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Given integer values x and y, C and C++ both return as the quotient q = x/y the floor of the floating point equivalent. I'm interested in a method of returning the ceiling instead. For example, ceil(10/5)=2 and ceil(11/5)=3.
The obvious approach involves something like:
q = x / y; if (q * y < x) ++q; This requires an extra comparison and multiplication; and other methods I've seen (used in fact) involve casting as a float or double. Is there a more direct method that avoids the additional multiplication (or a second division) and branch, and that also avoids casting as a floating point number?
985
Using simple maths, we can add the denominator to the numerator and subtract 1 from it and then divide it by denominator to get the ceiling value.
Integer division yields an integer result. For example, the expression 7 / 4 evaluates to 1 and the expression 17 / 5 evaluates to 3. C provides the remainder operator, %, which yields the remainder after integer division.
C library function - ceil() The C library function double ceil(double x) returns the smallest integer value greater than or equal to x.
For positive numbers
unsigned int x, y, q; To round up ...
q = (x + y - 1) / y; or (avoiding overflow in x+y)
q = 1 + ((x - 1) / y); // if x != 0 If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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