C ++ Double to Binary Representation (Reinterpret)

I recently decided to create a program that would allow me to print out the exact pattern of an instance bit of any type in C ++. I start with primitive built-in types. I am facing a problem when printing the binary representation of a type double

.

Here's my code:

#include <iostream>
using namespace std;

void toBinary(ostream& o, char a)
{
    const size_t size = sizeof(a) * 8;
    for (int i = size - 1; i >= 0; --i){
        bool b = a & (1UL << i);
        o << b;
    }
}

void toBinary(ostream& o, double d)
{
    const size_t size = sizeof(d);
    for (int i = 0; i < size; ++i){
        char* c = reinterpret_cast<char*>(&d) + i;
        toBinary(o, *c);
    }
}

int main()
{
    int a = 5;
    cout << a << " as binary: "; 
    toBinary(cout, static_cast<char>(a));
    cout << "\n";

    double d = 5;
    cout << d << " as double binary: ";
    toBinary(cout, d);
    cout << "\n";
}

      

        
        
        
      

    

My output is as follows: 5 as binary: 00000101

5 as binary code: 0000000000000000000000000000000000000000000000000001010001000000

However, I know that 5 as a floating point representation: 01000000 00010100 00000000 00000000 00000000 00000000 00000000 00000000

Maybe I don’t understand anything here, but I didn’t write the line reinterpret_cast<char*>(&d) + i

I wrote to handle double*

like char*

so that adding i

to it would advance the pointer to sizeof(char)

instead sizeof(double)

. (What exactly do I want here)? What am I doing wrong?