Python闭包技巧：函数嵌套与状态保持解析

def make_adder(n):
    def add(x):
        return x + n  # 引用了外部变量 n
    return add  # 返回函数，不是 add(x)
<p>plus_5 = make_adder(5)
print(plus_5(3))  # 输出 8</p>

def counter():
    count = 0
    def inc():
        nonlocal count
        count += 1
        return count
    return inc
<p>c = counter()
print(c())  # 1
print(c())  # 2</p>

funcs = []
for i in range(3):
    funcs.append(lambda: i)  # 全部引用同一个 i
print([f() for f in funcs])  # [2, 2, 2]

funcs = []
for i in range(3):
    funcs.append(lambda x=i: x)  # x=i 在定义时就固定了 i 的当前值
print([f() for f in funcs])  # [0, 1, 2]

def make_func(x):
    return lambda: x
funcs = [make_func(i) for i in range(3)]

def retry(max_attempts=3):
    def decorator(func):
        def wrapper(*args, **kwargs):
            for i in range(max_attempts):
                try:
                    return func(*args, **kwargs)
                except Exception:
                    if i == max_attempts - 1:
                        raise
            return None
        return wrapper
    return decorator
<p>@retry(max_attempts=2)
def fetch_data():</p><h1>模拟可能失败的操作</h1><pre class="brush:php;toolbar:false"><code>pass</code>