哪个更快找到偶数if(n％2 == 0)或if(n& 1 == 0)? [英] which is faster to find the even number if(n%2==0) or if(n&1==0)?

问题描述

以更快的速度找到偶数if(n％2 == 0)或if(n& 1 == 0)吗?

def isEven(n):
    if n%2 == 0:
        return True
    return False
OR

def isEven(n):
    if n&1 == 1:
        return False
    return True

推荐答案

这看起来像Python，所以我在Python(3.6)中尝试过:

This looks like Python, so I tried it in Python(3.6):

from timeit import timeit
a = timeit(stmt='for i in range(100): i%2')
b = timeit(stmt='for i in range(100): i&1')
print(a, b)

时间变化很大(谢谢，垃圾收集器！)，但是总的来说，这让我获得 i％2 大约4.7秒，获得 i& 1 大约6.3秒.我想这可能不是您期望的答案.

Times vary somewhat wildly (thanks, garbage collector!) but in general this gets me around 4.7 seconds for i%2 and 6.3 seconds for i&1, which I would guess is probably not the answer you would expect.

我使用dis检查了字节码，唯一的区别是运行BINARY_MODULO与BINARY_AND的一行，所以我不确定为什么会有如此大的时间差异.

I checked the bytecode using dis, and the only difference was the one line running BINARY_MODULO vs BINARY_AND, so I'm not sure why there's such a huge time discrepancy.

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