Detecting if a bit is set (at compile time)

Question

How to detect if bit at position n is set in constant variable?

In a *variable*? At *compile time*? Can you give an example? — johnsyweb, Apr 15 '11 at 11:33
You can't - a *variable* changes at *runtime*. Do you mean a *constant* perhaps? — Erik, Apr 15 '11 at 11:33
@There: you're right, they're all wrong. In C++, when we do `const int n = 21;`, `n` is indeed called a *variable* even though it can't vary (3/4: "A variable is introduced by the declaration of an object"). The fact that C++ terminology doesn't match precisely with the English language is something C++ programmers need to get used to, because if they don't they will eventually misinterpret something or other in the standard, for example thinking that something said of "variables" doesn't apply to `n` because it's (in their opinion but not the standard's) "a constant not a variable". — Steve Jessop, Apr 15 '11 at 12:05

ildjarn · Accepted Answer · 2011-04-15T15:31:47.963

6

template<std::uint64_t N, std::uint8_t Bit>
struct is_bit_set
{
    static bool const value = !!(N & 1u << Bit);
};

!! is used to succinctly coerce the value into a bool and avoid data truncation compiler warnings.

edited Apr 15 '11 at 15:31

answered Apr 15 '11 at 11:35

ildjarn

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this answer is nicest in my opinion. – There is nothing we can do Apr 15 '11 at 11:43

score 4 · Answer 2 · answered Apr 15 '11 at 11:34

4

int const a = 4;
int const bitset = !!((1 << 2) & a);

Now, bitset is 1. It would be 0 if you for example stored a 3. Yes, a is a variable.

answered Apr 15 '11 at 11:34

Johannes Schaub - litb

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Why not just `int bitset = ((1 << 2) & a) != 0);`? – sharptooth Apr 15 '11 at 11:35
@RedX yes `bitset` is a constant expression. You could use it for array dimensions like `int array[bitset + 1];`. – Johannes Schaub - litb Apr 15 '11 at 11:37
@Johannes thanks for your answer, would you mind explaining **why** this works? – There is nothing we can do Apr 15 '11 at 11:37
@johannes Ok. Other question: Why do you need the double negation at the beginning? – RedX Apr 15 '11 at 11:41
@RedX : The first coerces into a `bool` (value of `0` or `1` only); the second gives the value actually desired. – ildjarn Apr 15 '11 at 11:42
@RedX because otherwise `bitset` contains `4` and not `1`. @There, http://en.wikipedia.org/wiki/Bitwise_operation, lookup "AND" and "bitshifts". – Johannes Schaub - litb Apr 15 '11 at 11:42
@Johannes +1, as this is good answer, but I will accept ildjarn's as his is "more C++" way. – There is nothing we can do Apr 15 '11 at 11:45
@Johnsyweb only if you have imagination ;) – There is nothing we can do Apr 15 '11 at 11:51
Also, my answer will yield undefined behavior if you shift it in a way that makes the number negative. It's a fair choice, since he obviously put more thought into his answer. – Johannes Schaub - litb Apr 15 '11 at 11:52
@Johannes I'm glad that you agree with me on that. – There is nothing we can do Apr 15 '11 at 11:57

score 2 · Answer 3 · edited May 23 '17 at 12:18

2

The same thing as user ildjarn suggests in his answer, but with so-called "enum trick" that guarantees the compiler will do all the computation in compile time:

template<std::uint64_t N, std::uint8_t Bit>
struct is_bit_set
{
    enum { value = ( N & (1u << Bit) ) != 0 };
};

edited May 23 '17 at 12:18

Community

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answered Apr 15 '11 at 11:50

sharptooth

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I don't *think* this is necessary on any C++03/C++11 compliant compiler, but I'm too tired to go digging through the standard at the moment. In any case, you should probably change it to `!!(N & (1u << Bit))` so the value will be strictly constrained to `0` or `1`. – ildjarn Apr 15 '11 at 14:14
@ildjarn: Thank you, I've edited it in an equivalent fashion. – sharptooth Apr 15 '11 at 14:20

Detecting if a bit is set (at compile time)

3 Answers3