Python 中的简单素数生成器

有人可以告诉我这段代码做错了什么吗？无论如何，它只是打印“计数”。我只想要一个非常简单的素数生成器（没什么花哨的）。

import math

def main():
    count = 3
    one = 1
    while one == 1:
        for x in range(2, int(math.sqrt(count) + 1)):
            if count % x == 0: 
                continue
            if count % x != 0:
                print count

        count += 1

有一些问题：

• 为什么在未除以 x 时打印出 count ？这并不意味着它是素数，它只是意味着这个特定的 x 不能整除它
• continue移动到下一个循环迭代 - 但你真的想使用停止它break

这是经过一些修复的代码，它仅打印出素数：

import math

def main():
    count = 3
    
    while True:
        isprime = True
        
        for x in range(2, int(math.sqrt(count) + 1)):
            if count % x == 0: 
                isprime = False
                break
        
        if isprime:
            print count
        
        count += 1

对于更有效的素数生成，请参阅埃拉托斯特尼筛法，正如其他人建议的那样。这是一个很好的、优化的实现，有很多评论：

# Sieve of Eratosthenes
# Code by David Eppstein, UC Irvine, 28 Feb 2002
# http://code.activestate.com/recipes/117119/

def gen_primes():
    """ Generate an infinite sequence of prime numbers.
    """
    # Maps composites to primes witnessing their compositeness.
    # This is memory efficient, as the sieve is not "run forward"
    # indefinitely, but only as long as required by the current
    # number being tested.
    #
    D = {}
    
    # The running integer that's checked for primeness
    q = 2
    
    while True:
        if q not in D:
            # q is a new prime.
            # Yield it and mark its first multiple that isn't
            # already marked in previous iterations
            # 
            yield q
            D[q * q] = [q]
        else:
            # q is composite. D[q] is the list of primes that
            # divide it. Since we've reached q, we no longer
            # need it in the map, but we'll mark the next 
            # multiples of its witnesses to prepare for larger
            # numbers
            # 
            for p in D[q]:
                D.setdefault(p + q, []).append(p)
            del D[q]
        
        q += 1

请注意，它返回一个生成器。