Revisions to Fast ceiling of an integer division in C / C++

added 4 characters in body

edited Oct 25, 2021 at 12:05

There's a solution for both positive and negative x but only for positive y with just 1 division and without branches:

int ceildiv_ceil(int x, int y) { return x / y + (x % y > 0); }

Note, if x is positive then division is towards zero, and we should add 1 if reminder is not zero.

If x is negative then division is towards zero, that's what we need, and we will not add anything because x % y is not positive

There's a solution for both positive and negative x but only for positive y with just 1 division and without branches:

int ceil(int x, int y) { return x / y + (x % y > 0); }

Note, if x is positive then division is towards zero, and we should add 1 if reminder is not zero.

If x is negative then division is towards zero, that's what we need, and we will not add anything because x % y is not positive

There's a solution for both positive and negative x but only for positive y with just 1 division and without branches:

int div_ceil(int x, int y) { return x / y + (x % y > 0); }

Note, if x is positive then division is towards zero, and we should add 1 if reminder is not zero.

If x is negative then division is towards zero, that's what we need, and we will not add anything because x % y is not positive

added 18 characters in body

edited Jun 13, 2015 at 23:26

There's a solution for both positive and negative x andbut only for positive y with just 1 division and without branches:

int ceil(int x, int y) { return x / y + (x % y > 0); }

Note, if x is positive then division is towards zero, and we should add 1 if reminder is not zero.

If x is negative then division is towards zero, that's what we need, and we will not add anything because x % y is not positive

There's a solution for both positive and negative x and y with just 1 division and without branches:

int ceil(int x, int y) { return x / y + (x % y > 0); }

Note, if x is positive then division is towards zero, and we should add 1 if reminder is not zero.

If x is negative then division is towards zero, that's what we need, and we will not add anything because x % y is not positive

There's a solution for both positive and negative x but only for positive y with just 1 division and without branches:

int ceil(int x, int y) { return x / y + (x % y > 0); }

Note, if x is positive then division is towards zero, and we should add 1 if reminder is not zero.

If x is negative then division is towards zero, that's what we need, and we will not add anything because x % y is not positive

answered Jun 13, 2015 at 23:06

There's a solution for both positive and negative x and y with just 1 division and without branches:

int ceil(int x, int y) { return x / y + (x % y > 0); }

Note, if x is positive then division is towards zero, and we should add 1 if reminder is not zero.

If x is negative then division is towards zero, that's what we need, and we will not add anything because x % y is not positive

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