I know this is outside the scope of the community but I had nowhere else to turn to. I need screenshots of Visual Studio after running the following programs for an assignment. My PC died and it is due in a few hours. There are two MASM programs and they are both interactive, any logical variables will do. You can send the link of the images to:mihijek116#vapaka.com (attachments will not work) if the question is closed. Thanks in advance
INCLUDE Irvine32.inc
.data
fname BYTE 16 Dup(?)
age dword ?
np BYTE 'Enter Name:',0dh, 0ah,0
ap BYTE 'Enter Age:',0dh, 0ah,0
yn BYTE 'Your name is ',0
yy BYTE ' born in ',0
cy dword 2022
dob dword ?
.code
main PROC
mov edx, offset np
call WriteString
mov edx, offset fn
mov ecx, Sizeof fn
call readstring
call Crlf
mov edx, offset ap
call WriteString
call readInt
mov age, eax
mov eax, cy
sub eax, age
mov yob, eax
mov edx, offset yn
call WriteString
mov edx, offset fn
call WriteString
mov edx, offset yy
call WriteString
mov eax, dob
call WriteDec
call Crlf
call WaitMsg
exit
main ENDP
END main
INCLUDE Irvine32.inc
.data
two dword 2 ; store two
three dword 3 ; store three
c dword 1 ; store 1
n dword ? ; n variable
answer dword ? ; answer variable
a1 dword ? ; store 2n^2
a2 dword ? ; store 3n
pr BYTE 'Enter n:',0dh, 0ah,0 ; n prompt
eq BYTE 'ans= ',0 ; ans
.code
n2n PROC ; 2*n*n
mov eax, two
mov ecx, n
mul ecx
mov ecx,n
mul ecx
mov a1,eax
ret
n2n ENDP
n3 PROC ; 3n
mov eax, three
mov ecx, n
mul ecx
mov a2,eax
ret
3n ENDP
main PROC
mov edx, offset pr
call WriteString ; Ask user for n
call readInt
mov n, eax ; Store user input in n
call n2n ; call procedure n2n
call 3n ; call procedure n3
mov ebx, a1
add ebx, a2
add ebx, c
mov answer, ebx ; store result in answer
mov edx, offset eq
call WriteString
mov eax, answer
call WriteDec
call WaitMsg
exit
main ENDP
END main
Related
; Input x and y, output min of the two numbers
.586
.MODEL FLAT
INCLUDE io.h
.STACK 4096
.DATA
number DWORD ?
array DWORD 20, 15, 62, 40, 18
nbrElts DWORD 5
prompt BYTE "Enter value:", 0
string BYTE 80 DUP (?)
resultLbl BYTE "Position", 0
result BYTE 11 DUP (?), 0
.CODE
_MainProc PROC
input prompt, string, 20 ; read ASCII characters
atod string ; convert to integer
mov number, eax ; store in memory
push nbrElts ; 3rd parameter (# of elements in array)
lea eax, array ; 2nd parameter (address of array)
push eax
push number ; 1st parameter (value from user)
call searchArray ; searchArray(number, array, 5)
add esp, 12
dtoa result, eax ; convert to ASCII characters
output resultLbl, result ; output label and result
mov eax, 0 ; exit with return code 0
ret
_MainProc ENDP
; searchArray(int x, array, int y)
;
searchArray PROC
push ebp ; save base pointer
mov ebp, esp ; establish stack frame
push eax ; save registers
push ebx
push esi
push ecx
push edx
mov ebx, [ebp+8] ; x, value from user
mov esi, [ebp+12] ; address of array
mov ecx, [ebp+16] ; y, number of elements
mov edx, 1
mov ecx, 5
forLoop:
mov eax, [esi] ; a[i]
cmp eax, ebx ; eax = ebx ?
je isEqual
;cmp eax, ebx
add esi, 4
inc edx
loop forLoop
;mov eax, 0
cmp edx, 6
je notEqual
isEqual:
mov eax, edx
jmp exitCode
notEqual:
mov eax, 0
jmp exitCode
exitCode:
mov eax, edx
pop edx ; restore EBP
pop ecx ; restore EAX
pop esi
pop ebx
pop ebp
ret ; return
searchArray ENDP
END ; end of source code
The pops at the end of the function need to match the pushes at the beginning of the function. If they don't match, the stack pointer ends up in the wrong place and the ret returns to the wrong place.
In your case, you have an extra push without a corresponding pop.
The reason to push registers at the beginning and pop them at the end is to preserve their values. But you don't want to preserve the value of eax. You want to return a different value, the result of the function. So there is absolutely no reason to push eax.
I want to call a function that will perform upper to lower case conversion to a user typed string, preserving the especial characters. This part works, but only for the first 4 characters, everything after that just gets truncated. I believe it is because I have defined the parameters as DWORD:
I have tried using PAGE, PARA and BYTE. The first two don't work and with byte says type missmatch.
upperToLower proc, source:dword, auxtarget:dword
mov eax, source ;Point to string
mov ebx, auxtarget ; point to destination
L1:
mov dl, [eax] ; Get a character from buffer
cmp byte ptr [eax], 0 ; End of string? (not counters)
je printString ; if true, jump to printString
cmp dl, 65 ; 65 == 'A'
jl notUpper ; if less, it's not uppercase
cmp dl, 90 ; 90 == 'Z'
jg notUpper ; if greater, it's not uppercase
xor dl, 100000b ; XOR to change upper to lower
mov [ebx], dl ; add char to target
inc eax ; Move counter up
inc ebx ; move counter up
jmp L1 ; loop
notUpper: ; not uppercase
mov [ebx], dl ; copy the letter
inc eax ;next letter
inc ebx
jmp L1
printString:
invoke WriteConsoleA, consoleOutHandle, auxtarget, sizeof auxtarget, bytesWritten, 0
ret
upperToLower endp
The PROTO:
upperToLower PROTO,
source: dword,
auxtarget: dword
Invoke:
invoke upperToLower, offset buffer, offset target
The buffer parameter is: buffer db 128 DUP(?)
How can I get printed the whole string, and not just the first 4 characters?
Why are only 4 characters being printed? You write the string to the console with:
invoke WriteConsoleA, consoleOutHandle, auxtarget, sizeof auxtarget, bytesWritten, 0
The sizeof auxtarget parameter is the size of auxtarget which is a DWORD (4 bytes) thus you are asking to only print 4 bytes. You need to pass the length of the string. You can easily do so by taking the ending address in EAX and subtracting the source pointer from it. The result would be the length of the string you traversed.
Modify the code to be:
printString:
sub eax, source
invoke WriteConsoleA, consoleOutHandle, auxtarget, eax, bytesWritten, 0
A version of your code that follows the C call convention, uses both a source and destination buffer, tests for the pointers to make sure they aren't NULL, does the conversion using a similar method described by Peter Cordes is as follows:
upperToLower proc uses edi esi, source:dword, dest:dword
; uses ESI EDI is used to tell assembler we are clobbering two of
; the cdecl calling convetions non-volatile registers. See:
; https://en.wikipedia.org/wiki/X86_calling_conventions#cdecl
mov esi, source ; ESI = Pointer to string
test esi, esi ; Is source a NULL pointer?
jz done ; If it is then we are done
mov edi, dest ; EDI = Pointer to string
test edi, edi ; Is dest a NULL pointer?
jz done ; If it is then we are done
xor edx, edx ; EDX = 0 = current character index into the strings
jmp getnextchar ; Jump into loop at point of getting next character
charloop:
lea ecx, [eax - 'A'] ; cl = al-'A', and we do not care about the rest
; of the register
cmp cl, 25 ; if(c >= 'A' && c <= 'Z') c += 0x20;
lea ecx, [eax + 20h] ; without affecting flags
cmovna eax, ecx ; take the +0x20 version if it was in the
; uppercase range to start with
mov [edi + edx], al ; Update character in destination string
inc edx ; Go to next character
getnextchar:
movzx eax, byte ptr [esi + edx]
; mov al, [esi + edx] leaving high garbage in EAX is ok
; too, but this avoids a partial-register stall
; when doing the mov+sub
; in one instruction with LEA
test eax, eax ; Is the character NUL(0) terminator?
jnz charloop ; If not go back and process character
printString:
; EDI = source, EDX = length of string
invoke WriteConsoleA, consoleOutHandle, edi, edx, bytesWritten, 0
mov edx, sizeof buffer
done:
ret
upperToLower endp
A version that takes one parameter and changes the source string to upper case could be done this way:
upperToLower proc, source:dword
mov edx, source ; EDX = Pointer to string
test edx, edx ; Is it a NULL pointer?
jz done ; If it is then we are done
jmp getnextchar ; Jump into loop at point of getting next character
charloop:
lea ecx, [eax - 'A'] ; cl = al-'A', and we do not care about the rest
; of the register
cmp cl, 25 ; if(c >= 'A' && c <= 'Z') c += 0x20;
lea ecx, [eax + 20h] ; without affecting flags
cmovna eax, ecx ; take the +0x20 version if it was in the
; uppercase range to start with
mov [edx], al ; Update character in string
inc edx ; Go to next character
getnextchar:
movzx eax, byte ptr [edx] ; mov al, [edx] leaving high garbage in EAX is ok, too,
; but this avoids a partial-register stall
; when doing the mov+sub in one instruction with LEA
test eax, eax ; Is the character NUL(0) terminator?
jnz charloop ; If not go back and process character
printString:
sub edx, source ; EDX-source=length
invoke WriteConsoleA, consoleOutHandle, source, edx, bytesWritten, 0
done:
ret
upperToLower endp
Observations
A generic upperToLower function that does the string conversion would normally not do the printing itself. You'd normally call upperToLower to do the conversion only, then you'd output the string to the display in a separate call.
My goal in this code is to find the smallest number in the list. I used bubble sort method in this case; unfortunately, the code is not giving me the smallest/minimum number. Please take a look, Thanks:
include irvine32.inc
.data
input byte 100 dup(0)
stringinput byte "Enter any string: ",0
totallength byte "The total length is: ",0
minimum byte "The minimum value is: ",0
.code
stringLength proc
push ebp
mov ebp, esp
push ebx
push ecx
mov eax, 0
mov ebx, [ebp+8]
L1:
mov ecx, [ebx] ;you can use ecx, cx, ch, cl
cmp ecx, 0 ;you can use ecx, cx, ch, cl
JE L2
add ebx, 1
add eax, 1
jmp L1
L2:
pop ecx
pop ebx
mov ebp, esp
pop ebp
ret 4
stringLength endp
BubbleSort PROC uses ECX
push edx
xor ecx,ecx
mov ecx, 50
OUTER_LOOP:
push ecx
xor ecx,ecx
mov ecx,14
mov esi, OFFSET input
COMPARE:
xor ebx,ebx
xor edx,edx
mov bl, byte ptr ds:[esi]
mov dl, byte ptr ds:[esi+1]
cmp bl,dl
jg SWAP1
CONTINUE:
add esi,2
loop COMPARE
mov esi, OFFSET input
pop ecx
loop OUTER_LOOP
jmp FINISHED
SWAP1:
xchg bl,dl
mov byte ptr ds:[esi+1],dl
mov byte ptr ds:[esi],bl
jmp CONTINUE
FINISHED:
pop edx
ret 4
BubbleSort ENDP
main proc
call clrscr
mov edx, offset stringinput
call writeString
mov edx, offset input
call writeString
call stringLength
mov edx, offset input
mov ecx, sizeof input
call readstring
call crlf
mov edx,offset totallength
call writestring
call writedec
call crlf
mov edx, offset minimum
call crlf
call writeString
push offset input
call BubbleSort
mov edx, offset input
call writeString
call crlf
exit
main endp
end main
I haven't looked over your code, because sorting is an over complicated method for what you want to do. Not only that, but most of us don't pay too much attention to uncommented code. Just takes to long to figure out what you're trying to do.
Simply iterate through the entire list and start with 255 (FFH) in AL let's say. Each time you come across a number that is smaller than the one in AL, then replace it with that value and then when loop is finished, AL will have the lowest value.
If you need to know where it is in the list, you could maybe use AH which would be the difference between start address and current address. Knowledge of the instruction set is essential as finding the length of the string can be simplified by;
mov di, input ; Point to beginning of buffer
mov cx, -1 ; for a maximum of 65535 characters
xor al, al ; Looking for NULL
rep scasb
neg cx
dec cx ; CX = length of string.
Remember, ES needs to point to #DATA
I am looking for a way to print an integer in assembler (the compiler I am using is NASM on Linux), however, after doing some research, I have not been able to find a truly viable solution. I was able to find a description for a basic algorithm to serve this purpose, and based on that I developed this code:
global _start
section .bss
digit: resb 16
count: resb 16
i: resb 16
section .data
section .text
_start:
mov dword[i], 108eh ; i = 4238
mov dword[count], 1
L01:
mov eax, dword[i]
cdq
mov ecx, 0Ah
div ecx
mov dword[digit], edx
add dword[digit], 30h ; add 48 to digit to make it an ASCII char
call write_digit
inc dword[count]
mov eax, dword[i]
cdq
mov ecx, 0Ah
div ecx
mov dword[i], eax
cmp dword[i], 0Ah
jg L01
add dword[i], 48 ; add 48 to i to make it an ASCII char
mov eax, 4 ; system call #4 = sys_write
mov ebx, 1 ; file descriptor 1 = stdout
mov ecx, i ; store *address* of i into ecx
mov edx, 16 ; byte size of 16
int 80h
jmp exit
exit:
mov eax, 01h ; exit()
xor ebx, ebx ; errno
int 80h
write_digit:
mov eax, 4 ; system call #4 = sys_write
mov ebx, 1 ; file descriptor 1 = stdout
mov ecx, digit ; store *address* of digit into ecx
mov edx, 16 ; byte size of 16
int 80h
ret
C# version of what I want to achieve (for clarity):
static string int2string(int i)
{
Stack<char> stack = new Stack<char>();
string s = "";
do
{
stack.Push((char)((i % 10) + 48));
i = i / 10;
} while (i > 10);
stack.Push((char)(i + 48));
foreach (char c in stack)
{
s += c;
}
return s;
}
The issue is that it outputs the characters in reverse, so for 4238, the output is 8324. At first, I thought that I could use the x86 stack to solve this problem, push the digits in, and pop them out and print them at the end, however when I tried implementing that feature, it flopped and I could no longer get an output.
As a result, I am a little bit perplexed about how I can implement a stack in to this algorithm in order to accomplish my goal, aka printing an integer. I would also be interested in a simpler/better solution if one is available (as it's one of my first assembler programs).
One approach is to use recursion. In this case you divide the number by 10 (getting a quotient and a remainder) and then call yourself with the quotient as the number to display; and then display the digit corresponding to the remainder.
An example of this would be:
;Input
; eax = number to display
section .data
const10: dd 10
section .text
printNumber:
push eax
push edx
xor edx,edx ;edx:eax = number
div dword [const10] ;eax = quotient, edx = remainder
test eax,eax ;Is quotient zero?
je .l1 ; yes, don't display it
call printNumber ;Display the quotient
.l1:
lea eax,[edx+'0']
call printCharacter ;Display the remainder
pop edx
pop eax
ret
Another approach is to avoid recursion by changing the divisor. An example of this would be:
;Input
; eax = number to display
section .data
divisorTable:
dd 1000000000
dd 100000000
dd 10000000
dd 1000000
dd 100000
dd 10000
dd 1000
dd 100
dd 10
dd 1
dd 0
section .text
printNumber:
push eax
push ebx
push edx
mov ebx,divisorTable
.nextDigit:
xor edx,edx ;edx:eax = number
div dword [ebx] ;eax = quotient, edx = remainder
add eax,'0'
call printCharacter ;Display the quotient
mov eax,edx ;eax = remainder
add ebx,4 ;ebx = address of next divisor
cmp dword [ebx],0 ;Have all divisors been done?
jne .nextDigit
pop edx
pop ebx
pop eax
ret
This example doesn't suppress leading zeros, but that would be easy to add.
I think that maybe implementing a stack is not the best way to do this (and I really think you could figure out how to do that, saying as how pop is just a mov and a decrement of sp, so you can really set up a stack anywhere you like by just allocating memory for it and setting one of your registers as your new 'stack pointer').
I think this code could be made clearer and more modular if you actually allocated memory for a c-style null delimited string, then create a function to convert the int to string, by the same algorithm you use, then pass the result to another function capable of printing those strings. It will avoid some of the spaghetti code syndrome you are suffering from, and fix your problem to boot. If you want me to demonstrate, just ask, but if you wrote the thing above, I think you can figure out how with the more split up process.
; Input
; EAX = pointer to the int to convert
; EDI = address of the result
; Output:
; None
int_to_string:
xor ebx, ebx ; clear the ebx, I will use as counter for stack pushes
.push_chars:
xor edx, edx ; clear edx
mov ecx, 10 ; ecx is divisor, devide by 10
div ecx ; devide edx by ecx, result in eax remainder in edx
add edx, 0x30 ; add 0x30 to edx convert int => ascii
push edx ; push result to stack
inc ebx ; increment my stack push counter
test eax, eax ; is eax 0?
jnz .push_chars ; if eax not 0 repeat
.pop_chars:
pop eax ; pop result from stack into eax
stosb ; store contents of eax in at the address of num which is in EDI
dec ebx ; decrement my stack push counter
cmp ebx, 0 ; check if stack push counter is 0
jg .pop_chars ; not 0 repeat
mov eax, 0x0a
stosb ; add line feed
ret ; return to main
; eax = number to stringify/output
; edi = location of buffer
intToString:
push edx
push ecx
push edi
push ebp
mov ebp, esp
mov ecx, 10
.pushDigits:
xor edx, edx ; zero-extend eax
div ecx ; divide by 10; now edx = next digit
add edx, 30h ; decimal value + 30h => ascii digit
push edx ; push the whole dword, cause that's how x86 rolls
test eax, eax ; leading zeros suck
jnz .pushDigits
.popDigits:
pop eax
stosb ; don't write the whole dword, just the low byte
cmp esp, ebp ; if esp==ebp, we've popped all the digits
jne .popDigits
xor eax, eax ; add trailing nul
stosb
mov eax, edi
pop ebp
pop edi
pop ecx
pop edx
sub eax, edi ; return number of bytes written
ret
I think I'm having trouble with my swap, and how I'm accessing the elements in my array. Right now, all of the code runs, but the list does not change after the sort. He's the high level sort I'm trying to implement
for(k=0; k<request-1; k++) {
i = k;
for(j=k+1; j<request; j++) {
if(array[j] > array[i])
i = j;
}
exchange(array[k], array[i]);
}
Here's the assembly code. Note: the assignment is about pushing and popping elements on/off the stack, so I can't change the parameters.
;this is a library with macros for things like printing numbers and strings
INCLUDE Irvine32.inc
MIN = 10 ;lower range limit
MAX = 200 ;upper range limit
LRANGE = 100
HRANGE = 999
.data
;title, intro, and prompts
title_1 BYTE "PROGRAMMING ASSIGNMENT 5: RANDOM GEN/SORT", 0dh, 0ah, 0
intro_1 BYTE "This program generates random numbers in the range (100 - 999),", 0dh, 0ah
BYTE "displays the original list, sorts the list, and calculates the median value.", 0dh, 0ah
BYTE "Finally, it displays the sorted list in descending order.", 0dh, 0ah, 0
prompt_1 BYTE "How many numbers should be generated? (10 - 200): ", 0
error_1 BYTE "Out of range.", 0dh, 0ah, 0
display_1 BYTE "List of random numbers: ", 0dh, 0ah, 0
display_2 BYTE "The median is: ", 0
display_3 BYTE "The sorted list: ", 0dh, 0ah, 0
;placeholders for user entries and calculated data
randArray DWORD MAX DUP(?)
userNum DWORD ? ;integer to be entered by user
;strings for posting results
goodBye_1 BYTE "Thank you for using the Gen/sort-ulator! Good-bye!", 0
.code
main PROC
call Randomize
;Title Screen
push OFFSET title_1
push OFFSET intro_1
call Intro
;Get and validate user numbers
push OFFSET error_1
push OFFSET prompt_1
push OFFSET userNum
call GetData
;Fill Array with random numbers
push OFFSET randArray
push userNum
call FillArray
;display unsorted results
push OFFSET randArray
push userNum
push OFFSET display_1
call DisplayList
;sort the results
push OFFSET randArray
push userNum
call SortList
;display the median
push OFFSET randArray
push userNum
push OFFSET display_2
call median
;display sorted results
push OFFSET randArray
push userNum
push OFFSET display_3
call DisplayList
;Say "goodbye"
push OFFSET goodBye_1
call Goodbye
exit ; exit to operating system
main ENDP
;-------------------------------------------------------
;Gives an Intro to the program
; Receives parameters on the system stack (in the order pushed):
; Address of the title
; Address of the intro
;post: intro displayed
;registers: none
;-------------------------------------------------------
Intro PROC
pushad
mov ebp, esp
mov edx, [ebp+40]
call writeString
call CrLf
mov edx, [ebp+36]
call writeString
call CrLf
popad
ret 8
Intro ENDP
;-------------------------------------------------------
;Prompts user for an integer, int stores in userNum
; Receives parameters on the system stack (in the order pushed):
; Address of the error message
; Address of the prompt
; Address of return value
;Post: userNum
;registers: none
;-------------------------------------------------------
GetData PROC
pushad
;setup stack and prompt for entry
mov ebp, esp
reenter:
mov edx, [ebp+40]
mov ebx, [ebp+36]
call WriteString
call ReadInt
;validate entry
cmp eax, MIN ;if eax < LOWER
jl badEntry ;jump to summary
cmp eax, MAX ;if eax > UPPER
jg badEntry ;reprompt
jmp goodEntry ;else jump to end, we have a good value
;bad entry reprompt
badEntry:
mov edx, [ebp+44]
call WriteString
jmp reenter
goodEntry:
call CrLf
mov [ebx], eax
popad
ret 12
GetData ENDP
;-------------------------------------------------------
;Fills array with a number of random integers within RANGE
;Recieves parameters on the system stack (in order pushed)
; array
; userNum
;Post: array is filled with userNum number of randoms
;Registers used: none
;-------------------------------------------------------
FillArray PROC
pushad
mov ebp, esp
mov ecx, [ebp+36] ;initialize loop counter with user entry
mov edi, [ebp+40] ;setup array offset
fillLoop:
call nextRand
add edi, 4
loop fillLoop
popad
ret 8
FillArray ENDP
;-------------------------------------------------------
; Procedure nextRand
; adapted from check lecture 20 solutions
; Procedure to get the next random number in the range specified by the user.
; Preconditions: LRANGE < HRANGE
; Registers used: eax, edi
;-------------------------------------------------------
nextRand PROC
mov eax, HRANGE
sub eax, LRANGE
inc eax ;add 1 to get the number of integers in range
call RandomRange
add eax, LRANGE ;eax has value in [LOW - HIGH]
mov [edi],eax
ret
nextRand ENDP
;-------------------------------------------------------
;Sorts the contents of an integer array
; Receives parameters on the system stack (in order pushed)
; Array
; Array Size
;registers: none
;-------------------------------------------------------
sortList PROC
pushad
mov ebp, esp
mov ecx, [ebp+36]
mov edi, [ebp+40]
dec ecx ;ecx < request-1
mov ebx, 0 ;ebx=k
;for(k=0; k<request-1; k++)
outerLoop:
mov eax, ebx ;eax=i=k
mov edx, eax
inc edx ;edx=j=k+1
push ecx
mov ecx, [ebp+36] ;ecx < request
;for(j=k+1; j<request; j++)
innerLoop:
mov esi, [edi+edx*4]
cmp esi, [edi+eax*4]
jle skip
mov eax, edx
skip:
inc edx
loop innerLoop
;swap elements
lea esi, [edi+ebx*4]
push esi
lea esi, [edi+eax*4]
push esi
call exchange
pop ecx
inc ebx
loop outerLoop
popad
ret 8
sortList ENDP
;-------------------------------------------------------
; Exchange k and i
; Receives parameters on the system stack (in order pushed)
; array[k]
; array[i]
;registers: none
;-------------------------------------------------------
Exchange PROC
pushad
mov ebp,esp
mov eax, [ebp+40] ;array[k] low number
mov ecx, [eax]
mov ebx, [ebp+36] ;array[i] high number
mov edx, [ebx]
mov [eax], edx
mov [ebx], ecx
popad
ret 8
Exchange ENDP
;-------------------------------------------------------
;Displays the median of an integer array
; Receives parameters on the system stack (in order pushed)
; Array
; Array Size
; display string
;registers: none
;-------------------------------------------------------
Median PROC
pushad
mov ebp, esp
mov edi, [ebp+44]
;display string
mov edx, [ebp+36]
call writeString
;calculate median element
mov eax, [ebp+40]
cdq
mov ebx, 2
div ebx
shl eax, 2
add edi, eax
cmp edx, 0
je isEven
;Array size is odd, so display the middle value
mov eax, [edi]
call writeDec
call CrLf
call CrLf
jmp endMedian
isEven:
;Array size is even so average the two middle values
mov eax, [edi]
add eax, [edi-4]
cdq
mov ebx, 2
div ebx
call WriteDec
call CrLf
call CrLf
endMedian:
popad
ret 12
Median ENDP
;-------------------------------------------------------
;Displays the contents of an integer array, 10 per row
; Receives parameters on the system stack (in order pushed)
; Array
; Array Size
; display string
;registers: none
;-------------------------------------------------------
DisplayList PROC
pushad
mov ebp, esp
;display string
mov edx, [ebp+36]
call writeString
call CrLf
mov ecx, [ebp+40]
mov edi, [ebp+44]
mov ebx, 0
;display array contents
listloop:
inc ebx ;counter for 10 items per row
mov eax, [edi]
call writeDec
add edi, 4
cmp ebx, 10
jne noReturn ;jump if 10 items are not yet printed
call CrLf
mov ebx, 0
jmp noTab ;this skips adding a tab on a new row
noReturn:
mov al, TAB
call writeChar
noTab:
loop listloop
call CrLf
popad
ret 12
DisplayList ENDP
;-------------------------------------------------------
;Says good-bye to the user
; Receives parameters on the system stack:
; Address of string
;registers: none
;-------------------------------------------------------
Goodbye PROC
pushad
mov ebp, esp
mov edx, [ebp+36]
call writeString
call CrLf
popad
ret 4
Goodbye ENDP
END main
In order to exchange elements you should be passing their address (in other words, a pointer to each). What you did was just swapping the values passed as parameters, which were also immediately freed. Your code is the equivalent of this C function:
void Exchange(int x, int y)
{
int eax = x;
int ebx = y;
x = ebx;
y = eax;
}
You need something like:
void Exchange(int* x, int* y)
{
int eax = *x;
int ebx = *y;
*x = ebx;
*y = eax;
}
In asm, that may look like:
Exchange PROC
pushad
mov eax, [esp+40]
mov ecx, [eax]
mov ebx, [esp+36]
mov edx, [ebx]
mov [eax], edx
mov [ebx], ecx
popad
ret 8
Exchange ENDP
To call this function you would use the following:
lea esi, [edi+ebx*4]
push esi
lea esi, [edi+eax*4]
push esi
call Exchange
Note your Exchange function had an instruction mov eax, [edi] which I couldn't make sense of.
Update: Yes, you can emulate the functionality of LEA by doing the calculations manually. For example, lea esi, [edi+ebx*4] becomes:
mov esi, ebx
shl esi, 2 ; esi=ebx*4
add esi, edi ; esi=edi+ebx*4