问题
Let we have this code:
def big_function():
def little_function():
.......
.........
The Python documentation says about def
statement:
A function definition is an executable statement. Its execution binds the function name...
So, the question is:
Does def little_function()
execute every time when big_function
is invoked?
Question is about def
statement exactly, not the little_function()
body.
回答1:
You can check the bytecode with the dis
module:
>>> import dis
>>> def my_function():
... def little_function():
... print "Hello, World!"
...
...
>>> dis.dis(my_function)
2 0 LOAD_CONST 1 (<code object little_function at 0xb74ef9f8, file "<stdin>", line 2>)
3 MAKE_FUNCTION 0
6 STORE_FAST 0 (little_function)
9 LOAD_CONST 0 (None)
12 RETURN_VALUE
As you can see the code for the inner function is compiled only once. Every time you call my_function
it is loaded and a new function object is created(in this sense the def little_function
is executed every time my_function
is called), but this doesn't add much overhead.
回答2:
The code in the inner function is compiled only once, so there shouldn't be a significant runtime penalty.
Only inner's function closure gets updated each time the outer function is called. See here, for example, for more details about closures.
Here's a quick demonstration, examining the closure:
def f(x):
a = []
b = x + 1
def g():
print a, b
return g
In [28]: y = f(5)
In [29]: y
Out[29]: <function __main__.g>
In [30]: y.func_closure
Out[30]:
(<cell at 0x101c95948: list object at 0x101c3a3f8>,
<cell at 0x101c958a0: int object at 0x100311aa0>)
In [31]: y.func_closure[1].cell_contents
Out[31]: 6
来源:https://stackoverflow.com/questions/16031092/function-inside-function-every-time