Question

Check whether a number x is nonzero using the legal operators except !.

Examples: isNonZero(3) = 1, isNonZero(0) = 0

Legal ops: ~ & ^ | + << >>

• Note : Only bitwise operators should be used. if, else, for, etc. cannot be used.
• Edit1 : No. of operators should not exceed 10.
• Edit2 : Consider size of int to be 4 bytes.

int isNonZero(int x) {
return ???;
}

Using ! this would be trivial , but how do we do it without using ! ?

Answer 1

The logarithmic version of the adamk function:

int isNotZero(unsigned int n){
  n |= n >> 16;
  n |= n >> 8;
  n |= n >> 4;
  n |= n >> 2;
  n |= n >> 1;
  return n & 1;
};

And the fastest one, but in assembly:

xor eax, eax
sub eax, n  // carry would be set if the number was not 0
xor eax, eax
adc eax, 0  // eax was 0, and if we had carry, it will became 1

Something similar to assembly version can be written in C, you just have to play with the sign bit and with some differences.

EDIT: here is the fastest version I can think of in C:

1) for negative numbers: if the sign bit is set, the number is not 0.

2) for positive: 0 - n will be negaive, and can be checked as in case 1. I don’t see the - in the list of the legal operations, so we’ll use ~n + 1 instead.

What we get:

int isNotZero(unsigned int n){ // unsigned is safer for bit operations
   return ((n | (~n + 1)) >> 31) & 1;
}