci sequence to begin with 0, 1, 1, 2, 3, 5, 8, 13, 21, ...
Fibonacci himself started his sequence from 1, 2, 3, 5, 8, ...
See the Wikipedia article https://en.wikipedia.org/wiki/Fibonacci_number
I didn't follow your pseudo code too much as it has flaws of its own!
Why your assembly program fails
You say your "code is stuck at an infinite loop", but that's not really the case. It is just that your loop executes an extra 50 iterations. The reason is that the DOS.GetCharacter function 01h gives you an ASCII code, that you have to convert into the digit that the pressed key represents. eg. If you press 3, DOS gives you AL=51, and you need to subtract 48 to obtain the inputted digit which is 3.
But wait, don't use this number 3 as your loop counter already! Since the 1st and 2nd Fibonacci numbers are known from the start, calculating the 3rd Fibonacci number requires just 1 iteration of the loop. Account for this and subtract 2 beforehand.
Once your program has found the answer you simply move the result from BX to DX, and expect the DOS.PrintString function 09h to output the number. It can't do that. It's a function that outputs a series of characters (so a string beginning at the address in DX), however your result is still a number in a register. You have to convert it into its textual representation. Displaying numbers with DOS has all the fine details about this conversion!
Next code allows the user to input a single-digit from 1 to 9
...
mov ah, 01h ; DOS.GetCharacter
int 21h ; -> AL = ["1","9"]
sub al, 48 ; -> AL = [1,9]
cbw ; -> AH = 0
mov cx, ax ; -> CX = [1,9]
xor ax, ax ; -> AX = 0
dec cx
jz PrintIt ; 1st Fib is 0
inc ax ; -> AX = 1
dec cx
jz PrintIt ; 2nd Fib is 1
cwd ; -> DX = 0
CalcIt: ; 3rd Fib and others
xchg ax, dx
add ax, dx
loop CalcIt
PrintIt: ; AX is at most 21 (because of the limited input)
aam
add ax, 3030h ; Conversion into text
xchg al, ah
cmp al, '0'
mov dh, 02h ; DOS.PrintCharacter
xchg ax, dx
je Ones
int 21h
Ones:
mov dl, dh
int 21h
Because in your program the output is very limited, I used a special code to display at most 2 digits. For the general case of outputting numbers see this Q/A.
I think this should be good:
.MODEL SMALL
.STACK 100h
.DATA
STRING0 DB 'Enter INDEX $'
STRING1 DB 'OUTPUT: $'
.CODE
MAIN PROC
MOV AX,#DATA
MOV DS,AX
LEA DX, STRING0
MOV AH,9
INT 21H
MOV AH, 2
MOV DL,0AH ;NEW LINE
INT 21H
MOV DL,0DH
INT 21H
MOV AH,1
INT 21H
SUB CX,CX
SUB AL,30H ;To convert char into digit value
MOV CL,AL
MOV SI,0
MOV AX,0
MOV BX,1
LOOP1:
PUSH BX
ADD BX,AX
POP AX
INC SI
LOOP LOOP1
MOV AH, 2
MOV DL,0AH ;NEW LINE
INT 21H
MOV DL,0DH
INT 21H
LEA DX, STRING1
MOV AH,9
INT 21H
;Print the result
MOV AX,BX
MOV SI,0
;Push digits from right to left into stack
LOOP2:
MOV DL,10
DIV DL
PUSH AX
MOV AH,0
INC SI
CMP AL,0
JNE LOOP2
;Pop digits from stack and print them
LOOP3:
POP DX
MOV DL,DH
MOV DH,0
ADD DL,30H ;To convert digit to char
MOV AH,2
INT 21H
DEC SI
CMP SI,0
JNE LOOP3
HLT
MAIN ENDP
END MAIN
I write the following code to read some numbers ranging from -15 to 15 from the user and the user may define how many numbers to enter. Then I bubble sort the array to get the smallest number. (Bubble sort because I will need to print other information) However, the code is not working. Here is my code.
// oops.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
int _tmain(int argc, _TCHAR* argv[])
{
char message0[] = "How many numbers do you want to enter? \n";
char message1[] = "Enter the current reading: \n";
char message2[] = "Error!\n";
char message3[] = "The smallest number is: \n";
char format1[] = "%d";
char format2[] = "%s";;
int myarray[10000];
int No;
int counter;
int *p;
p = myarray - 1;
_asm{
lea eax, message0
push eax
call printf
add esp, 4
//read how many numbers the user would like to input
lea eax,counter
push eax
lea eax, format1
push eax
call scanf_s
add esp,8
mov No, 1
mov ecx, counter
mov ebx, 0
//read user's input
Input: push ecx
push No
lea eax, message1
push eax
call printf
add esp, 8
lea eax, myarray[ebx]
push eax
lea eax, format1
push eax
call scanf_s
add esp,8
//judge if the number is in the range of -15 to 15
JudgeInput: mov eax, myarray[ebx]
cmp eax,-15
jl Illegal
cmp eax,15
jle Legal
Illegal: lea eax,message2
push eax
call printf
add esp,4
pop ecx
jmp Input
Legal: add ebx,4
inc No
pop ecx
loop Input
//bubble sort
mov esi, p
mov ecx, counter
outer : mov edx, ecx
inner : cmp edx, ecx
jz exchangeNo
mov eax, [esi + ecx * 4]
mov ebx, [esi + edx * 4]
cmp eax, ebx
jnb exchangeNo
mov[esi + ecx * 4], ebx
mov[esi + edx * 4], eax
exchangeNo :
dec edx
jnz inner
loop outer
finish:
smallest: //print the smallest number
mov ebx,0
lea eax,message3
push eax
lea eax, format2
push eax
call printf
mov eax,0
lea ebx,myarray
sub ebx,4
add ebx,No
lea eax, [ebx]
push eax
lea eax,format1
call printf
add esp,16
}
return 0;
}
It would not return the smallest number. Sometimes it returns strange characters. I get really confusing. Additionally, when I enter negative numbers, the bubble sort seems not working well.
I have solved the problem. Here is my updated code:
int _tmain(int argc, _TCHAR* argv[])
{
char message0[] = "How many numbers do you want to enter? \n";
char message1[] = "Enter the current reading: \n";
char message2[] = "Error!\n";
char message3[] = "\nThe smallest number is: ";
char format1[] = "%d";
char format2[] = "%s";;
int myarray[10000];
int No;
int counter;
int *p;
p = myarray - 1;
_asm{
lea eax, message0
push eax
call printf
add esp, 4
lea eax,counter
push eax
lea eax, format1
push eax
call scanf_s
add esp,8
//get the user's input into the array
mov No, 1
mov ecx, counter
mov ebx, 0
Input:
push ecx
push No
lea eax, message1
push eax
call printf
add esp, 8
lea eax, myarray[ebx]
push eax
lea eax, format1
push eax
call scanf_s
add esp,8
//judge if the input is between -15 and 15
JudgeInput:
mov eax, myarray[ebx]
cmp eax,-15
jl Illegal
cmp eax,15
jle Legal
//if not, print out error message
Illegal:
lea eax,message2
push eax
call printf
add esp,4
pop ecx
jmp Input
//if yes, loop again
Legal:
add ebx,4
inc No
pop ecx
loop Input
//bubble sort
mov esi, p
mov ecx, counter
//the outer loop
outer : mov edx, ecx
//the inner loop
inner : cmp edx, ecx
je exchangeNo
mov eax, [esi + ecx * 4]
mov ebx, [esi + edx * 4]
cmp eax, ebx
jge exchangeNo
mov[esi + ecx * 4], ebx
mov[esi + edx * 4], eax
exchangeNo :
dec edx
jge inner
loop outer
finish:
//find out the smallest number
smallest :
lea eax, message3
push eax
lea eax, format2
push eax
call printf
lea ebx, myarray
mov eax, [ebx]
push eax
lea eax, format1
push eax
call printf
add esp, 16
}
}
I have a program in assembly for the Linux terminal that's supposed to work through a series of mathematical manipulations, compare the final value to 20, and then using if logic, report <, > or = relationship. Code is:
segment .data
out_less db "Z is less than 20.", 10, 0
out_greater db "Z is greater than 20.", 10, 0
out_equal db "Z is equal to 20.", 10, 0
segment .bss
segment .text
global main
extern printf
main:
mov eax, 10
mov ebx, 12
mov ecx, eax
add ecx, ebx ;set c (ecx reserved)
mov eax, 3
mov ebx, ecx
sub ebx, eax ;set f (ebx reserved)
mov eax, 12
mul ecx
add ecx, 10 ;(a+b*c) (ecx reserved)
mov eax, 6
mul ebx
mov eax, 3
sub eax, ebx
mov ebx, eax ;(d-e*f) (ebx reserved) reassign to ebx to free eax
mov eax, ecx
div ebx
add ecx, 1 ;(a+b*c)/(d-e*f) + 1
cmp ecx, 20
jl less
jg greater
je equal
mov eax, 0
ret
less:
push out_less
call printf
jmp end
greater:
push out_greater
call printf
jmp end
equal:
push out_equal
call printf
jmp end
end:
mov eax, 0
ret
Commands for compiling in terminal using nasm and gcc:
nasm -f elf iftest.asm
gcc -o iftest iftest.o
./iftest
Equivalent C code would be:
main() {
int a, b, c, d, e, f, z;
a = 10;
b = 12;
c = a + b;
d = 3;
e = 6;
f = c - d;
z = ((a + b*c) / (d - e*f)) + 1;
if (z < 20) {
printf("Z (%d) is less than 20.\n", z);
}
else if (z > 20) {
printf("Z is greater than 20.\n");
}
else {
printf("Z is equal to 20.\n");
}
}
The current output is incorrect. The C code will report that z = -1, and therefore less than 20, but the assembly code outputs Z is greater than 20. I've tried printing out the final value, but I run into this issue where printing the value somehow changes it. I've checked and rechecked the math logic and I can't see why it shouldn't give the correct value, unless I'm using the math operators incorrectly. Any and all help is appreciated.
I think the problem is here:
div ebx
add ecx, 1 ;(a+b*c)/(d-e*f) + 1
The result of the div instruction is not in ecx.
I'm new at programming assembly. Now I'm trying to write a program that converts number from decimal to binary. But I got stuck with one program while trying to input. After i output msg2 and get into loop, program doesn't turn off. I can input a lot of numbers and program doesn't turn off. I guess problem is in convertnumber: cmp si,cx (si is how many numbers I have to input, cx- how many numbers I have already written), but I am not sure about that. Where have I made a mistake and how could I correct it?
.MODEL small
.Stack 100h
.DATA
msg0 db 'how many numbers will include your input number(example. 123 is 3 numbers)? $'
msg1 db 'Now input number from 0 to 65535: $'
number db 255, ?, 256 dup ('$')
numberinAscii db 255, ?, 256 dup ('$')
enterbutton db 13,10,'$'
.CODE
start:
mov ax, #data
mov ds,ax
mov ah,09h
mov dx, offset msg0 ; first message output
int 21h
xor ah,ah ; function 00h of
int 16h ; int 16h gets a character (ASCII translation in AL)
int 3
mov bl,al
mov dl,al
mov ah,02h ; function 02h - display character
int 21h ; call DOS service
mov ah,09h
mov dx, offset enterbutton
int 21h
mov ah, 09h
mov dx, offset msg1 ; output second message
int 21h
jmp covertHowMany ; converting number that we entered
next:
xor si,si
mov si, ax ; number that we entered now is in si
xor cx,cx
mov cx,0 ;cx=0
enterfirstnumber: ;entering first number (example 123, first number is 1)
xor ah,ah
int 16h ; int 16h gets a one character
int 3
mov bl,al
mov dl,al
mov ah,02h ; function 02h - display character
int 21h ;
jmp convertnumber ; converting this number
input: ;converting number from ascii char to ascii integer
mov ax,bx
mov dx,10
mul dx ; ax:=ax*10
mov bx,ax ; number that I try to convert is in bx now
xor ah,ah
int 16h ; int 16h gets a character (ASCII translation in AL)
int 3
mov bl,al
mov dl,al
mov ah,02h ; function 02h - display character
int 21h
jmp convertnumber
convertHowMany:
sub al,30h ; convert from ascii character to ascii number
jmp next
convertnumber:
sub al,30h
add bx,ax
inc cx
cmp cx, si
jne input
jmp ending
ending:
mov ax,04C00h
int 21h
end start
I see at least two problems with your code:
The first is that when you reach convertHowMany you assume that AL still contains the character that the user typed in. That will not be the case, since both INT 21h/AH=02h and INT 21h/AH=09h modify AL. You'll have to save and restore the value of AL somehow (e.g. by pushing and popping AX).
The second problem is how you initialize SI before the loop. You're moving the value of AX into SI, which means both AL and AH. AH is not zero at that point, because you've just used INT 21h/AH=09h.
You could change the sequence xor si,si / mov si,ax into something like mov si,ax / and si,0FFh.
I am finishing up an assembly program that replaces characters in a string with a given replacement character. The assembly code calls C functions and the assembly program itself is called from main in my .c file. However, when trying to finish and return a final int value FROM the assembly program TO C, I get segfaults. My .asm file is as follows:
; File: strrepl.asm
; Implements a C function with the prototype:
;
; int strrepl(char *str, int c, int (* isinsubset) (int c) ) ;
;
;
; Result: chars in string are replaced with the replacement character and string is returned.
SECTION .text
global strrepl
_strrepl: nop
strrepl:
push ebp ; set up stack frame
mov ebp, esp
push esi ; save registers
push ebx
xor eax, eax
mov ecx, [ebp + 8] ;load string (char array) into ecx
jecxz end ;jump if [ecx] is zero
mov al, [ebp + 12] ;move the replacement character into esi
mov edx, [ebp + 16] ;move function pointer into edx
firstLoop:
xor eax, eax
mov edi, [ecx]
cmp edi, 0
jz end
mov edi, ecx ; save array
movzx eax, byte [ecx] ;load single byte into eax
push eax ; parameter for (*isinsubset)
mov edx, [ebp + 16]
call edx ; execute (*isinsubset)
mov ecx, edi ; restore array
cmp eax, 0
jne secondLoop
add esp, 4 ; "pop off" the parameter
mov ebx, eax ; store return value
add ecx, 1
jmp firstLoop
secondLoop:
mov eax, [ebp+12]
mov [ecx], al
mov edx, [ebp+16]
add esp, 4
mov ebx, eax
add ecx, 1
jmp firstLoop
end:
pop ebx ; restore registers
pop esi
mov esp, ebp ; take down stack frame
pop ebp
mov eax, 9
push eax ;test
ret
and my c file is:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
//display *((char *) $edi)
// These functions will be implemented in assembly:
//
int strrepl(char *str, int c, int (* isinsubset) (int c) ) ;
int isvowel (int c) {
if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
return 1 ;
if (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U')
return 1 ;
return 0 ;
}
int main(){
char *str1;
int r;
str1 = strdup("ABC 123 779 Hello World") ;
r = strrepl(str1, '#', &isdigit) ;
printf("str1 = \"%s\"\n", str1) ;
printf("%d chararcters were replaced\n", r) ;
free(str1) ;
return 0;
}
In my assembly code, you can see in end
mov eax, 9
push eax
I am simply trying to return the value 9 to the value "r" which is an int in the C file. This is just a test to see if I can return an int back to r in the c file. Eventually I will be returning the number of characters that were replaced back to r. However, I need to figure out why the following code above is segfaulting. Any ideas?
mov eax, 9
push eax ; NOT a good idea
ret
That is a big mistake. It's going to return based on the lowest thing on the stack and you've just pushed something on to the stack that's almost certainly not a valid return address.
Most functions return a code by simply placing it into eax (this depends on calling convention of course but that's a pretty common one), there's generally no need to push it on to the stack, and certainly plenty of downside to doing so.
Return values are normally stored in EAX on X86 32 bit machines. So your pushing it on the stack after storing it in EAX is wrong, because the function it is returning to will try to use what is in EAX as a value for IP (instruction pointer)
Ret with no argument pops the return address off of the stack and jumps to it.
source