3 ms·
ah sorry. i did import dis dis.dis("a = foo.bar(b)") which gave 1 0 LOAD_NAME 0 (foo) 2 LOAD_ATTR
by PyComfy 9y ago
ah sorry. i did
import dis
dis.dis("a = foo.bar(b)")
which gave
1 0 LOAD_NAME 0 (foo)
2 LOAD_ATTR 1 (bar)
4 LOAD_NAME 2 (b)
6 CALL_FUNCTION 1
8 STORE_NAME 3 (a)
10 LOAD_CONST 0 (None)
12 RETURN_VALUE
- tom_mellior 9y agoAh, OK. Yes, those LOAD_NAMEs are slower than LOAD_FAST. If you put the code into a function, you get this: >>> def f(foo, b): ... a = foo.bar(b) ... >>> dis.dis(f) 2 0 LOAD_FAST 0 (foo) 3 LOAD_ATTR 0 (bar) 6 LOAD_FAST 1 (b) 9 CALL_FUNCTION 1 (1 positional, 0 keyword pair) 12 STORE_FAST 2 (a) 15 LOAD_CONST 0 (None) 18 RETURN_VALUE LOAD_FAST is the normal case for locals inside a function. Not sure off the top of my head where LOAD_NAME would be generated in normal usage (i.e. where you don't evaluate code from a string). Edit: Also, I'm talking about Python 3. Maybe you aren't.