Implementing ceil function without using if-else

Question

I just wanted to know that is there a way of implementing ceil function without using if-else? With if-else (for a/b) it can be implemented as:

if a%b == 0:
    return(a/b)
else:
    return(a//b + 1)

Answer 1

Like this should work if they are integers (I guess you have a rational number representation):

a/b + (a%b!=0)

Otherwise, replace a/b with int(a/b), or, better, as suggested below a//b.

Answer 2

Simplest would be.

a//b + bool(a%b)

And just for safety,

b and (a//b + bool(a%b))

Cheers.

Answer 3

-(-a//b)

Perhaps the simplest?

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*Edited per @Gilles's comment:

for an integer n,

floor(x)=n for x in [n, n+1)
ceil(y)=n+1 for y in (n, n+1]

So, floor(-y)=-n-1 for -y in [-n-1, -n),

and ceil(y)=-floor(-y)=n+1 for y in (n, n+1]

In Python, floor(a/b) = a//b. Thus ceil(a/b) = -(-a//b)