Getting values into an array via scanf - arrays

.data
arr: .long 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
arr_size: .long 10
in: .string "%d"
.text
.global main
main:
#scanf
movl $0, %ecx # index i
scanloop:
cmpl arr_size, %ecx
je endscan # if ecx == 10 -> break
leal arr(, %ecx, 4), %eax # load the address of arr[i] into eax
pushl (%eax) # push the value that is being pointed at onto the stack
pushl $in
call scanf
addl $8, %esp
incl %ecx
jmp scanloop
#endscanf
endscan:
#...
I don't know how to "scanf" values into an array. I thought it'd work if I calculate the address of each index, push the value that the address points to onto the stack and call scanf. But I get a memory access error.
So, is there a way to do it like that?

For reasons I cannot tell it works if I use %edi instead of %ecx as my index.
I still don't know how to use a debugger, so I printed out my index in every iteration to see what's wrong. The thing was that %ecx never got beyond 1, which made the procedure stuck in an infinite loop.
With %edi, however, incrementation works perfectly fine.
PS: Out of curiosity I've tried other registers as well, namely %esi and %ebx. It also works for them. If I ever find out the reasons for this, I'll refer back to this post.
Edit: Ok, the mistake was a silly one. I didn't obey to the calling conventions.
So scanf, like printf, apparently changes register values. By convention, %eax, %ecx and %edx are caller-save, whereas %ebx, %edi and %esi are callee-save.
Because I, as the caller, didn't save %ecx prior to the call to scanf, I got an unwanted value in %ecx.
It worked for %edi, %esi and %ebx because these are callee-save, hence guaranteed to be not changed by the called function. So if I had tried %eax or %edx instead, it most likely wouldn't have worked either.
And yes, it works perfectly fine if I push %ecx on the stack before calling scanf.
Edit2:
Oh, and solved the issue with the access problem by making myself a global variable in which I scan a number, and move that number into the desired index:
.data
num: .long 0
#...
scanloop:
cmpl arr_size, %ecx
je endscan # if ecx == 10 -> break
pushl %ecx #caller-save
pushl $num
pushl $in
call scanf
addl $8, %esp
popl %ecx #restore
movl num, %eax #in order to avoid two memory arguments to movl
leal arr(, %ecx, 4), %edx #calculate address of index
movl %eax, (%edx) #store num in that index
incl %ecx
jmp scanloop
#endscanf
#...

Related

Update array in assembly

Just starting to learn assembly, so I am having trouble with some basic. For example, I am trying to modify a file that normally runs to find the maximum value.
What I am trying to do is iterate through the data_items and modify it by changing the list item, while still maintaining its ability to execute to find the maximum value.
I have tried modifying the start_loop to include an addl data_items, 0x01 but it errors out since it only pulls the first value.
I then adjusted it to do something like addl data_items(,%esi,4), 0x01 and iterate through adjusting it but that didn't work either. I know this is probably very simplistic, but I can't seem to find a method that works.
#
.section .data
data_items: #These are the data items trying to modify and store
.long 3,67,34,222,45,75,54,34,44,33,22,11,66,0
.section .text
.globl _start
_start:
movl $0, %edi
movl data_items(,%edi,4), %eax
movl %eax, %ebx
start_loop:
cmpl $0, %eax
je loop_exit
incl %edi
movl data_items(,%edi,4), %eax
cmpl %ebx, %eax
jle start_loop
movl %eax, %ebx
jmp start_loop
loop_exit:
movl $1, %eax
int $0x80
Once you have an array element in %eax just increment the memory that was last used to read %eax from:
start_loop:
cmpl $0, %eax ; Not actual element if this is 0
je loop_exit
incl data_items(,%edi,4) <<<<<<<<<<<<<<<<<
incl %edi
movl data_items(,%edi,4), %eax
This does not mess with the current finding of the max value. Only next time will the max value be 1 greater.

Storing Local variables in Assembly

So I've been working on a problem (and before you ask, yes, it is homework, but I've been putting in faithful effort!) where I have some assembly code and want to be able to convert it (as faithfully as possible) to C.
Here is the assembly code:
A1:
pushl %ebp
movl %esp, %ebp
subl $16, %esp
movl $0, -4(%ebp)
jmp .L2
.L4:
movl -4(%ebp), %eax
sall $2, %eax
addl 8(%ebp), %eax
movl (%eax), %eax
cmpl 12(%ebp), %eax
jg .L6
.L2:
movl -4(%ebp), %eax
cmpl 16(%ebp), %eax
jl .L4
jmp .L3
.L6:
nop
.L3:
movl -4(%ebp), %eax
leave
ret
And here's some of the C code I wrote to mimic it:
int A1(int a, int b, int c) {
int local = 0;
while(local < c) {
if(b > (int*)((local << 2) + a)) {
return local;
}
}
return local;
}
I have a few questions about how assembly works.
First, I notice that in L4, the body of the while loop, nothing is ever assigned to local. It's initialized to be 0 at the start of the function, and then never modified again. Looking at the C code I made for it, though, that seems odd, considering that the loop will go on indefinitely if the if-condition fails. Am I missing something there? I was under the impression that you'd need a snippet of code like:
movl %eax, -4(%ebp)
in order to actually assign anything to the local variable, and I don't see anything like that in the body of the while loop.
Secondly, you'll see that in the assembly code, the only local variable that's declared is "local". Hence, I have to use a snippet of code like:
if(b > (int*)((local << 2) + a))
The output of this line doesn't look much like the assembly code, though, and I think I might have made a mistake. What did I do wrong here?
And finally (thanks for your patience!), on a related note, I understand that the purpose of this if-loop in the while loop is to break out if the condition is fulfilled, and then to return local. Hence L6 and "nop" (which is basically saying nothing). However, I don't know how to replicate this in my program. I've tried "break", and I've tried returning local as you see here. I understand the functionality - I just don't know how to replicate it in C (short of using goto, but that kind of defeats the purpose of the exercise...).
Thank you for your time!
This is my guess:
int A1 (int *a, int value, int size)
{
int i = 0;
while (i<size)
{
if (a[i] <= value)
break;
}
return i;
}
Which, compiled back to assembly, gives me this code:
A1:
.LFB0:
pushl %ebp
movl %esp, %ebp
subl $16, %esp
movl $0, -4(%ebp)
jmp .L2
.L4:
movl -4(%ebp), %eax
leal 0(,%eax,4), %edx
movl 8(%ebp), %eax
addl %edx, %eax
movl (%eax), %eax
cmpl 12(%ebp), %eax
jg .L2
jmp .L3
.L2:
movl -4(%ebp), %eax
cmpl 16(%ebp), %eax
jl .L4
.L3:
movl -4(%ebp), %eax
leave
ret
Now this seems to be identical to your original ASM code, just the code starting at L4 is not the same, but if we anotate both codes:
ORIGINAL
movl -4(%ebp), %eax ;EAX = local
sall $2, %eax ;EAX = EAX*4
addl 8(%ebp), %eax ;EAX = EAX+a, hence EAX=a+local*4
ASM-C-ASM
movl -4(%ebp), %eax ;EAX = i
leal 0(,%eax,4), %edx ;EDX = EAX*4
movl 8(%ebp), %eax ;EAX = a
addl %edx, %eax ;EAX = EAX+EDX, hence EAX=a+i*4
Both codes continue with
movl (%eax), %eax
Because of this, I guess a is actually a pointer to some variable type that uses 4 bytes. By the comparison between the second argument and the value read from memory, I guess that type must be either int or long. I choose int solely by convenience.
Of course this also means that this code (and the original one) does not make any sense. It lacks the i++ part somewhere. If this is so, then a is an array, and the third argument is the size of the array. I've named my local variable i to keep with the tradition of naming index variables like this.
This code would scan the array searching for a value inside it that is equal or less than value. If it finds it, the index to that value is returned. If not, the size of the array is returned.

Y86 Stopped in 1 Steps Exception HLT

I translated a simple c program into IA32 and then transliterated it into Y86 but I am getting an error I don't understand or know how to debug since I am just learning Y86. The error is:
Stopped in 1 steps at PC = 0x1. Exception 'HLT', CC Z=1 S=0 O=0
Changes to registers:
Changes to memory:
The program is supposed to initialize i to 0 and then proceed through a for loop until i is greater than or equal to 5 and increment i each time. Inside the for loop I set j equal to i*2 and k is equal to j+1. My Y86 code is as follows:
main:
irmovl $0, %ebx
jmp L2
halt
L3:
rrmovl %ebx, %eax
addl %eax, %eax
rrmovl %eax, %ecx
rrmovl %ecx, %eax
irmovl $1, %esi
addl %esi, %eax
rrmovl %eax, %edx
addl %esi, %ebx
L2:
irmovl $4, %edi
subl %edi, %ebx
jle L3
I can provide the C code and IA32 code I transliterated from if it would help you answer my problem I really need some help thanks.
You forgot to add the extra NL (CR) in the source file. YAS is flawed. When it assembles it wraps the assembly around if there is no NL (CR) at the end of the source (ys) file, overwriting the beginning of the created object code (yo) file.
The result... well it is bad. In your case, YAS probably inserted a HLT instruction in the first character of the object file.

Loop Assembly x86

I am a beginner in Assembly(x86 ATT Syntax). I am working on an assignment where I have to go through each index of a 2d array and find the number of 1's. The method takes in 2d int array,int w, int h. How do I implement the loop to go go from 0 to w and bounce out with loop instruction. I know how to do it with a jmp instruction but loop just gives me errors/segFaults. This is my attempt with a jump statement and it works. How would I go about converting the inner loop using a loop instruction?
pushl %ebp
movl %esp, %ebp
movl $0, -4(%ebp)
movl $0, -8(%ebp)
movl $0, -12(%ebp)
movl $0, -4(%ebp) #outside loop
jmp .L10
.L14: #Inner Loop
movl $0, -8(%ebp)
jmp .L11
.L13:
movl -4(%ebp), %eax
leal 0(,%eax,4), %edx
movl 8(%ebp), %eax
addl %edx, %eax
movl (%eax), %eax
movl -8(%ebp), %edx
sall $2, %edx
addl %edx, %eax
movl (%eax), %eax
cmpl $1, %eax
jne .L12
addl $1, -12(%ebp)
.L12:
addl $1, -8(%ebp)
.L11: #check inner loop
movl -8(%ebp), %eax
cmpl 12(%ebp), %eax
jl .L13
addl $1, -4(%ebp)
.L10: #check outside loop
movl -4(%ebp), %eax
cmpl 16(%ebp), %eax
jl .L14
movl -12(%ebp), %eax
leave
ret
Generally using loop has no advantages except maybe smaller code. It's usually slower and less flexible, thus not recommended.
That said, if you still want to use it you should know that it employs the ecx register for counting down to zero. So you need to restructure your code to accommodate that. In your case, that means loading ecx with the value of w and letting it count down. You will also need to apply an offset of -1 during indexing, since your current loop variable goes from 0 to w-1, but ecx will go from w down to 1 (inclusive).
Furthermore, the loop instruction is used after the loop body, that is it implements a do-while loop. To skip the loop body if the count is zero a companion instruction, JECXZ, can be used.
You can use lodsl (which moves %esi up by default) and loop (which moves %ecx down) in tandem. I am not sure if it is more efficient than what gcc generates from c, which is basically your code, but it looks prettier.
What I have done here doesn't answer your question precisely -- rather than use loop for the inner loop I have assumed the whole array is stored contiguously and then there is only a single loop to worry about. When compiling from c on my machine, it is stored contiguously, but I'm not sure you should rely on that. Hopefully what I have done gives you enough to understand how loop and lodsl work, and you can modify your code to use them just in the inner loop.
.data
x:
.long 6
y:
.long 5
array:
.long 1,1,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,1,0,0,1,1
.text
.global _start
_start:
# set up for procedure call
push x
push y
push $array
# call and cleanup
call cnt
add $0xc, %esp
# put result in %ebx and finish up
#(echo $? gives value in bash if <256)
mov %eax, %ebx
mov $1, %eax
int $0x80
# %ebx will hold the count of 1s
# %ecx will hold the number of elements to check
# %esi will hold the address of the first element
# Assumes elements are stored contiguously in memory
cnt:
# do progogue
enter $0, $1
# set %ebx to 0
xorl %ebx, %ebx
# grab x and y parameters from stack and
# multiply together to get the number of elements
# in the array
movl 0x10(%ebp), %eax
movl 0xc(%ebp), %ecx
mul %ecx
movl %eax, %ecx
# get address of first element in array
movl 0x8(%ebp), %esi
getel:
# grab the value at the address in %esi and increment %esi
# it is put in %eax
lodsl
# if the value in %eax is 1, increment %ebx
cmpl $1, %eax
jne ne
incl %ebx
ne:
# decrement %ecx and if it is greater than 0, keep going
loop getel
# %ecx is zero so we are done. Put the count in %eax
movl %ebx, %eax
# do epilogue
leave
ret
It really is prettier without all the comments.
.data
x:
.long 6
y:
.long 5
array:
.long 1,1,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,1,0,0,1,1
.text
.global _start
_start:
push x
push y
push $array
call cnt
add $0xc, %esp
mov %eax, %ebx
mov $1, %eax
int $0x80
cnt:
enter $0, $1
xorl %ebx, %ebx
movl 0x10(%ebp), %eax
movl 0xc(%ebp), %ecx
mul %ecx
movl %eax, %ecx
movl 0x8(%ebp), %esi
getel:
lodsl
cmpl $1, %eax
jne ne
incl %ebx
ne:
loop getel
movl %ebx, %eax
leave
ret

Infinite loop in my simple assembly program

I am learning loops and jumps in assembly and I try to make a simple loop. I want the printf command to be called 10 times. I have set the counter variable to 1. I have also set %edx to 1 and then I increment it for every iteration. If it is equal to 10, then we should exit the loop. But now the loop is infinite. I have debugged with gdb and %edx seems to be overwritten in the printf function. That is why I push %edx to the stack and the pop it back after the printf call, but it doesn't work. What have I missed?
.section .data
output:
.asciz "Value is %d\n"
val1:
.int 123
counter:
.int 1
.section .text
.globl _start
_start:
nop
movl counter, %edx # start at 1
gohere:
movl val1, %ebx # move value 123 to %ebx
pushl %edx # push %edx to stack
pushl %ebx # push %ebx to stack
pushl $output
call printf # call printf
popl %edx # pop %edx value
inc %edx
cmp $10, %edx # if %edx is less than 10...
jl gohere # ... go to gohere, otherwise exit
movl $0, %ebx
movl $1, %eax
int $0x80
you pushed output as the last push so the first pop will pop output. it is Stack and it is LIFO. in your code output will be in edx after you pop it.
to solve it put two pops before popl edx:
popl output
popl ebx

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