python - 令人困惑的范围变化 - 发生了什么事？

Question

def Test(value):
    def innerFunc():
        print value
    innerFunc()

def TestWithAssignment(value):
    def innerFunc():
        print value
        value = "Changed value"
    innerFunc()

Test("Hello 1")
# Prints "Hello 1"

TestWithAssignment("Hello 2")
# Throws UnboundLocalError: local variable 'value' referenced before assignment
# on the "print value" line

为什么第二个失败，因为唯一的区别是在打印语句之后的赋值？我对此很困惑。

Answer 1

问题在于 Python 希望您是显式的，而您希望是隐式的。 execution model Python 使用将名称绑定(bind)到最近的可用 封闭 范围。

def Test(value):
    # Local Scope #1
    def innerFunc():
        # Local Scope #2
        print value 
        # No immediate local in Local Scope #2 - check up the chain
        # First find value in outer function scope (Local Scope #1).
        # Use that.
    innerFunc()

def TestWithAssignment(value):
    # Local Scope #1
    def innerFunc():
        # Local Scope #2
        print value 
        # Immediate local variable found in Local Scope #2.
        # No need to check up the chain.
        # However, no value has been assigned to this variable yet.
        # Throw an error.
        value = "Changed value"
    innerFunc()

(据我所知)没有一种方法可以在 Python 2.x 中遍历范围 - 你有 globals() 和 locals() -但是全局和本地范围之间的任何范围都无法访问(如果这不是真的，我希望得到纠正)。

但是，您可以将局部变量 value 传递到您的内部局部作用域中:

def TestWithAssignment(value):
    def innerFunc(value):
        print value 
        # Immediate local **and assigned now**.
        value = "Changed value"
        # If you need to keep the changed value
        # return value
    innerFunc(value)
    # If you need to keep the changed value use:
    # value = innerFunc(value)

在 Python 3 中你有新的 nonlocal可用于引用包含范围的语句(感谢@Thomas K)。

def TestWithAssignment(value):
    def innerFunc():
        nonlocal value
        print value 
        value = "Changed value"
    innerFunc()