Floating Point to Binary Value(C++)

Use union and bitset:

#include <iostream>
#include <bitset>
#include <climits>

int main()
{
    union
    {
        float input; // assumes sizeof(float) == sizeof(int)
        int   output;
    } data;

    data.input = 2.25125;

    std::bitset<sizeof(float) * CHAR_BIT> bits(data.output);
    std::cout << bits << std::endl;

    // or
    std::cout << "BIT 4: " << bits[4] << std::endl;
    std::cout << "BIT 7: " << bits[7] << std::endl;
}

It may not be an array but you can access bits with [] operator as if you were using an array.

Output

$ ./bits
01000000000100000001010001111011
BIT 4: 1
BIT 7: 0

int fl = *(int*)&floatVar; //assuming sizeof(int) = sizeof(float)

int binaryRepresentation[sizeof(float) * 8];

for (int i = 0; i < sizeof(float) * 8; ++i)
    binaryRepresentation[i] = ((1 << i) & fl) != 0 ? 1 : 0;

Explanation

(1 << i) shifts the value 1, i bits to the left. The & operator computes the bitwise and of the operands.

The for loop runs once for each of the 32 bits in the float. Each time, i will be the number of the bit we want to extract the value from. We compute the bitwise and of the number and 1 << i:

Assume the number is: 1001011, and i = 2

1<<i will be equal to 0000100

  10001011
& 00000100
==========
  00000000

if i = 3 then:

  10001011
& 00001000
==========
  00001000

Basically, the result will be a number with ith bit set to the ith bit of the original number and all other bits are zero. The result will be either zero, which means the ith bit in the original number was zero or nonzero, which means the actual number had the ith bit equal to 1.

other approach, using stl

#include <iostream>
#include <bitset>

using namespace std;
int main()
{
    float f=4.5f;
    cout<<bitset<sizeof f*8>(*(long unsigned int*)(&f))<<endl;
    return 0;
}