python文档中关于默认参数是这样说的:
Default parameter values are evaluated when the function definition is executed. This means that the expression is evaluated once, when the function is defined, and that the same “pre-computed” value is used for each call.
因此,如下代码:
def function(data=[]):
print id(data), data
data.append(1)
for i in range(3):
function()
输出结果为:
4462873272 []
4462873272 [1]
4462873272 [1, 1]
一切正像文档中说的那样,data只计算一次,之后的调用返回的是已经计算出来的值(因此,id(data)没有变化,每次append均针对同一个data)。
但是我们带参数调用function时,即:
for i in range(3):
function([])
输出结果为:
4462872984 []
4462872984 []
4462872984 []
data的值是符合预期的,但是三次调用function,id(data)的值竟然也一样,这个是为什么呢?带参数传递时,每次调用都应该是新建data的吧。
Read the documentation carefully: https://docs.python.org/2/library/functions.html#id.
Because:
The implementation of id in CPython is conceptually equivalent to the address of the object in memory. Then please look at this code:
There are two different list objects before and after, but since the former object is destroyed after use, the latter list may reuse the same memory space, so the address returned by the id is the same.
LZ please pay attention to the difference between constants and variables. If you want a different id, you have to use list().
I just learned python a while ago. I saw a part about the default parameter list and wanted to share my understanding.
When calling function f() three times in a row,
The first time f() will find the default parameter data = []
And its address in memory can be addressed through something like 'f.data'.
Next, calling f() again will repeat the above operation. Because no new parameter is entered, the previous data will be found as the parameter list by default.
f()
f()
In addition, you can also call the f.defaults attribute to see what the default parameters of the current situation of function f are
For example, after defining f,
So the default parameters are variable during the execution of f() several times.
And if you call f([]), it will not affect the default parameters,
You can try it if you are interested
f(data = [] ) iterates :)
You can refer to here. I think it is very clear about the parameters of functions in Python
The question is like this...
This is considered a relatively advanced content of Python..Beginners will basically encounter this problem
Okay, let’s start the text
The introspection of the function results in that your parameter is a attribute of the function type (although there is no way to access it with
.
).This means that when you define a function, your parameters have become part of the function object. Similar to:
You can find that the
"a"
just printed out..The same is true here for the function. The defined parameter attributes are shared
But the above rules are only for variable data types, such as lists
Because the immutable data type is not modified in the current memory block... but points to another memory block..
Extended reading:
Two python features:
1. python dynamic type problem:
Dynamic type means: your variable is just a something similar to a pointer. It’s just that he can point at random... what he points to is a type. This The type can be integer, string, class, etc.
One of your variables can change the type it points to. For example:
a = 1
ina
points to an integer type. The value of this integer type is1
(of course there are other components, such as There are also count and type declarations, etc.)But when you re-point to , for example:
a = "hello world"
, what happens is that the variablea
points to a string type, and the change to the integer type of1
is count-1, the memory will not be reclaimed until count reaches 0.This is the legendary dynamic type
2. Introspection of functions
Functions are also types in python.
For example, you
Then a function type is generated and assigned to the variable foo..
And everything in this type can be operated through the domain operator
.
.Agree ls