Macro to compute number of bits needed to store a number n

Question

Let's say I need to write C macro that returns number of bits(1..32) needed to store unsigned 32-bit integer. (Result equals ceiling(log2(n)).

I need it as compile-time computed macro, not a function.

I could do

 #define NBITS(n) ((n)&(1<<31)?32:(n)&(1<<30)?31:...

it works, but is rather long. (Speed does not matter here, computation is at compile time).

Is there shorter way to write this macro ? Shortest ?

Answer 1

#define NBITS2(n) ((n&2)?1:0)
#define NBITS4(n) ((n&(0xC))?(2+NBITS2(n>>2)):(NBITS2(n)))
#define NBITS8(n) ((n&0xF0)?(4+NBITS4(n>>4)):(NBITS4(n)))
#define NBITS16(n) ((n&0xFF00)?(8+NBITS8(n>>8)):(NBITS8(n)))
#define NBITS32(n) ((n&0xFFFF0000)?(16+NBITS16(n>>16)):(NBITS16(n)))
#define NBITS(n) (n==0?0:NBITS32(n)+1)
#include <iostream>
using namespace std;

int main(){
    cout << NBITS(0) << endl;
    cout << NBITS(1) << endl;
    cout << NBITS(2) << endl;
    cout << NBITS(3) << endl;
    cout << NBITS(4) << endl;
    cout << NBITS(1023) << endl;
    cout << NBITS(1024) << endl;
}

it is good?