Use Python five points to get the scope method

1. Block-level scope

Think about whether there will be output when running the following program at this time? Will the execution be successful?

#块级作用域
 
if 1 == 1:
    name = "lzl"
 
print(name)
 
 
for i in range(10):
    age = i
 
print(age)

Let’s take a look at the execution results first

C:/Users/L/PycharmProjects/s14/preview/Day8/作用域/main.py
lzl
9

Process finished with exit code 0

The code was executed successfully, no problem; in Java/C# , executing the above code will prompt that name and age are not defined, but it can be executed successfully in Python. This is because there is no block-level scope in Python. The variable# in the code block ##, it can be called externally, so it can run successfully;

2. Local scope

Looking back at the knowledge learned before, when we learn

function, the function is Separate scope, there is no block-level scope in Python, but there is local scope; look at the following code

#local scope

def  func():
    name = "lzl"
 
print(name)

Run this code and think about it. Will there be output?

Traceback (most recent call last):
  File "C:/Users/L/PycharmProjects/s14/preview/Day8/作用域/main.py", line 23, in <module>
    print(name)
NameError: name 'name' is not defined

An error is reported when running. I believe everyone can understand this. The name variable only takes effect inside the func() function, so it cannot be called globally. Make a simple adjustment to the above code and look at it again. What's the result?

#局部作用域
 
def  func():
    name = "lzl"
 
func()          #执行函数
print(name)

Added a code to the previous code. Before the variable name is printed, a function is executed. Will the printing change at this time?

Traceback (most recent call last):
  File "C:/Users/L/PycharmProjects/s14/preview/Day8/作用域/main.py", line 23, in <module>
    print(name)
NameError: name 'name' is not defined

The execution still reports an error, let’s go back to the sentence just now: even if the function is executed, the scope of name is only inside the function, and it still cannot be called from outside; remember the first two knowledge points, Next, let’s start the amplification trick

3. Scope chain

Adjust the function and see what the execution result of the following code is?

#作用域链
 
name = "lzl"
def f1():
    name = "Eric"
    def f2():
        name = "Snor"
        print(name)
    f2()
f1()

Having learned functions, you must know that Snor will be output after f1() is executed; let’s remember a concept first. There is a scope chain in Python. Variables will be found from the inside to the outside. Go to your own scope first. I didn’t go to the superiors to look for it until I couldn’t find it and reported an error

4. Ultimate Edition Scope

Okay, enough of the foreshadowing, the Ultimate Edition is here~~

#终极版作用域
 
name = "lzl"
 
def f1():
    print(name)
 
def f2():
    name = "eric"
    f1()
 
f2()

Think about whether the final execution result of f2() is to print "lzl" or "eric"? Remember your answer. Instead of posting the answer now, take a look at the following code:

#终极版作用域
 
name = "lzl"
 
def f1():
    print(name)
 
def f2():
    name = "eric"
    return f1
 
ret = f2()
ret()
 
#输出：lzl

The execution result is "lzl". Analyze the above code. The execution result of f2() is function f1. The memory address, that is, ret=f1; executing ret() is equivalent to executing f1(). When executing f1(), it has nothing to do with f2(). name="lzl" and f1() are in the same scope chain, inside the function If there is no variable, it will look outside, so the value of the variable name is "lzl" at this time; if you understand this, then you also know the answer to the ultimate code that you did not give the answer just now

#终极版作用域
 
name = "lzl"
 
def f1():
    print(name)
 
def f2():
    name = "eric"
    f1()
 
f2()
 
# 输出：lzl

Yes, the output It is "lzl". Remember that before the function is executed, the scope has been formed and the scope chain has also been generated.

5. Sina interview question

li = [lambda :x for x in range(10)]

Determine the type of li? What type are the elements in li?

print(type(li))
print(type(li[0]))
 
# <class &#39;list&#39;>
# <class &#39;function&#39;>

You can see that li is a list type and the elements in the list are functions. Then print the return value of the first element in the list. What is the return value at this time?

#lambada 面试题
 
li = [lambda :x for x in range(10)]
 
res = li[0]()
print(res)
 
#输出：9

liThe return value

of the first function is 9, not 0. Remember: the internal code will not be executed before the function is executed; you can practice the code in the blog by yourself, Deepen your impression

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