Python's five points to fix the scope

1. Block-level scope

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

2. Local scope

Review the knowledge learned before, When we learn functions, functions are separate scopes. 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 any 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 &#39;name&#39; is not defined

Running error, 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 then look What's the result?

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

Add a line of code to the previous code. Before printing the variable name, execute the function. 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 &#39;name&#39; is not defined

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; put the first two Remember the knowledge points, and you will start to expand your skills next time

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, first go to your own scope I didn’t go looking for it until I couldn’t find it and reported an error

4. Ultimate version scope

Okay, enough of the foreshadowing, the ultimate version is here~~

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

​

Think about it In the end, does the execution result of f2() 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 of 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 at this time is "lzl"; if you understand this, then you also know the answer to the ultimate code that the answer was not given 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 questions

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 to deepen your impression