this is a code to reverse a string
I am really struggling to finish this code. I could not find what's wrong with it, however the output is wrong.
mov esi, OFFSET source
mov edi, OFFSET target
add edi, SIZEOF target-2
mov ecx, SIZEOF source-1
L1:
mov al, [esi]
mov [edi], al
inc esi
dec edi
loop L1
It's a bit odd that you pass two lengths to the function. All sorts of bad stuff might happen if the two mismatch.
Better to pass the length of the string explicitly, or have the code figure out the length.
As getting the length of a old-skool c-string is non-trivial to code, but trivial to google, I'm going to just pass it as a parameter.
'...the output is wrong.'
The problem is that your strings need to be zero terminated, but you are not putting the terminating zero on your destination string.
First off, if you want to have the string be a valid c-style string, make sure you add a terminating zero, like so: source db "test test",0
mov esi, OFFSET source ;Start of source
mov edi, OFFSET target ;start of dest
;length EQU SIZEOF source ;we are reversing source
mov ecx, SIZEOF source ;Length of the string
;(includes the terminating 0)
Setup:
;//a c-string must have a terminating 0!
xor eax,eax ;al=0, put the terminating zero in first
L1:
mov [edi+ecx-1], al ;if length(ecx)=1, then write to [edi] directly.
mov al, [esi]
inc esi
loop L1
Remarks on the code
There is no need to keep three counters in flight (edi,esi,ecx), two is enough. esi counts up, ecx counts down.
The x86 has lots of really helpful addressing modes, which are mostly free performance wise.
The last iteration will read the terminating zero in al, we don't need to reverse this and you've already written it at the beginning, so it is (silently) dropped.
Because of the terminating zero, length is at least 1. This is good because if somehow you were to feed 0 into loop it will loop 'forever'; not good ('forever' being 4+ billion times).
Note that your code does not take account of Unicode, so it will not work for UTF8, but lets assume it's just a learning exercise.
If you follow an ABI, then you can just pass the parameters in registers, meaning you can skip on some of the initialization. Given that your code is not going to win any prizes for raw speed, I've skipped this step.
Related
Theoretical question. But let's say I have written an assembly program. I have "labelx:" I want the program to read at this memory address and only this size and print to stdout.
Would it be something like
jmp labelx
And then would i then use the Write syscall , making sure to read from the next instruction from labelx:
mov rsi,rip
mov rdi,0x01
mov rdx,?
mov rax,0x01
syscall
to then output to stdout.
However how would I obtain the size to read itself? Especially if there is a
label after the code i want to read or code after. Would I have to manually
count the lines?
mov rdx,rip+(bytes*lines)
And then syscall with populated registers for the syscall to write to from rsi to rdi. Being stdout.
Is this Even possible? Would i have to use the read syscall first, as the write system call requires rsi to be allocated memory buffer. However I assumed .text is already allocated memory and is read only. Would I have to allocate onto the stack or heap or a static buffer first before write, if it's even possible in the first place?
I'm using NASM syntax btw. And pretty new to assembly. And just a question.
Yes, the .text section is just bytes in memory, no different from section .rodata where you might normally put msg: db "hello", 10. x86 is a Von Neumann architecture (not Harvard), so there's no distinction between code pointers and data pointers, other than what you choose to do with them. Use objdump -drwC -Mintel on a linked executable to see the machine-code bytes, or GDB's x command in a running process, to see bytes anywhere.
You can get the assembler to calculate the size by putting labels at the start/end of the part you want, and using mov edx, prog_end - prog_start in the code at the point where you want that size in RDX.
See How does $ work in NASM, exactly? for more about subtracting two labels (in the same section) to get a size. (Where $ is an implicit label at the start of the current line, although $ isn't likely what you want here.)
To get the current address into a register, you need a RIP-relative LEA, not mov, because RIP isn't a general-purpose register and there's no special form of mov that reads it.
here:
lea rsi, [rel here] ; with DEFAULT REL you could just use [here]
mov edi, 1 ; stdout fileno
mov edx, .end - here ; assemble-time constant size calculation
mov eax, 1 ; __NR_write
syscall
.end:
This is fully position-independent, unlike if you used mov esi, here. (How to load address of function or label into register)
The LEA could use lea rsi, [rel $] to assemble to the same machine-code bytes, but you want a label there so you can subtract them.
I optimized your MOV instructions to use 32-bit operand-size, implicitly zero-extending into the full 64-bit RDX and RAX. (And RDI, but write(int fd, void *buf, size_t len) only looks at EDI anyway for the file descriptor).
Note that you can write any bytes of any section; there's nothing special about having a block of code write itself. In the above example, put the start/end labels anywhere. (e.g. foo: and .end:, and mov edx, foo.end - foo taking advantage of how NASM local labels work, by appending to the previous non-local label, so you can reference them from somewhere else. Or just give them both non-dot names.)
I'm trying to increment 1 to a variable in IA32 Assembly in Linux
section .data
num: dd 0x1
section .text
global _start
_start:
add dword [num], 1
mov edx, 1
mov ecx, [num]
mov ebx,1
mov eax,4
int 0x80
mov eax,1
int 0x80
Not sure if it's possible to do.
In another literature I saw the follow code:
mov eax, num
inc eax
mov num, eax
Is it possible to increment a value to a var without moving to a register?
If so, do I have any advantage moving the value to a register?
Is it possible to increment a value to a var without moving to a register?
Certainly: inc dword [num].
Like practically all x86 instructions, inc can take either a register or memory operand. See the instruction description at http://felixcloutier.com/x86/inc; the form inc r/m32 indicates that you can give an operand which is either a 32-bit register or 32-bit memory operand (effective address).
If you're interested in micro-optimizations, it turns out that add dword [num], 1 may still be somewhat faster, though one byte larger, on certain CPUs. The specifics are pretty complicated and you can find a very extensive discussion at INC instruction vs ADD 1: Does it matter?. This is partly related to the slight difference in effect between the two, which is that add will set or clear the carry flag according to whether a carry occurs, while inc always leaves the carry flag unchanged.
If so, do I have any advantage moving the value to a register?
No. That would make your code larger and probably slower.
Yesterday I was looking at some 32 bit code generated by VC++ 2010 (most probably; don't know about the specific options, sorry) and I was intrigued by a curious recurring detail: in many functions, it zeroed out ebx in the prologue, and it always used it like a "zero register" (think $zero on MIPS). In particular, it often:
used it to zero out memory; this is not unusual, as the encoding for a mov mem,imm is 1 to 4 bytes bigger than mov mem,reg (the full immediate value size has to be encoded even for 0), but usually (gcc) the necessary register is zeroed out "on demand", and kept for more useful purposes otherwise;
used it for compares against zero - as in cmp reg,ebx. This is what stroke me as really unusual, as it should be exactly the same as test reg,reg, but adds a dependency to an extra register. Now, keep in mind that this happened in non-leaf functions, with ebx being often pushed (by the callee) on and off the stack, so I would not trust this dependency to be always completely free. Also, it also used test reg,reg in the exact same fashion (test/cmp => jg).
Most importantly, registers on "classic" x86 are a scarce resource, if you start having to spill registers you waste a lot of time for no good reason; why waste one through all the function just to keep a zero in it? (still, thinking about it, I don't remember seeing much register spillage in functions that used this "zero-register" pattern).
So: what am I missing? Is it a compiler blooper or some incredibly smart optimization that was particularly interesting in 2010?
Here's an excerpt:
; standard prologue: ebp/esp, SEH, overflow protection, ... then:
xor ebx, ebx
mov [ebp+4], ebx ; zero out some locals
mov [ebp], ebx
call function_1
xor ecx, ecx ; ebx _not_ used to zero registers
cmp eax, ebx ; ... but used for compares?! why not test eax,eax?
setnz cl ; what? it goes through cl to check if eax is not zero?
cmp ecx, ebx ; still, why not test ecx,ecx?
jnz function_body
push 123456
call throw_something
function_body:
mov edx, [eax]
mov ecx, eax ; it's not like it was interested in ecx anyway...
mov eax, [edx+0Ch]
call eax ; virtual method call; ebx is preserved but possibly pushed/popped
lea esi, [eax+10h]
mov [ebp+0Ch], esi
mov eax, [ebp+10h]
mov ecx, [eax-0Ch]
xor edi, edi ; ugain, registers are zeroed as usual
mov byte ptr [ebp+4], 1
mov [ebp+8], ecx
cmp ecx, ebx ; why not test ecx,ecx?
jg somewhere
label1:
lea eax, [esi-10h]
mov byte ptr [ebp+4], bl ; ok, uses bl to write a zero to memory
lea ecx, [eax+0Ch]
or edx, 0FFFFFFFFh
lock xadd [ecx], edx
dec edx
test edx, edx ; now it's using the regular test reg,reg!
jg somewhere_else
Notice: an earlier version of this question said that it used mov reg,ebx instead of xor ebx,ebx; this was just me not remembering stuff correctly. Sorry if anybody put too much thought trying to understand that.
Everything you commented on as odd looks sub-optimal to me. test eax,eax sets all flags (except AF) the same as cmp against zero, and is preferred for performance and code-size.
On P6 (PPro through Nehalem), reading long-dead registers is bad because it can lead to register-read stalls. P6 cores can only read 2 or 3 not-recently-modified architectural registers from the permanent register file per clock (to fetch operands for the issue stage: the ROB holds operands for uops, unlike on SnB-family where it only holds references to the physical register file).
Since this is from VS2010, Sandybridge wasn't released yet, so it should have put a lot of weight on tuning for Pentium II/III, Pentium-M, Core2, and Nehalem where reading "cold" registers is a possible bottleneck.
IDK if anything like this ever made sense for integer regs, but I don't know much about optimizing for CPUs older than P6.
The cmp / setz / cmp / jnz sequence looks particularly braindead. Maybe it came from a compiler-internal canned sequence for producing a boolean value from something, and it failed to optimize a test of the boolean back into just using the flags directly? That still doesn't explain the use of ebx as a zero-register, which is also completely useless there.
Is it possible that some of that was from inline-asm that returned a boolean integer (using a silly that wanted a zero in a register)?
Or maybe the source code was comparing two unknown values, and it was only after inlining and constant-propagation that it turned into a compare against zero? Which MSVC failed to optimize fully, so it still kept 0 as a constant in a register instead of using test?
(the rest of this was written before the question included code).
Sounds weird, or like a case of CSE / constant-hoisting run amok. i.e. treating 0 like any other constant that you might want to load once and then reg-reg copy throughout the function.
Your analysis of the data-dependency behaviour is correct: moving from a register that was zeroed a while ago essentially starts a new dependency chain.
When gcc wants two zeroed registers, it often xor-zeroes one and then uses a mov or movdqa to copy to the other.
This is sub-optimal on Sandybridge where xor-zeroing doesn't need an execution port, but a possible win on Bulldozer-family where mov can run on the AGU or ALU, but xor-zeroing still needs an ALU port.
For vector moves, it's a clear win on Bulldozer: handled in register rename with no execution unit. But xor-zeroing of an XMM or YMM register still needs an execution port on Bulldozer-family (or two for ymm, so always use xmm with implicit zero-extension).
Still, I don't think that justifies tying up a register for the duration of a whole function, especially not if it costs extra saves/restores. And not for P6-family CPUs where register-read stalls are a thing.
I wrote a simple assembly program that gets two integers from the user via a prompt, multiplies them together and prints that out. I wanted to do this directly with sys_read and not scanf so I could manually convert the input to an integer after removing the LF.
Here's the full source: http://pastebin.com/utnjTvNZ
In particular, what I want to do now is manually add a newline to the result of the multiplication that is now converted back to it's ASCII char equivalent. Initially, I thought I could just left shift 16 bits and add 0xA leaving me with, for example, 0x0034000A on the stack for 2*2 (0x0034 is "4" in ASCII chars), followed by a null terminator and a LF. However, the LF is printing before the result. I figured this was an endianess thing, so I tried the reverse (0x000A0034) and that just printed some other ASCII char instead.
So, simply, how do I properly push a newline to the stack so that this is printed with a newline following the number when using sys_write? What I'm missing is how strings are stored on the stack... which I can't test because normally you just create a variable and push the address onto the stack.
I'm aware some things in here could be done better, cleaner and up-to-standards and whatnot. I understand things intuitively so it's something I just need to do to better understand the stack and Linux system calls in general.
Okay, so to answer my own question thanks to the help of Jester, to add a newline to the 32-bit word I was displaying in memory, I had to understand endianness. Since I compiled for 32-bit, my program is functioning on 32-bit words. These words' bytes are written into memory "backwards". The words themselves are still stored in "normal" order. For example 0x0A290028 0x0A293928 prints (NULL)LF(9)LF. The bytes are backwards but the words are not. Sys_write, since it just uses a void *buf and isn't aware of strings, just reads bytes in endian-order from the buffer and spits them out.
What I was able to do was simply left-shift my single-digit product, for example, 0x00000034 by 8-bits. This left me with 0x00003400. To that, I could add 0x000A0000. This would result in 0x000A3400, and the number "4" being printed followed by a newline.
So, the new procedure looks like this:
multprint:
mov eax, sys_write ;4
mov ebx, stdout ;1
mov ecx, resultstr
mov edx, resultstrLen
dec edx
int 0x80
pop eax ;multiplican't
pop ebx ;multiplicand
mul ebx
add eax, '0'
shl eax, 8 ;make room for () and LF
add eax, 0x0A290028
push eax
mov ecx, esp
;mov [num], eax ;use these two lines if I don't want to use the stack
;mov ecx, num
mov eax, sys_write
mov ebx, stdout
mov edx, 4
int 0x80
I have this ASM function, that takes 4 arguments. The first two arguments are passed by value, the last two are passed by reference. So I'm using this:
PUSH EBP
MOV EBP, ESP
SUB ESP, 20
MOV EAX, [EBP+8]
MOV EBX, [EBP+12]
LEA ECX, [EBP+16]
LEA EDX, [EBP+20]
PUSH EDX
PUSH ECX
PUSH EBX
PUSH EAX
CALL Function
LEAVE
RETN 20
(Note that I'm using this code inside C++ using the VC's __asm statement).
But while searching about the use of LEA to pass arguments as pointers (aka by reference) I found:
[...] Note there are NO brackets in this line. Putting the square brackets around
something means "get the contents of", so you were effectively defeating the
LEA op. [...]
I want to pass both arguments at EBP+16 and EBP+20 by reference, but how can I do that if can't use brackets? If I don't put them, then the compiler throws an error (C2424).
Thanks in advance.
Try
MOV ECX, EBP+16
MOV EDX, EBP+20
lea has one type of operation, so use the syntax that makes your compiler glad (for example, fasm requires brackets always, while masm requires its absence around label arguments).
Note, that:
Windows calling conventions require you to preserve ebx during calls;
I doubt taking addresses of 3rd and 4th arguments was really your intention;
You can use push dword [ebp+xxx] instruction.