Login | Register   
RSS Feed
Download our iPhone app
Browse DevX
Sign up for e-mail newsletters from DevX

By submitting your information, you agree that devx.com may send you DevX offers via email, phone and text message, as well as email offers about other products and services that DevX believes may be of interest to you. DevX will process your information in accordance with the Quinstreet Privacy Policy.

Tip of the Day
Language: Web
Expertise: All
Apr 18, 2000



Application Security Testing: An Integral Part of DevOps

A Smart Ceiling for a Ratio of Two Integers

I often use the following trick in my code.

Let "a" and "b" be real positive numbers. You want to find the smallest integer greater than or equal to "a/b". Perhaps your code looks like:
#include <math.h>
double a, b;
long c;
c = (long) ceil( a / b );
Now, the question remains the same, but your numbers "a" and "b" happen to be integers. You can write:
#include <math.h>
long a, b, c;
c = (long) ceil( (double)a/(double)b );
However, there is a more elegant and efficient solution:
long a, b, c;
c = (a - 1) / b + 1;
Eugene Zak
Comment and Contribute






(Maximum characters: 1200). You have 1200 characters left.



We have made updates to our Privacy Policy to reflect the implementation of the General Data Protection Regulation.
Thanks for your registration, follow us on our social networks to keep up-to-date