【debug】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

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