CODEWITHSUNDEEP
c++c++11std-bitset
Abdul Rehman
Abdul Rehman

Reputation: 1727

split std::bitset in two halves?

I am implementing DES algorithm and I need to split std::bitset<56> permutationKey in two halves.

std::bitset<56> permutationKey(0x133457799BBCDF);
std::bitset<28> leftKey;
std::bitset<28> rightKey;

std::bitset<56> divider(0b00000000000000000000000000001111111111111111111111111111);

rightKey = permutationKey & divider;
leftKey = (permutationKey >> 28) & divider;

I tried to typecast bitset<56> to bitset<28> but it didn't work.

Other way to achieve the same thing is to iterate and assign each bit individually. I want to achieve it without using loops there must be another way.

I was able to do it with primitive types

uint64_t key = 0b0001010101010101110110001100001110000011111100000000011111000000;
                    //00010101.01010101.11011000.11000011---|---10000011.11110000.00000111.11000000
uint32_t right = (uint32_t)key;
uint32_t left = key >> 32;

How can I split bitset like this?

Upvotes: 5

Views: 3327

Answers (2)

Andrushenko Alexander
Andrushenko Alexander

Reputation: 1973

The proposed solution in previous answer is nice but not general: how do we split a bitset that is to big to be converted to u_long without loss of bits? Intuitively we look here for something like iterators and ranges:

vector<int> vec({1,2,3,4});
vector<int> vec1(vec.begin(), vec.begin() + vec.size() / 2); // {1,2}
vector<int> vec2(vec.begin() + vec.size() / 2, vec.end()); // // {3,4}

Unfortunately bitsets do not have iterators and ranges. If you do not care much about speed of conversion I would propose following simple and general solution:

const size_t size = ...  // we assume size is even
bitset<size> bss;
.......
const string str = bss.to_string();
bitset<size/2>bss1(str.substr(0, str.size() / 2));          // first half
bitset<size/2>bss2(str.substr(str.size() / 2, str.size())); // second half

Upvotes: 2

Richard Hodges
Richard Hodges

Reputation: 69882

std::bitset<56> permutationKey(0x133457799BBCDF);
std::bitset<56> divider(0b00000000000000000000000000001111111111111111111111111111);

auto rightKey = std::bitset<28> ( (permutationKey & divider).to_ulong() );
auto leftKey = std::bitset<28> ( ((permutationKey >> 28) & divider).to_ulong() );

Upvotes: 6

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