Java rounding up to an int using Math.ceil

Question

int total = (int) Math.ceil(157/32);

Why does it still return 4? 157/32 = 4.90625, I need to round up, I\'ve looked around and this se

Answer 1

In Java adding a .0 will make it a double...

int total = (int) Math.ceil(157.0 / 32.0);

Answer 2

int total = (157-1)/32 + 1

or more general

(a-1)/b +1 

Answer 3

Also to convert a number from integer to real number you can add a dot:

int total = (int) Math.ceil(157/32.);

And the result of (157/32.) will be real too. ;)

Answer 4

int total = (int) Math.ceil( (double)157/ (double) 32);

Answer 5

Nobody has mentioned the most intuitive:

int x = (int) Math.round(Math.ceil((double) 157 / 32));

This solution fixes the double division imprecision.

Answer 6

Java provides only floor division / by default. But we can write ceiling in terms of floor. Let's see:

Any integer y can be written with the form y == q*k+r. According to the definition of floor division (here floor) which rounds off r,

floor(q*k+r, k) == q  , where 0 ≤ r ≤ k-1

and of ceiling division (here ceil) which rounds up r₁,

ceil(q*k+r₁, k) == q+1  , where 1 ≤ r₁ ≤ k

where we can substitute r+1 for r₁:

ceil(q*k+r+1, k) == q+1  , where 0 ≤ r ≤ k-1

Then we substitute the first equation into the third for q getting

ceil(q*k+r+1, k) == floor(q*k+r, k) + 1  , where 0 ≤ r ≤ k-1

Finally, given any integer y where y = q*k+r+1 for some q,k,r, we have

ceil(y, k) == floor(y-1, k) + 1

And we are done. Hope this helps.