有一些问题：
- 当计数不除以x时，为什么要打印计数？这并不意味着它是素数，仅意味着该特定x不会将其除
continue
移至下一个循环迭代-但你确实想使用停止它break
这是你的代码，其中包含一些修复程序，它仅输出质数：
import mathdef 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
有关更有效的质子生成，请参见其他人的建议，参见戊二烯筛。这是一个不错的，经过优化的实现，带有很多注释：
# Sieve of Eratosthenes# Code by David Eppstein, UC Irvine, 28 Feb 2002# http://pre.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。
