UnboundLocalError: local variable 'a' referenced before assignment

（1）下面这种情况是不会报错的：

>>> x = 10
>>> def bar():
...     print(x)
>>> bar()
10

（2）但是这种情况就会报UnboundLocalError的错误：

>>> x = 10
>>> def foo():
...     print(x)
...     x += 1

该代码会提示UnboundLocalError: local variable ‘x’ referenced before assignment。

Python FAQ给出的回答是:

This is because when you make an assignment to a variable in a scope, that variable becomes local to that scope and shadows any similarly named variable in the outer scope. Since the last statement in foo assigns a new value to x, the compiler recognizes it as a local variable. Consequently when the earlier print(x) attempts to print the uninitialized local variable and an error results.

即如果给一个变量分配一个值，则这个变量会被认为属于当前的代码块，会屏蔽外部代码块（全局）的相同名字的变量。因为（2）中的代码中x+=1，相当于给x分配了一个值，则编译器意识到x是一个局部变量，因为x在print后面被赋值，则在之前调用的print(x)试图访问一个未被初始化的局部变量而失败。

（3）解决方法：将变量声明为global的

>>> x = 10
>>> def foobar():
...     **global x**
...     print(x)
...     x += 1
>>> foobar()
10

（4）在嵌套的代码块中用nonlocal 关键字

>>> def foo():
...    x = 10
...    def bar():
...        global x   #nonlocal x
...        print(x)
...        x += 1
...    bar()
...    print(x)
>>> foo()
NameError: name 'x' is not defined

这是因为在全局的区域没有定义x变量，x仅仅出现在上一级区域中，因此可以用nonlocal来替换掉global，则x是上一级区域的变量。