Python 3：我的修饰函数运行两次

我有一些简单的斐波那契序列函数，我正在使用Travis CI/Docker进行单元测试和构建：

fib_recursive.py：

from fib.fib import benchmark, fib_rec_memo

@benchmark
def print_fib(n):
    for x in range(0, n):
        print(fib_rec_memo(x))

print_fib(100)

以下是fib.fib导入源代码：

from time import time
from functools import wraps

def benchmark(func):
    @wraps(func)
    def wrapper(*args, **kwargs):
        t = time()
        func(*args, **kwargs)
        print(func.__name__, 'took:', time() - t)
        return func(*args, **kwargs)
    return wrapper


def fib_rec_memo(n, hash = {0:1, 1:1}):
    if n not in hash:
        hash[n] = fib_rec_memo(n-1) + fib_rec_memo(n-2)
    return hash[n]


@benchmark
def fib_standard(num):
    a, b = 0, 1
    c = []
    while a < num:            # First iteration:
        c.append(a)            # yield 0 to start with and then
        a, b = b, a + b    # a will now be 1, and b will also be 1, (0 + 1)
    return c

有人知道为什么吗

---

谢谢。

您在

包装器中调用装饰函数两次

函数：

func(*args, **kwargs)  # here ...
print(func.__name__, 'took:', time() - t)
return func(*args, **kwargs)  # ... and here again

通过将结果存储到变量并在计时输出后返回存储的结果，可以避免这种情况：

rval = func(*args, **kwargs)  # call it once and store result ...
print(func.__name__, 'took:', time() - t)
return rval  # ... then return result

“出于某种原因”——因为你真的叫了两次？

rval = func(*args, **kwargs)  # call it once and store result ...
print(func.__name__, 'took:', time() - t)
return rval  # ... then return result