 # Programming-Idioms # Idiom #263 Integer logarithm in base 2

Write two functions log2d and log2u, which calculate the binary logarithm of their argument n rounded down and up, respectively. n is assumed to be positive. Print the result of these functions for numbers from 1 to 12.

``````program main
implicit none
integer :: i
do i=1,12
print *,i,log2d(i),log2u(i)
end do
contains
integer function log2d (n)
integer, intent(in) :: n
log2d = bit_size(n) - 1 - leadz(n)
end function log2d

integer function log2u (n)
integer, intent(in) :: n
end function log2u
end program main``````
``````log2d :: Double -> Integer
log2d = floor . logBase 2

log2u :: Double -> Integer
log2u = ceiling . logBase 2

main :: IO ()
main = print \$ [log2d, log2u] <*> [1..12]``````
``````function log2d(n: uint32): integer;
var
temp: uint32;
begin
Result := 0;
temp := 1;
while (temp < n) do
begin
Inc(Result);
temp := 1 shl Result;
end;
if (temp > n) then
Dec(Result);
end;

function log2u(n: uint32): integer;
begin
Result := log2d(n);
if (1 shl Result < n) then
Inc(Result);
end;

var
i: integer;
begin
for i := 1 to 16 do
writeln(i,': log2d = ',log2d(i),', log2u = ',log2u(i));
end.``````
``use POSIX qw( log2 floor ceil );``
``````sub log2d { floor log2 shift };

sub log2u { ceil log2 shift };``````
``import math``
``````def log2d(n):
return math.floor(math.log2(n))

def log2u(n):
return math.ceil(math.log2(n))

for n in range(1, 13):
print(n, log2d(n), log2u(n))``````
``````def log2d(n) =  n.bit_length - 1

def log2u(n) = 2 ** log2d(n) == n ? log2d(n) : log2d(n) + 1

(1..12).each{|n| puts "#{n}  #{log2d(n)}  #{log2u(n)}" }``````
``````fn log2d(n: f64) -> f64 {
n.log2().floor()
}

fn log2u(n: f64) -> f64 {
n.log2().ceil()
}

fn main() {
for n in 1..=12 {
let f = f64::from(n);
println!("{} {} {}", n, log2d(f), log2u(f));
}
}``````

tkoenig