python - 生成器方法 throw 的返回值

Question

throw 的帮助说:

builtins.generator 实例的 throw(...) 方法 throw(typ[,val[,tb]]) -> 在生成器中引发异常， 返回下一个产生的值或引发 StopIteration

“返回下一个产生的值”这是我的问题!

让我们看一下这个例子:

def gen():
    counter = 1
    while True:
        try:
            yield counter
            counter += 1
        except Exception as e:
            print("Exception and tpye: ", e, type(e))
            

x = gen()
print("Result of first next call: ", next(x))
res = x.throw(ValueError("Whatever!"))
print("result of throw call: ", res)
print("Result of first next call: ", next(x))

输出看起来像这样:

Result of first next call:  1
Exception and tpye:  Whatever! <class 'ValueError'>
result of throw call:  1
Result of first next call:  2

throw的返回值不应该是2而不是1吗？帮助文件说“return NEXT yielded value2”不是“return again previously yielded value”

我担心我完全错了？

Answer 1

当您使用 gen.throw(...) 时，异常会在 yield 语句中引发。

当您捕获它时，不会执行以下计数器增量。您可以将它移到 finally: 子句中，以确保无论如何，您始终会增加 counter:

def gen():
    counter = 1
    while True:
        try:
            yield counter
        except Exception as e:
            print("Exception and type: ", e, type(e))
        finally:
            counter += 1