python "send"方法不会更改"next"的值？

Question

我正在尝试生成器的发送函数，我期望发送会改变正在生成的值，所以我在 ipython 中尝试了：

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

好吧，我想我对"发送"的真正作用有错误的理解，如何纠正它？

Answer 1

你产生了i但你没有将语句yield返回值分配给它。如果分配返回值，您将看到预期的输出：

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，此时值将返回给调用方。这就是为什么第一次调用时看不到value is:文本的原因。第二次调用send或next时，将从yield恢复执行。如果调用send则由yield语句返回给定的参数，否则yield返回None。