Unable to understand x86 instruction(s)

Question

Recently I have been given a set of assembly instructions which I must learn to understand. I know an extremely basic level of x86 assembly but I'm starting to come across more nuances that are very difficult for me to 'just google'. So I have to ask you lovely people.

Below are the instructions

mov     ecx, [esp+10h+arg_0]
...
...
mov     [esp+10h+arg_8], eax
...
...
and     al, 0E0h

Are arg_0 and arg_8 arguments supplied to the program at runtime or arguments supplied to a function within the program? Also please help me understand what 0E0h is in the last instruction. Is that a function, memory address, or hex value? Thank you.

Answer 1

the first instruction is a read
the second instruction is a write
the third instruction is a compare

arg0 , arg_8 are arguments provided to the function and is provided by the caller like

call xyz(arg0,arg1....argn);

normally arguments provided to a function are not modified unless the argument is a pass by reference

for understanding opcodes operation you must understand

1. how an opcode works
2. what are its operands ,
3. how many operands does it take
4. what is result of the operation
5. which flags are affected
6. what exceptions are raised
7. what is the mode in which this opcode can be used
8. are there any bitwise exceptions
9. are there modifiers to the opcode
10. Compatibility etc these are all described in intel / amd manuals
    or you can use any online reference like reference for and opcode

if you go through the linked reference you will understand that the operand can be an immediate (constant value , [imm8 , imm16 , imm32 )
or register (al...,r8 or ax...,r16 or eax...,r32)
or memory (byte,m8 or word , m16 or dword , m32)

based on the description infer the type of your operand 0E0h