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0 pointssaagarjha7y ago0 comments

> Why and how is argv[1] at [esp+8]? I realize this is a Linux-specific implementation detail but would like to understand.

The kernel puts them there when the program starts; I believe this is part of the System V calling convention so it's not Linux-specific.

> argv[1] is "gotten" by reading from esi. You then proceed to poke eax and ecx. TIL that SI, AX and CX were implicitly linked. What specifically is going on here?

They are not linked, the lodsb instruction loads a byte from the address pointed at by esi into ax.

> How does the LEA usage here work? You're putting the address of "[ecx+eax-'0']" into ecx. First question, how is ecx not clobbered? Secondly, how does that dereference work? I see similar semantics used in the 2nd and 3rd instructions, it seems AX and CX are linked in some way (in this situation).

lea just does math, it doesn't dereference anything. ecx is being clobbered, but that's what we want: the lea is doing the equivalent of ecx += eax - '0'.

> So the jecxz is bailing out (to an assumed label, that's okay) if ecx is 0. Cool. But why do you now zero and then increment eax?

This will be the return value of main and the exit code of the program. Since you errored, you want to return a nonzero value.

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exikyut7y ago

>> argv[1] is "gotten" by reading from esi. You then proceed to poke eax and ecx. TIL that SI, AX and CX were implicitly linked. What specifically is going on here?

> Where do you see this?

Here:

  mov esi, [esp+8]  ; get argv[1]                value written to esi; okay
  xor eax, eax      ; clear the upper bits       \ how is writing to eax
  xor ecx, ecx      ; will hold converted value  / and ecx now relevant?!

> lea just does math, it doesn't dereference anything.

Right; I said "LEA usage"; the instruction itself I understand, but "[ecx+eax-'0']" is throwing me for a loop.

T-hawk7y ago

LEA does math, but with a particular structure to the inputs: a pointer plus an index plus a constant. That's meant to be used as a C compiler would, where the pointer would be to a struct array, the index would be the position within the array, and the constant would be the byte offset within the struct.

Calculating a memory address in that way is a common operation, so x86 processors give it special support with hardware that can calculate that sum all in one step.

Most often the computed address is used in a MOV instruction. LEA provides access to the address-computation hardware without doing the actual data move. That's useful because some math operations can be done faster through that hardware than with general-purpose math instructions. LEA requires that pattern of register + register*shift + constant (we aren't using the shift here).

    imul ecx, ecx, 10
    lea ecx, [ecx+eax-'0']

Those two instructions multiply ecx by 10 and then add the next digit (stored as ASCII in eax) to it. Subtracting '0' converts that digit from ASCII to its numerical value. It just so happens that LEA can execute the add and conversion fast and together in one step, because that combined operation matches the pattern of computing an address.

willvk7y ago

I didn't know you could do that with lea. That's pretty cool!

saagarjhaOP7y ago

Sorry, I'm actively editing my answer as I get a better sense of your questions and I can answer more of them. Please see my edits; also, the specific lines you pointed out are not related to reading in argv[1]. They're just zeroing out eax and ecx; eax because we will only be writing to the lower bits with the lodsb and don't want out arithmetic to be fouled by anything left over in the higher bits, and ecx because we are continuously adding to it to in atoi, so it needs to start off at zero.

userbinator7y ago

They are not linked, the lodsb instruction loads a byte from the address pointed at by esi into ax.

lodsb (b for Byte) loads into al, not ax. It's lodsw (Word) which uses ax, and of course there's lodsd for eax and even lodsq for rax (64-bit only).

This will be the return value of main and the exit code of the program. Since you errored, you want to return a nonzero value.

Actually that's the non-error path; eax is set to 1 and edx to 0 as initial values for computing fib(n) in the fibloop.

saagarjhaOP7y ago

Oops, you're right. I should have read that more carefully…

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0 pointssaagarjha7y ago0 comments

> Why and how is argv[1] at [esp+8]? I realize this is a Linux-specific implementation detail but would like to understand.

The kernel puts them there when the program starts; I believe this is part of the System V calling convention so it's not Linux-specific.

> argv[1] is "gotten" by reading from esi. You then proceed to poke eax and ecx. TIL that SI, AX and CX were implicitly linked. What specifically is going on here?

They are not linked, the lodsb instruction loads a byte from the address pointed at by esi into ax.

lea just does math, it doesn't dereference anything. ecx is being clobbered, but that's what we want: the lea is doing the equivalent of ecx += eax - '0'.

> So the jecxz is bailing out (to an assumed label, that's okay) if ecx is 0. Cool. But why do you now zero and then increment eax?

This will be the return value of main and the exit code of the program. Since you errored, you want to return a nonzero value.

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exikyut7y ago

>> argv[1] is "gotten" by reading from esi. You then proceed to poke eax and ecx. TIL that SI, AX and CX were implicitly linked. What specifically is going on here?

> Where do you see this?

Here:

  mov esi, [esp+8]  ; get argv[1]                value written to esi; okay
  xor eax, eax      ; clear the upper bits       \ how is writing to eax
  xor ecx, ecx      ; will hold converted value  / and ecx now relevant?!

> lea just does math, it doesn't dereference anything.

Right; I said "LEA usage"; the instruction itself I understand, but "[ecx+eax-'0']" is throwing me for a loop.

T-hawk7y ago

Calculating a memory address in that way is a common operation, so x86 processors give it special support with hardware that can calculate that sum all in one step.

    imul ecx, ecx, 10
    lea ecx, [ecx+eax-'0']

willvk7y ago

I didn't know you could do that with lea. That's pretty cool!

saagarjhaOP7y ago

userbinator7y ago

They are not linked, the lodsb instruction loads a byte from the address pointed at by esi into ax.

lodsb (b for Byte) loads into al, not ax. It's lodsw (Word) which uses ax, and of course there's lodsd for eax and even lodsq for rax (64-bit only).

This will be the return value of main and the exit code of the program. Since you errored, you want to return a nonzero value.

Actually that's the non-error path; eax is set to 1 and edx to 0 as initial values for computing fib(n) in the fibloop.

saagarjhaOP7y ago

Oops, you're right. I should have read that more carefully…

j / k navigate · click thread line to collapse