Python 程序如何使用break语句进行控制

有人能解释一下这个程序和它的输出吗？我对if声明有疑问。我无法理解break语句在以下情况下是如何工作的：

for n in range(2, 10):
    for x in range(2, n):
        if n % x == 0:
            print n, 'equals', x, '*', n/x
            break
    else:
        # loop fell through without finding a factor
        print n, 'is a prime number'

输出：

2 is a prime number
3 is a prime number
4 equals 2 * 2
5 is a prime number
6 equals 2 * 3
7 is a prime number
8 equals 2 * 4
9 equals 3 * 3

---

break

语句离开循环而不输入

else

子句。如果循环终止时未到达

中断

，则将输入

else

子句。换句话说，循环搜索一个可能的除数；如果找到一个，它将打印它并使用

break

离开循环。如果未找到除数，for循环将“正常”终止，并因此进入

else

子句（然后在该子句中打印已找到素数）。

我将添加一些注释：

for n in range(2, 10): #Loops from 2 to 9, inclusive. Call this Loop A.
    for x in range(2, n): #Loops from 2 to n-1, inclusive. Call this Loop B.
        if n % x == 0: #If n is divisible by x, execute the indented code
            print n, 'equals', x, '*', n/x #print the discovered factorization
            break #Break out of loop B, skipping the "else" statement
    else: #If the loop terminates naturally (without a break) this will be executed
        # loop fell through without finding a factor
        print n, 'is a prime number' 

---

显然，这个程序试图识别素数。一个素数，除了1（显然！）和它本身之外，没有因子（即，当你把一个素数除以x时，总是有一个余数）。因此，我们需要在测试之前测试从2（即不是1）到数字的每个数字，看看它是否是测试数字的一个因子

正在运行的测试的步骤如下：

# S1 is a set of numbers, and we want to identify the prime numbers within it.
S1 = [2, 3, 4, 5, 6, 7, 8, 9]

# test a whether n is PRIME:
for n in S1:
    # if n / x has no remainder, then it is not prime
    for x in range(2, n):
        if...
            I have NO REMAINDER, then x is a factor of n, and n is not prime
            -----> can "BREAK" out of test, because n is clearly not PRIME
            --> move on to next n, and test it
        else:
            test next x against n
            if we find NO FACTORS, then n is PRIME

---

Break直接离开最内部的循环，进入外部for循环的下一步。

请格式化代码并输出。发布真实的代码，但没有…谢谢。这真的帮助我了解发生了什么。