如何访问python3中的上限值？

Question

在JavaScript中，此代码返回4：

let x = 3; let foo = () => { console.log(x); } let bar = () => { x = 4; foo(); } bar();

Answer 1

要分配给全局x，需要在bar函数中声明global x

Answer 2

很明显，程序、机器正在映射工作

bar()

# in bar function you have x, but python takes it as a private x, not the global one
def bar():
  x = 4
  foo()

# Now when you call foo(), it will take the global x = 3 
# into consideration, and not the private variable [Basic Access Modal]
def foo():
   print(x)

# Hence OUTPUT
# >>> 3

现在，如果要打印4，而不是3，这是全局的，您需要在foo（）中传递私有值，并使foo（）接受一个参数

def bar():
  x = 4
  foo(x)

def foo(args):
   print(args)

# OUTPUT
# >>> 4

或

Use global inside your bar(), so that machine will understand that x inside in bar, is global x, not private

def bar():
  # here machine understands that this is global variabl
  # not the private one
  global x = 4
  foo()

Answer 3

如果变量名在全局范围内定义，并且在函数的局部范围内使用，则会发生两种情况：

• 您正在进行读取操作（例如：简单地打印它），那么变量引用的值与全局对象相同

x = 3

def foo():
  print(x)

foo()

# Here the x in the global scope and foo's scope both point to the same int object

• 您正在执行写入操作（例如：为变量赋值），然后在函数的局部范围内使用它引用创建一个新对象。这不再指向全局对象

x = 3

def bar():
  x = 4

bar()

# Here the x in the global scope and bar's scope points to two different int objects

但是，如果要使用全局范围中的变量并在局部范围内对其进行写操作，则需要将其声明为global

x = 3

def bar():
  global x
  x = 4

bar()

# Both x points to the same int object