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1200字范文 > 无限循环python代码_python - 代码遇到意外的无限循环? - SO中文参考 - www.soinside.com...

无限循环python代码_python - 代码遇到意外的无限循环? - SO中文参考 - www.soinside.com...

时间:2023-05-09 18:30:31

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无限循环python代码_python - 代码遇到意外的无限循环? - SO中文参考 - www.soinside.com...

调试(使用ipython):

In [2]: s = 'asdfasdf asdf asdf asd sdfa'

In [4]: def count_words(sentence):

...: count = 0

...: last_position = 0

...:

...: while sentence.find(" ", last_position) != -1:

...: count += 1

...: print(f'count: {count}, position: {last_position}')

...: last_position = sentence.find(" ", last_position)

...: print(f'new position: {last_position}')

...: if count > 4:

...: break

...: return count

...:

In [5]: count_words(s)

count: 1, position: 0

new position: 8

count: 2, position: 8

new position: 8

count: 3, position: 8

new position: 8

count: 4, position: 8

new position: 8

count: 5, position: 8

new position: 8

Out[5]: 5

看看原因:

In [6]: s.find?

Docstring:

S.find(sub[, start[, end]]) -> int

Return the lowest index in S where substring sub is found,

such that sub is contained within S[start:end]. Optional

arguments start and end are interpreted as in slice notation.

Return -1 on failure.

Type: builtin_function_or_method

定影:

In [7]: def count_words(sentence):

...: count = 0

...: last_position = 0

...:

...: while sentence.find(" ", last_position) != -1:

...: count += 1

...: last_position = sentence.find(" ", last_position+1)

...: return count

...:

In [8]: count_words(s)

Out[8]: 5

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