Python 流控爆发混乱

Question

我试图理解一个复杂的流程控制，但我无法开始工作，所以简化了它，但也不理解它的工作原理。

简化的流量控制为：

smurf = True
print('smurf status: ', smurf)
jackets = list(range(5))
print(jackets)

for j in jackets:
    print('la la la la', j, '\n')
    while smurf is True:
        print('jacket is :', j)
        if j == jackets[-3]:
            smurf = False
            break
            print('smurf is FALSIOOOO')
        j += 1
        print('smurf with jacket ',j-1, ' be ok')
print('END OF SMURFGATE')

输出为：

smurf status:  True
[0, 1, 2, 3, 4]
la la la la 0 

jacket is : 0
smurf with jacket  0  be ok
jacket is : 1
smurf with jacket  1  be ok
jacket is : 2
la la la la 1 

la la la la 2 

la la la la 3 

la la la la 4 

END OF SMURFGATE

我认为会发生（但没有发生）的是，一旦第 j 次迭代达到测试条件，while 循环将停止，流程控制将跳到脚本的最后一行并打印 'END OF SMURFGATE'。

期望的输出：

smurf status:  True
[0, 1, 2, 3, 4]
la la la la 0 

jacket is : 0
smurf with jacket  0  be ok
la la la la 1
jacket is : 1
smurf with jacket  1  be ok
la la la la 2
jacket is : 2


END OF SMURFGATE

Answer 1

您错过了 for 循环的一个额外中断，现在您只中断了 while 循环。试试这个：

smurf = True
print('smurf status: ', smurf)
jackets = list(range(5))
print(jackets)

for j in jackets:
    print('la la la la', j, '\n')
    while smurf is True:
        print('jacket is :', j)
        if j == jackets[-3]:
            smurf = False
            break
            print('smurf is FALSIOOOO')
        j += 1
        print('smurf with jacket ',j-1, ' be ok')
    break # To escape the for-loop
print('END OF SMURFGATE')

输出：

smurf status:  True
[0, 1, 2, 3, 4]
la la la la 0 

jacket is : 0
smurf with jacket  0  be ok
jacket is : 1
smurf with jacket  1  be ok
jacket is : 2
END OF SMURFGATE

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