Print bit representation of a string

Question

How to print the bit representation of a string

std::string = "\x80";

void print (std::string &s) {

    //How to implement this
}

Answer 1

Try:

#include <iostream>

using namespace std;

void print(string &s) {
  string::iterator it; 
  int b;

  for (it = s.begin(); it != s.end(); it++) {
    for (b = 128; b; b >>= 1) {
      cout << (*it & b ? 1 : 0); 
    }   
  }
}

int main() {
  string s = "\x80\x02";
  print(s);
}

Answer 2

Little-endian or big-endian?

for (int i = 0; i < s.length(); i++)
    for (char c = 1; c; c <<= 1) // little bits first
        std::cout << (s[i] & c ? "1" : "0");
for (int i = 0; i < s.length(); i++)
    for (unsigned char c = 0x80; c; c >>= 1) // big bits first
        std::cout << (s[i] & c ? "1" : "0");

Since I hear some grumbling about portability of assuming that a char is a 8-bit byte in the comments of the other answers...

for (int i = 0; i < s.length(); i++)
    for (unsigned char c = ~((unsigned char)~0 >> 1); c; c >>= 1)
        std::cout << (s[i] & c ? "1" : "0");

This is written from a very C-ish standpoint... if you're already using C++ with STL, you might as well go the whole way and take advantage of the STL bitset functionality instead of playing with strings.

Answer 3

I am sorry I marked this as a duplicate. Anyway, to do this:

void printbits(std::string const& s) {
   for_each(s.begin(), s.end(), print_byte());
}

struct print_byte {
     void operator()(char b) {
        unsigned char c = 0, byte = (unsigned char)b;
        for (; byte; byte >>= 1, c <<= 1) c |= (byte & 1);
        for (; c; c >>= 1) cout << (int)(c&1);
    }
};

Answer 4

expanding on Stephan202's answer:

#include <algorithm>
#include <iostream>
#include <climits>

struct print_bits {
    void operator()(char ch) {
     for (unsigned b = 1 << (CHAR_BIT - 1); b != 0; b >>= 1) {
      std::cout << (ch & b ? 1 : 0); 
     }
    }
};

void print(const std::string &s) {
    std::for_each(s.begin(), s.end(), print_bits());
}

int main() {
    print("\x80\x02");
}

Answer 5

I'd vote for bitset:

void pbits(std::string const& s) { 
    for(std::size_t i=0; i<s.size(); i++) 
        std::cout << std::bitset<CHAR_BIT>(s[i]) << " "; 
} 

int main() {
    pbits("\x80\x70"); 
}

Answer 6

Easiest solution is next:

const std::string source("test");
std::copy( 
 source.begin(), 
 source.end(), 
 std::ostream_iterator< 
  std::bitset< sizeof( char ) * 8 > >( std::cout, ", " ) );

• Some stl implementations allow std::setbase() manipulator for base 2.
• You could write your own manipulator if want most flexible solution than existing.

EDIT:
Oops. Someone already posted similar solution.

Answer 7

If you want to do it manually, you can always use a lookup table. 256 values in a static table is hardly a lot of overhead:

static char* bitValues[] = 
{
"00000000",
"00000001",
"00000010",
"00000011",
"00000100",
....
"11111111"
};

Then printing is a simple matter of:

for (string::const_iterator i = s.begin(); i != s.end(); ++i)
{
    cout << bitValues[*i];
}