mov instruction in x86 assembly

From what I've read about mov, it copies the second argument into the first argument. Then, what does this do?

movl    8(%ebp),    %edx

It copies whatever is in edx to the first parameter of the function (since an offset of +8 from ebp is a parameter)?

I feel like what this really means is moving the first parameter into the edx register, but I read on Wikipedia that it is the other way around?

Answer

movl 8(%ebp), %edx

is in "AT&T Syntax"; in this syntax, the source comes first and the destination second. So yes, your belief is correct. Most documentation uses the "Intel Syntax", which has the reverse ordering. This is a source of considerable confusion for people new to x86 assembly.

In Intel Syntax, your instruction would be written:

mov edx, [ebp + 8]

Note the absence of % before the register names, and the use of square brackets instead of parentheses for the address, and the lack of an l suffix on the instruction. These are dead giveaways to know which form of assembly you are looking at.