How to round up the result of integer division?
I'm thinking in particular of how to display pagination controls, when using a language such as C# or Java.
If I have x items which I want to display in chunks of y per page, how many pages will be needed?
Found an elegant solution:
int pageCount = (records + recordsPerPage - 1) / recordsPerPage;
Source: Number Conversion, Roland Backhouse, 2001
Converting to floating point and back seems like a huge waste of time at the CPU level.
Ian Nelson's solution:
int pageCount = (records + recordsPerPage - 1) / recordsPerPage;
Can be simplified to:
int pageCount = (records - 1) / recordsPerPage + 1;
AFAICS, this doesn't have the overflow bug that Brandon DuRette pointed out, and because it only uses it once, you don't need to store the recordsPerPage specially if it comes from an expensive function to fetch the value from a config file or something.
Ie this might be inefficient, if config.fetch_value used a database lookup or something:
int pageCount = (records + config.fetch_value('records per page') - 1) / config.fetch_value('records per page');
This creates a variable you don't really need, which probably has (minor) memory implications and is just too much typing:
int recordsPerPage = config.fetch_value('records per page')
int pageCount = (records + recordsPerPage - 1) / recordsPerPage;
This is all one line, and only fetches the data once:
int pageCount = (records - 1) / config.fetch_value('records per page') + 1;
This should give you what you want. You will definitely want x items divided by y items per page, the problem is when uneven numbers come up, so if there is a partial page we also want to add one page.
int x = number_of_items;
int y = items_per_page;
// with out library
int pages = x/y + (x % y > 0 ? 1 : 0)
// with library
int pages = (int)Math.Ceiling((double)x / (double)y);
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