C/C++ check if one bit is set in, i.e. int variable

Question

In C, if you want to hide bit manipulation, you can write a macro:

#define CHECK_BIT(var,pos) ((var) & (1<<(pos)))

and use it this way to check the n^th bit from the right end:

CHECK_BIT(temp, n - 1)

In C++, you can use std::bitset.

Check if bit N (starting from 0) is set:

temp & (1 << N)

There is no builtin function for this.

I would just use a std::bitset if it's C++. Simple. Straight-forward. No chance for stupid errors.

typedef std::bitset<sizeof(int)> IntBits;
bool is_set = IntBits(value).test(position);

or how about this silliness

template<unsigned int Exp>
struct pow_2 {
    static const unsigned int value = 2 * pow_2<Exp-1>::value;
};

template<>
struct pow_2<0> {
    static const unsigned int value = 1;
};

template<unsigned int Pos>
bool is_bit_set(unsigned int value)
{
    return (value & pow_2<Pos>::value) != 0;
} 

bool result = is_bit_set<2>(value);

What the selected answer is doing is actually wrong. The below function will return the bit position or 0 depending on if the bit is actually enabled. This is not what the poster was asking for.

#define CHECK_BIT(var,pos) ((var) & (1<<(pos)))

Here is what the poster was originally looking for. The below function will return either a 1 or 0 if the bit is enabled and not the position.

#define CHECK_BIT(var,pos) (((var)>>(pos)) & 1)

Yeah, I know I don't "have" to do it this way. But I usually write:

    /* Return type (8/16/32/64 int size) is specified by argument size. */
template<class TYPE> inline TYPE BIT(const TYPE & x)
{ return TYPE(1) << x; }

template<class TYPE> inline bool IsBitSet(const TYPE & x, const TYPE & y)
{ return 0 != (x & y); }

E.g.:

IsBitSet( foo, BIT(3) | BIT(6) );  // Checks if Bit 3 OR 6 is set.

Amongst other things, this approach:

• Accommodates 8/16/32/64 bit integers.
• Detects IsBitSet(int32,int64) calls without my knowledge & consent.
• Inlined Template, so no function calling overhead.
• const& references, so nothing needs to be duplicated/copied. And we are guaranteed that the compiler will pick up any typo's that attempt to change the arguments.
• 0!= makes the code more clear & obvious. The primary point to writing code is always to communicate clearly and efficiently with other programmers, including those of lesser skill.
• While not applicable to this particular case... In general, templated functions avoid the issue of evaluating arguments multiple times. A known problem with some #define macros.
  E.g.: #define ABS(X) (((X)<0) ? - (X) : (X))
        ABS(i++);

According to this description of bit-fields, there is a method for defining and accessing fields directly. The example in this entry goes:

struct preferences {
    unsigned int likes_ice_cream : 1;
    unsigned int plays_golf : 1;
    unsigned int watches_tv : 1;
    unsigned int reads_books : 1;
}; 

struct preferences fred;

fred.likes_ice_cream = 1;
fred.plays_golf = 1;
fred.watches_tv = 1;
fred.reads_books = 0;

if (fred.likes_ice_cream == 1)
    /* ... */

Also, there is a warning there:

However, bit members in structs have practical drawbacks. First, the ordering of bits in memory is architecture dependent and memory padding rules varies from compiler to compiler. In addition, many popular compilers generate inefficient code for reading and writing bit members, and there are potentially severe thread safety issues relating to bit fields (especially on multiprocessor systems) due to the fact that most machines cannot manipulate arbitrary sets of bits in memory, but must instead load and store whole words.