Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
612 views
in Technique[技术] by (71.8m points)

Python: 'break' outside loop

in the following python code:

narg=len(sys.argv)
print "@length arg= ", narg
if narg == 1:
        print "@Usage: input_filename nelements nintervals"
        break

I get:

SyntaxError: 'break' outside loop

Why?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Because break cannot be used to break out of an if - it can only break out of loops. That's the way Python (and most other languages) are specified to behave.

What are you trying to do? Perhaps you should use sys.exit() or return instead?


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...