a function to check if the nth bit is set in a byte

Question

I want a simple C function which will return true if the n-th bit in a byte is set to1. Otherwise it will return false.

This is a critical function in terms of execution time, so I am thinking of the most optimal way to do that.

Answer 1

The following function can do what you need:

int isNthBitSet (unsigned char c, int n) {
    static unsigned char mask[] = {128, 64, 32, 16, 8, 4, 2, 1};
    return ((c & mask[n]) != 0);
}

This assumes 8-bit bytes (not a given in C) and the zeroth bit being the highest order one. If those assumption are incorrect, it simply comes down to expanding and/or re-ordering the mask array.

No error checking is done since you cited speed as the most important consideration. Do not pass in an invalid n, that'll be undefined behaviour.

At insane optimisation level -O3, gcc gives us:

isNthBitSet:    pushl   %ebp
                movl    %esp, %ebp
                movl    12(%ebp), %eax
                movzbl  8(%ebp), %edx
                popl    %ebp
                testb   %dl, mask(%eax)
                setne   %al
                movzbl  %al, %eax
                ret
mask:           .byte   -128, 64, 32, 16, 8, 4, 2, 1

which is pretty small and efficient. And if you make it static and suggest inlining, or force it inline as a macro definition, you can even bypass the cost of a function call.

Just make sure you benchmark any solution you're given, including this one ^(a). The number one mantra in optimisation is "Measure, don't guess!"

If you want to know how the bitwise operators work, see here. The simplified AND-only version is below.

The AND operation & will set a bit in the target only if both bits are set in the tewo sources. The relevant table is:

AND | 0 1
----+----
 0  | 0 0
 1  | 0 1

For a given char value, we use the single-bit bit masks to check if a bit is set. Let's say you have the value 13 and you want to see if the third-from-least-significant bit is set.

Decimal  Binary
  13     0000 1101
   4     0000 0100 (the bitmask for the third-from-least bit).
         =========
         0000 0100 (the result of the AND operation).

You can see that all the zero bits in the mask result in the equivalent result bits being zero. The single one bit in the mask will basically let the equivalent bit in the value flow through to the result. The result is then zero if the bit we're checking was zero, or non-zero if it was one.

That's where the expression in the return statement comes from. The values in the mask lookup table are all the single-bit masks:

Decimal  Binary
  128    1000 0000
   64    0100 0000
   32    0010 0000
   16    0001 0000
    8    0000 1000
    4    0000 0100
    2    0000 0010
    1    0000 0001

---

^(a) I know how good I am, but you don't :-)

Answer 2

Just check the value of (1 << bit) & byte. If it is nonzero, the bit is set.

Answer 3

Let the number be num. Then:

return ((1 << n) & num);

Answer 4

bool isSet(unsigned char b, unsigned char n) { return b & ( 1 << n); }

Answer 5

Another approach would be

    bool isNthBitSet (unsigned char c, int n) {
      return (1 & (c >> n));
    }

Answer 6

#include<stdio.h>
int main()
{
   unsigned int n,a;
   printf("enter value for n\n");
   scanf("%u",&n);
   pintf("enter value for a:\n");
   scanf("%u",&a);
   a= a|(((~((unsigned)0))>>(sizeof(int)*8-1))<<n);
   printf("%u\n",a);
}   

Answer 7

#include<stdio.h>

int main() 
{
        int data,bit;
        printf("enter data:");
        scanf("%d",&data);
        printf("enter bit position to test:");
        scanf("%d",&bit);
        data&(1<<bit)?printf("bit is set\n"):printf("bit is clear\n");

   return 0;
}