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
log2u = bit_size(n) - leadz(n-1)
end function log2u
end program main
import static java.lang.Math.ceil;
import static java.lang.Math.floor;
import static java.lang.Math.log;
import static java.lang.System.out;
interface F {
double log2 = log(2);
double f(double n);
}
F log2d = x -> floor(log(x) / F.log2),
log2u = x -> ceil(log(x) / F.log2);
for (int i = 1; i <= 12; ++i) {
out.printf("log2d(%s) = %s%n", i, log2d.f(i));
out.printf("log2u(%s) = %s%n", i, log2u.f(i));
}
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.