Thread: Converting a string to binary

I have a couple of question about the Converting a string to binary article on this site.

First of all, CHAR_BIT - Is this a const int = 8 ? Im assumiong it's been declared in one of the headers.

Secondly, this code snip is found in the print_bits method

Code:

  for ( unsigned i = 0; i < n_bits; ++i ) {
    std::cout<< !!( val & 1 );
    val >>= 1;
  }

Whats the point of using two ! when doing the cout ?

whats does val >>=1 do ?

appreciate any response.

Thanks for your response.

Originally posted by Zach L.

>> whats does val >>=1 do ?

Bit shifting. Say you have the following byte:

1011 1011

Shifting to the right by one bit (>> and >>=) produces:

0101 1101

so >> and >>= means the same thing ?

Of course...

I've been playing around with the above example and couldnt this be an even simpler solution ?

Code:

#include <iostream> 
#include <climits> 
#include <string> 

template <typename T>
void print( T val )
{
   for( int bit=128; bit > 0; bit /= 2) {
       int sum = (val & bit);

       if (sum > 0)
          cout << "1";
       else
           cout << "0";
  }
}

int main()
{
  std::string s = "This is a test";
  std::string::iterator it;

  for ( it = s.begin(); it != s.end(); it++ ) {
        print ( *it );
        std::cout<<"  "<< *it <<std::endl;
    }
  system("PAUSE");
  return 0;
}

>>First of all, CHAR_BIT - Is this a const int = 8 ?
It's an implementation defined value, usually defined in <climits> like so (for octets)

#define CHAR_BIT 8

>>Whats the point of using two ! when doing the cout ?
It normalizes a value into either 0 or 1. If you only use one ! then the bits will be swapped, 0101 will be 1010 and that would be wrong. So the first ! changes the values to 0 and 1, and the second swaps them so that they hold the right 0 and 1 values.

>>whats does val >>=1 do ?
It shifts the bits in val right by one and assigns the result to val.

Originally posted by Casey

>>Whats the point of using two ! when doing the cout ?
It normalizes a value into either 0 or 1. If you only use one ! then the bits will be swapped, 0101 will be 1010 and that would be wrong. So the first ! changes the values to 0 and 1, and the second swaps them so that they hold the right 0 and 1 values.

Thanks for your feedback. Im not 100% sure i follow you on the above. Could you please give me an example where !! work but not using a ! doesnt work ?

Zach L. : A bit lazy on my part not to include CHAR_BIT + sizeof( val )

This should explain it

Code:

#include <iostream> 
#include <climits> 
#include <string> 

template <typename T>
void print_bits1 ( T val )
{
  unsigned int n_bits = sizeof ( val ) * CHAR_BIT;

  for ( unsigned i = 0; i < n_bits; ++i ) {
    std::cout<< !( val & 1 );
    val >>= 1;
  }
}

template <typename T>
void print_bits2 ( T val )
{
  unsigned int n_bits = sizeof ( val ) * CHAR_BIT;

  for ( unsigned i = 0; i < n_bits; ++i ) {
    std::cout<< !!( val & 1 );
    val >>= 1;
  }
}

int main()
{
    print_bits1(1U);
    std::cout<<std::endl;
    print_bits2(1U);
}

Originally posted by Zach L.

Since any non-zero number is true (say x != 0), then !x is false (0), and !(!x)=!!x =1 (since ! will return 1 for true, not just an arbitrary number).

Thanks a bunch, now I get it