CODEWITHSUNDEEP
c++c++11templatesconstexpr
Mozbi
Mozbi

Reputation: 1479

Initializing c++ std::bitset at compile time

I'm trying to initialize a std::bitset<256> at compile time with some of its indices, lets say 50-75 and 200-225 set to 1.

Based on http://en.cppreference.com/w/cpp/utility/bitset/bitset It looks like my 2 options are:

constexpr bitset();
constexpr bitset( unsigned long long val );

Could someone shed light on how I would initialize my bitset considering the second constructor wouldn't work for large indices?

Upvotes: 8

Views: 5376

Answers (2)

Stephan Lechner
Stephan Lechner

Reputation: 35154

Bitset constructor constexpr bitset( unsigned long long val ) is limited in the sense that it cannot pre-set bits from position 64 upwards (as unsigned long long is a 64 bit value). Any bit higher than 64 will be initialized with 0, as the following example shows (cf. bitset constructors):

std::bitset<70> bl(ULLONG_MAX); // [0,0,0,0,0,0,1,1,1,...,1,1,1] in C++11

If you want to initialize also the higher bits in the constructor, I think you won't get around to use an initializer of type string or char*, i.e. bitset constructors of type (3) or (4).

Of course, for a bitset of length 256, the initializer string gets long:

std::bitset<256> myBitset("111100000011111000001110000010101000100000100010010000100000000001111");
// I did not count the digits...

Upvotes: 1

Vivek
Vivek

Reputation: 330

You can use the constructor that takes a string as an argument and initialize it that way.

For example. the following line, will initialize b with 0000000101111001:

std::bitset<16> b (std::string("0101111001"));

Upvotes: 0

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