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？我对此非常困惑。

Answer 1

问题是Python希望你是明确的，你想要隐含。 Python使用的execution model将名称绑定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()

没有（据我所知）在Python 2.x中使用范围的方法 - 你有globals()和locals() - 但是全局范围和本地范围之间的任何范围无法访问（如果不是这样，我爱需要更正）。

但是，您可以将局部变量value传递到内部本地范围：

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，您有了新的nonlocal语句，可用于引用包含范围（Thanks @Thomas K）。

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

Answer 2

这应该解决它：

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

令人困惑的范围变化 - 发生了什么？

2 个答案: