None of the previous answers quite got to the bottom of my own confusion, so I'd like to add my own.
What I was missing is that lea operations treat the use of parentheses different than how mov does.
Think of C. Let's say I have an array of long that I call array. Now the expression array[i] performs a dereference, loading the value from memory at the address array + i * sizeof(long) [1].
On the other hand, consider the expression &array[i]. This still contains the sub-expression array[i], but no dereferencing is performed! The meaning of array[i] has changed. It no longer means to perform a deference but instead acts as a kind of a specification, telling & what memory address we're looking for. If you like, you could alternatively think of the & as "cancelling out" the dereference.
Because the two use-cases are similar in many ways, they share the syntax array[i], but the existence or absence of a & changes how that syntax is interpreted. Without &, it's a dereference and actually reads from the array. With &, it's not. The value array + i * sizeof(long) is still calculated, but it is not dereferenced.
The situation is very similar with mov and lea. With mov, a dereference occurs that does not happen with lea. This is despite the use of parentheses that occurs in both. For instance, movq (%r8), %r9 and leaq (%r8), %r9. With mov, these parentheses mean "dereference"; with lea, they don't. This is similar to how array[i] only means "dereference" when there is no &.
An example is in order.
Consider the code
movq (%rdi, %rsi, 8), %rbp
This loads the value at the memory location %rdi + %rsi * 8 into the register %rbp. That is: get the value in the register %rdi and the value in the register %rsi. Multiply the latter by 8, and then add it to the former. Find the value at this location and place it into the register %rbp.
This code corresponds to the C line x = array[i];, where array becomes %rdi and i becomes %rsi and x becomes %rbp. The 8 is the length of the data type contained in the array.
Now consider similar code that uses lea:
leaq (%rdi, %rsi, 8), %rbp
Just as the use of movq corresponded to dereferencing, the use of leaq here corresponds to not dereferencing. This line of assembly corresponds to the C line x = &array[i];. Recall that & changes the meaning of array[i] from dereferencing to simply specifying a location. Likewise, the use of leaq changes the meaning of (%rdi, %rsi, 8) from dereferencing to specifying a location.
The semantics of this line of code are as follows: get the value in the register %rdi and the value in the register %rsi. Multiply the latter by 8, and then add it to the former. Place this value into the register %rbp. No load from memory is involved, just arithmetic operations [2].
Note that the only difference between my descriptions of leaq and movq is that movq does a dereference, and leaq doesn't. In fact, to write the leaq description, I basically copy+pasted the description of movq, and then removed "Find the value at this location".
To summarize: movq vs. leaq is tricky because they treat the use of parentheses, as in (%rsi) and (%rdi, %rsi, 8), differently. In movq (and all other instruction except lea), these parentheses denote a genuine dereference, whereas in leaq they do not and are purely convenient syntax.
[1] I've said that when array is an array of long, the expression array[i] loads the value from the address array + i * sizeof(long). This is true, but there's a subtlety that should be addressed. If I write the C code
long x = array[5];
this is not the same as typing
long x = *(array + 5 * sizeof(long));
It seems that it should be based on my previous statements, but it's not.
What's going on is that C pointer addition has a trick to it. Say I have a pointer p pointing to values of type T. The expression p + i does not mean "the position at p plus i bytes". Instead, the expression p + i actually means "the position at p plus i * sizeof(T) bytes".
The convenience of this is that to get "the next value" we just have to write p + 1 instead of p + 1 * sizeof(T).
This means that the C code long x = array[5]; is actually equivalent to
long x = *(array + 5)
because C will automatically multiply the 5 by sizeof(long).
So in the context of this StackOverflow question, how is this all relevant? It means that when I say "the address array + i * sizeof(long)", I do not mean for "array + i * sizeof(long)" to be interpreted as a C expression. I am doing the multiplication by sizeof(long) myself in order to make my answer more explicit, but understand that due to that, this expression should not be read as C. Just as normal math that uses C syntax.
[2] Side note: because all lea does is arithmetic operations, its arguments don't actually have to refer to valid addresses. For this reason, it's often used to perform pure arithmetic on values that may not be intended to be dereferenced. For instance, cc with -O2 optimization translates
long f(long x) {
return x * 5;
}
into the following (irrelevant lines removed):
f:
leaq (%rdi, %rdi, 4), %rax # set %rax to %rdi + %rdi * 4
ret