[shakow]

That's what I would have thought as well, but looks like that on x86, both clang and gcc use variations of LEA. But if they're doing it this way, I'm pretty sure it must be faster, because even if you change the ×4 for a <<2, it will still generate a LEA.

https://godbolt.org/z/EKj58dx9T

[shaggie76]

Not only is LEA more flexible I believe it's preferred to SHL even for simple operations because it doesn't modify the flags register which can make it easier to schedule.

[fweimer]

It's more about the non-destructive destination part, which can avoid a move. Compilers tend to prefer SHL/SAL of LEA because its encoding is shorter: https://godbolt.org/z/9Tsq3hKnY

[Cold_Miserable]

shlx doesn't alter the flag register.

[fweimer]

SHLX does not support an immediate operand. Non-destructive shifts with immediate operands only arrive with APX, where they are among the most commonly used instructions (besides paired pushes/pops).

[adrian_b]

They use LEA for multiplying with small constants up to 9 (not only with powers of two, but also with 3, 5 and 9; even more values could be achieved with two LEA, but it may not be worthwhile).

For multiplying with powers of two greater or equal to 16, they use shift left, because LEA can no longer be used.

[Validark]

Using an lea is better when you want to put the result in a different register than the source and/or you don't want to modify the flags registers. shlx also avoids modifying flags, but you can't shift by an immediate, so you need to load the constant into a register beforehand. In terms of speed, all these options are basically equivalent, although with very slightly different costs to instruction caches and the register renaming in the scheduler. In terms of execution, a shift is always 1 cycle on modern hardware.