Common local variable ' x ' referenced before assignment problem

Def fun1 ():
    x = 5
    def fun2 ():
        x *= 2 return
        x return
    fun2 ()

As above code, call FUN1 ()

Run Error: unboundlocalerror:local variable ' x ' referenced before assignment.

This is because for the FUN1 function, x is a local variable, and for the FUN2 function, X is a non global external variable. When the X is modified in fun2, X is treated as a fun2 local variable, the definition of x in FUN1 is masked, and if X is read only in fun2, this error does not occur.

Workaround: Use the nonlocal keyword

Def fun1 ():
    x = 5
    def fun2 ():
        nonlocal x
        x *= 2 return
        x return
    fun2
() fun1 () out[14 ]: 10

With nonlocal x, X is no longer considered an internal variable of fun2 in Fun2 (), and the definition of x in the FUN1 function is not blocked.