2010/12/28 Lukas Lueg email@example.com
Consider the following code:
def foobar(x): for i in range(5): x[i] = i
The bytecode in python 2.7 is the following:
2 0 SETUP_LOOP 30 (to 33) 3 LOAD_GLOBAL 0 (range) 6 LOAD_CONST 1 (5) 9 CALL_FUNCTION 1 12 GET_ITER
>> 13 FOR_ITER 16 (to 32) 16 STORE_FAST 1 (i)
3 19 LOAD_FAST 1 (i) 22 LOAD_FAST 0 (x) 25 LOAD_FAST 1 (i) 28 STORE_SUBSCR 29 JUMP_ABSOLUTE 13
>> 32 POP_BLOCK >> 33 LOAD_CONST 0 (None) 36 RETURN_VALUE
Can't we optimize the LOAD_FAST in lines 19 and 25 to a single load and put the reference twice on the stack? There is no way that the reference of i might change in between the two lines. Also, the load_fast in lne 22 to reference x could be taken out of the loop as x will always point to the same object....
Yes, you can, but you need:
It's not that simple, and the results aren't guaranteed to be good.
Also, consider that Python, as a dynamic-and-not-statically-compiled language need to find a good trade-off between compilation time and execution.
Just to be clear, a C program is usually compiled once, then executed, so you can spend even hours to better optimize the final binary code.
With a dynamic language, usually the code is compiled and the executed as needed, in "realtime". So it isn't practical neither desirable having to wait too much time before execution begins (the "startup" problem).
Python stays in a "gray area", because modules are usually compiled once (when they are first used), and executed many times, but it isn't the only case.
You cannot assume that optimization techniques used on other (static) languages can be used/ported in Python.