Strings (Poetry Club Door Policy)
Did you know that I originally wrote this exercise? Story et al! Hopefully this will help me today.
I do my regular stuff with one change: I saw in a community solution you can do multiple global exports in one line:
global front_door_response, front_door_password, back_door_response, back_door_password
I can now run the tests and two pass without any implementation:
I also see Expected 83, Was 0. I think this means that my expectation of characters just being bytes, and likely being ASCII (aka there is no encoding) is correct. I do think it would be helpful to see in the error message both the byte and printable character (if any).
Gotta start reading!
It is possible to declare a string in NASM using the directive db , as usual for declared data
The usage of db reads as if it is special, but I don’t believe it is. I think I already used it before. I think it means “byte” size entries. Perhaps could use a tiny mention that this is just the same as with the arrays we learnt about before. To me, as “usual” doesn’t convey “it’s not special”, especially because of the next part:
Even though this defines an array, it is not necessary to use a comma (, ) to separate values:
Is this also the case for non-string arrays? I don’t know but I continue.
Reading the instructions further, I finally realise that d as in dword probably means double and q as in qword means quadruple. Maybe this was mentioned somewhere, but I do not recall.
As I read the string instructions (lods etc), I do not yet understand how to use them or why to use them, but abstractly I think I need this to copy characters out of the array or copy (parts of) arrays. We’ll see. Oh and yeah, again rcx being special! The note reads weirdly:
Note, however, that rcx is not an operand to those instructions. Its value must be adjusted before using them.
I think better would be something like:
rcx is the register rcx. Just like with loop instructions, you cannot set the value to rcx by an operand to the instructions, but it must be set before the instruction is used.
front_door_response
I need to get the first character in the string. The string is “an array” of bytes, and what’s passed in is the address to the first sentence (array/string). That means that that the first byte at that address is exactly what I need.
Let me first try to mov the first byte. I’d like to try out something like movs, but I find the instructions pretty unclear. The summary says it will copy from rsi to rdi, but the explanation earlier says it will “copy 1 byte, changing rsi and rdi by 1” and even earlier " They usually expect the source to be a memory location in rsi and/or the destination to be a memory location in rdi ."
The summary and first description tell me: rsi and rdi need to be memory locations and then it will copy 1 byte from left to right (or in reverse given the DF). So I don’t understand what this “changing x and y by 1” is. Is it automatically changing the memory addresses by 1 so you can keep copying? Unclear.
There are also no examples. So mov it is!
mov al, [rdi]
ret
So that worked. Next!
front_door_password
It must modify this string in-place, making it correctly capitalized.
Capitalization of ASCII characters is deceptively simple. For example the letter A is represented as 0x41 (or 65) and a is 0x61 (or 97). They are 32 apart. Capitalizing a letter is lowering its byte value by 32 and doesn’t need to happen if the value is already smaller than 97
So if I wanted to titleize a word (not uppercase), it would be:
mov al, [rdi] ; temporarily store a copy of the first character
cmp al, 0x61 ; check if already capitalized
jb .skip ; skip capitalization if already capitalized
sub al, 32 ; capitalize
mov [rdi], al ; write the character back to the memory
.skip:
ret
However, I want to explore how I could do this for each character. There is no length given for the string, but the exercise states each string is terminated by \0. That can be used to loop and break. (I’ll probably need this later anyway, so let’s try).
I think I can keep a counter to increase the memory address and break when the \0 is reached. I want to use loopne, but I also don’t want to give an arbitary size. I am wondering if I could set rcx to -1 and have it loop until loopne. Not going to try tho, I don’t need to. jne gives me that behaviour.
.upcase:
call upcase_one ; (there is no empty password)
inc rdi ; move memory by 1 byte
cmp byte [rdi], `\0` ; check the next bye
jne .upcase
ret
upcase_one:
mov al, [rdi]
cmp al, 0x61 ; check if already uppercase
jb .skip ; skip capitalization if already uppercase
sub al, 32 ; upcase
mov [rdi], al ; write the character back to the memory
.skip:
ret
I tried it out and it works!
Expected ‘Summer’ Was ‘SUMMER’. Combined letters are: SUMMER
Expected ‘Sophia’ Was ‘SOPHIA’. Combined letters are: sophia
Expected ‘Code’ Was ‘CODE’. Combined letters are: Code
Ah okay. These cases indicate that I could of course get capital letters at non-zero positions.
I need to:
- upcase the first letter (uppercase)
- downcase everything else (lowercase)
That should be easy now:
downcase_one:
mov al, [rdi]
cmp al, 0x61 ; check if already lower case
jge .skip ; skip capitalization if already lowercase
add al, 32 ; downcase
mov [rdi], al ; write the character back to the memory
.skip:
ret
Then combine it:
call upcase_one ; (there is no empty password)
inc rdi ; move memory by 1 byte
cmp byte [rdi], `\0` ; check the next bye
jne .downcase ; break if last character
ret
.downcase:
call downcase_one
inc rdi ; move memory by 1 byte
cmp byte [rdi], `\0` ; check the next bye
jne .downcase
ret
Rejoice!
back_door_response
What I can do here is:
- store the current byte if it is a printable character
- go to the next byte
- break if
\0
.store:
cmp byte [rdi], 0x41 ; check if uppercase A
jb .next
cmp byte [rdi], 0x7A ; check if lowercase z
jg .next
mov al, [rdi] ; store as return value
.next:
inc rdi ; move memory by 1 byte
cmp byte [rdi], `\0` ; check the next bye
jne .store
ret
I feel very good at my process trying to complete this.
back_door_password
I need to do two things:
- format the string
- add
", please."
My process will be:
- copy the second argument’s memory to the first location
- call front door password to capitalize it
- “add” the second string at the end (dunno how yet)
I re-read the extra instructions:
| instruction |
description |
| lods |
loads from the memory location in rsi into rax |
| stos |
stores rax into the memory location in rdi |
rsi happens to be the second argument passedrdi happens to be the first argument passed
So if I want to copy the string from rsi into the rdi location, I think I can do it by calling lods folowing by stos. However, I think this is what movs is supposed to be for: “copies between memory locations (rsi to rdi)”.
I’ll just have to deal with that “memory increase thing”.
I don’t think I can use rep here because I do not know the length of the string and I don’t think I should use repe or repne because cmps also moves “the pointers” and idk, feels weird. Let me just try to do it manually:
lea r8, [rdi] ; store address of destination so I can "rewind"
.move:
; lodsb
; stosb
movsb ; move one byte from rsi to rdi, inc. both
cmp byte [rsi], `\0` ; check if next byte rsi is end of string
jne .move
I will try this out first, because the test output will tell me if this is correct or not.
Expected ‘Work, please.’ Was ‘work\xFC\x7F’. Combined letters are: work
Expected ‘Horse, please.’ Was ‘horse\x7F’. Combined letters are: horse
\x7C is the delete (special) character, and \xFC I think is in the extended set? Idk. Doesn’t really matter. It copied my string, except I probably want to copy the \0 too! This is probably the issue because I am no longer correctly terminating the string (for this exercise).
I add movsb below the jne .move and all is golden:
Expected ‘Work, please.’ Was ‘work’. Combined letters are: work
Expected ‘Horse, please.’ Was ‘horse’. Combined letters are: horse
I can now rewind the memory address and call front_door_password:
mov rdi, r8
call front_door_password
Expected ‘Work, please.’ Was ‘Work’. Combined letters are: work
Expected ‘Horse, please.’ Was ‘Horse’. Combined letters are: horse
Sweet. Now need to add the final piece of info.
I think what I can do is store ", please.\0" as an array in memory .odata, and then copy it at the end of the string. Quickly counting the size: 10 bytes including the final terminator.
section .odata
politeness db `, please.`, 0
