convert uint8_t array to bitset in C++

Question

Is there a quick way to convert an array of uint8_t to a biteset.

uint8_t test[16]; 
// Call a function which populates test[16] with 128 bits
function_call(& test);
for(int i=0; i<16; i++)
  cout<<test[0]; // outputs a byte
cout<<endl;
std:: bitset<128> bsTest;

I tried this but does not work

bsTest(test);

Answer 1

I propose you a possible solution.

Not so good, not so quick, a little dirty but I hope it can help.

#include <bitset>
#include <iostream>

int main ()
 {
   uint8_t  test[16] = { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h',
                         'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p' };

   std::bitset<128> bsTest { } ;

   for ( unsigned ui = 0 ; ui < 16 ; ++ui )
    {
      bsTest <<= 8;

      std::bitset<128> bsTmp { (unsigned long) test[ui] };

      bsTest |= bsTmp;
    }

   std::cout << bsTest;

   return 0;
 }

The idea is initialize the bitset to zero

std::bitset<128> bsTest { } ;

and add a uint8_t at a time at the end of another bitset

std::bitset<128> bsTmp { (unsigned long) test[ui] };

then merge (bit or) the two bitsets

bsTest |= bsTmp;

and shift 8 bit the result

bsTest <<= 8;

p.s.: sorry for my bad English