python "send" 方法不会改变 "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，所以我尝试了 ipython:

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 "send" 的真正作用有错误的理解，如何纠正它？

Answer 1

您产生了 i 但您没有将 yield 语句中的 return 值分配给它。如果您分配 return 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：第一次调用send或next时，生成器从start开始执行到第一个yield语句此时值被 return 发送给调用者。这就是为什么 value is: 文本在第一次调用时看不到的原因。当第二次调用 send 或 next 时，执行从 yield 恢复。如果 send 被调用，则给它的参数由 yield 语句 return 编辑，否则 yield returns None.

python "send" 方法不会改变 "next" 的值？

python "send" method doesn't change the value of "next"?

python

yield

generator

return-value

send