Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
I have a 64-bit long that I want to round down to the nearest 10,000, so I am doing a simple:
long myLong = 123456789
long rounded = (myLong / 10000) * 10000; //rounded = 123450000
This appears to do what I expect, but as I’m not 100% on the internals of how integer types get divided, I am just slightly concerned that there may be situations where this doesn’t work as expected.
Will this still work at very large numbers / edge cases?
Yes, it will work, so long as no result, intermediate or otherwise, exceeds
long.MaxValue.To be explicit about your constants you could use the
Lspecifier at the end, e.g.123456789L.For straightforward calculations like this, can I suggest Pex from Microsoft ( http://research.microsoft.com/en-us/projects/pex/ ), which looks for edge cases and tests them. This is a clean-cut example, but if you were building up lots of logic based on things you are unsure of, it’s a great tool.