python "send"方法不会更改 "next"的值？

Question

我正在尝试生成器的发送功能，我期望发送会改变正在产生的值，所以我在 ipython 中尝试:

In [17]: def z(n):
    ...:     i=0
    ...:     while(i<n):
    ...:         val=yield i
    ...:         print "value is:",val
    ...:         i+=1
    ...:
In [24]: z1=z(10)
In [25]: z1.next()
Out[25]: 0

In [26]: z1.send(5) # I was expecting that after "send", output value will become "5"
value is: 5
Out[26]: 1

In [27]: z1.next()
value is: None # I was expecting that z1.next() will restart from "6" because I sent "5"
Out[27]: 2

嗯，我想我对“发送”的真正作用有错误的理解，如何纠正它？

Answer 1

您正在生成 i 但您没有将 yield 语句的返回值分配给它。如果您分配返回值，您将看到您期望的输出:

def z(n):
    print 'Generator started'
    i=0
    while(i<n):
        val=yield i
        print "value is:",val
        if val is not None:
            i = val
        i+=1

z1=z(10)
print 'Before start'
print z1.next()
print z1.send(5)
print z1.next()

输出:

Before start
Generator started
0
value is: 5
6
value is: None
7

更新:当第一次调用send或next时，生成器从开始执行到第一个yield 语句，此时值返回给调用者。这就是为什么 value is: text is not seen on first call。当 send 或 next 被第二次调用时，执行从 yield 恢复。如果 send 被调用，则 yield 语句返回给它的参数，否则 yield 返回 None。