Re: printing double in binary

James Kanze <>
Sun, 2 May 2010 15:34:34 -0700 (PDT)
On May 2, 7:15 pm, bauran <> wrote:

In Thinking in C++ by Bruce eckel, there is a program given to
print a double value in binary.(Chapter 3, page no. 189)

int main(int argc, char* argv[])
  if(argc != 2)
    cout << "Must provide a number" << endl;
  double d = atof(argv[1]);
  unsigned char* cp = reinterpret_cast<unsigned char*>(&d);
  for(int i = sizeof(double); i > 0 ; i -= 2)

assume printbinary() prints (const unsigned char) in binary
i.e. 8 bits. Moreover I haven't provided header files to
shorten code. Please assume proper header files and namespaces
are included.

Here while printing cp[i] when i=8(assuming double is of 8
bytes), wouldn't it be undefined behaviour?

Maybe. If sizeof double is 8, it's undefined behavior.

I'm not too sure what this code is supposed to do. Printing a
hexadecimal dump of objects is pretty standard, and usually done
with something like:

    template<typename T>
    dump( std::ostream& dest, T const& obj )
        IOSave state( dest );
        dest.setf( std::hex, std::base_field);
        dest.fill( '0' );
        unsigned char const* p = reinterpret_cast<unsigned char const
*>( &obj );
        for ( int i = 0; i != sizeof(T); ++ i) {
            if ( i != 0 ) {
                dest << ' ';
            dest << std::setw(2) << p[ i ];

I mean this code doesn't work as it doesn't print cp[0].
Bruce Eckel is a very renowned writer. How can he do this type
of mistake?

I've yet to meet a human being who didn't make mistakes.

James Kanze

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