Why is calling eval so slow?

[Serhiy Storchaka]

05.05.18 19:10, Steven D'Aprano пише:
> # calling a regular function
> python3.5 -m timeit -s "f = lambda: 99" "f()"
> # evaluating a function code object
> python3.5 -m timeit -s "f = (lambda: 99).__code__" "eval(f)"
> # estimate the overhead of the eval name lookup
> python3.5 -m timeit "eval"
> # evaluating a pre-compiled byte-code object
> python3.5 -m timeit -s "f = compile('99', '', 'eval')" "eval(f)"
> # evaluating a string
> python3.5 -m timeit "eval('99')"
> 
> 
> And the results on my computer:
> 
> # call a regular function
> 1000000 loops, best of 3: 0.245 usec per loop
> # evaluate the function __code__ object
> 1000000 loops, best of 3: 1.16 usec per loop
> # overhead of looking up "eval"
> 10000000 loops, best of 3: 0.11 usec per loop
> 
> 
> which means it takes four times longer to execute the code object alone,
> compared to calling the function object which executes the code object.
> 
> Why so slow?

1. Add an overhead of calling eval(), not just looking up its name. It 
is comparable with a time of calling a simple lambda.
2. Add an overhead of getting the globals dict and creating a locals dict.

$ python3.6 -m timeit -s "f = lambda: 99" "f()"
10000000 loops, best of 3: 0.0658 usec per loop
$ python3.6 -m timeit -s "f = (lambda: 99).__code__" "eval(f)"
1000000 loops, best of 3: 0.282 usec per loop
$ python3.6 -m timeit "eval"
10000000 loops, best of 3: 0.0226 usec per loop
$ python3.6 -m timeit -s "f = (lambda: 99).__code__; g = {}" "eval(f, g)"
10000000 loops, best of 3: 0.165 usec per loop

0.0658*2 + 0.0226 = 0.1542. It is comparable with 0.165.