Counting Bits To Left/Right in 32-bit Integer Using Bit Operations

[noblerare]

Homework Statement

I am wondering is there a way to count the number of bits to the left or right of a given 1 in a 32-bit integer?

For example, if I give the function the number 32 = 0b100000, there are 5 bits to the right of the 1 and hence, 26 bits to the left of the 1.

The catch is, is there a way to do this by simply using bit operations? i.e. no loops or conditionals? Bit operations include: & ^ | << >> ! and ~

(Note: I am coding this in C)

The Attempt at a Solution

Obviously, if I subtract 1 from the original number, I get all 1s to the right of the original number but I'm still stumped as to how I can count the actual number of bits without using loops or conditional statements.

 

[mgb_phys]

http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetKernighan

 

[noblerare]

Thanks, but the link you posted has the solution using a for loop.

Is there a way to do this without using loops or conditional statements? Simply using bit operations such as << >> ! ~ & | ^

 

[mgb_phys]

Not the kernighan solution - but the ones below it (that was the only entry that had an index position)

 

[Mark44]

Read on in the page that mgb_phys linked to. The very first routine uses a for loop, but there are other techniques on the page that don't use loops.