Floating point endianness · the case of IEEE 754 binary64

Floating point numbers have a standard – IEEE 754. But how exactly those numbers stored? Below is a piece of code that shows it:

#include <iostream>
#include <iomanip>
#include <set>

using namespace std;

int main(int argc, char**argv)
{
    std::multiset<double>* mset = new std::multiset<double>();
    mset->insert(3.14);
    mset->insert(6.28);
    mset->insert(2.71);
    mset->insert(0.61);
    mset->insert(1.61);
    mset->insert(9.89);
    mset->insert(0.5);
    for(std::multiset<double>::const_iterator it = mset->cbegin(); it != mset->cend(); it++) {
        double val = *it;
        cout << val << " = ";
        unsigned char* blobchar = reinterpret_cast<unsigned char*>(&val);
        cout << hex << uppercase << internal << setfill('0');
        for(unsigned i=0; i<sizeof(double); ++i) {
            cout << setw(2) << (unsigned)blobchar[i] << " ";
        };
        cout << dec << endl;
    };
    delete mset;
    return 0;
}

In my x86 Linux box, this is the output:

0.5 = 00 00 00 00 00 00 E0 3F 
0.61 = 85 EB 51 B8 1E 85 E3 3F 
1.61 = C3 F5 28 5C 8F C2 F9 3F 
2.71 = AE 47 E1 7A 14 AE 05 40 
3.14 = 1F 85 EB 51 B8 1E 09 40 
6.28 = 1F 85 EB 51 B8 1E 19 40 
9.89 = 48 E1 7A 14 AE C7 23 40

The first line, a round number in binary, is a clue that the double is indeed saved as little endian because the Intel x86 architecture is little endian. Take the second line as an example, we tie out the representation to validate that it is actually the case:

IEEE 754 binary64 format:

sign bit: 1 bit (bit 0)
exponent: 11 bits (bit 1 to bit 11)
- biased (0 to 2047 for representing \(e+1023\)) or unbiased (\(-1024\) to \(1023\) in two’s complement)
- exponent of \(-1023\) and \(+1024\) are reserved
mantissa: 53 bits with 52 bits explicitly stored (bit 12 to bit 63)
- with implicit integer bit of value 1 except for special values
- guaranteed precision of 15 significant decimal digits (\(53\log_{10}2 = 15.95\))

Reversing for big-endian order, 0.61 becomes

3F E3 85 1E B8 51 EB 85

sign bit: 0
exponent: 0x3FE (=011 1111 1110)
- 1022 in decimal, biased representation of \(1022-1023 = -1\)
mantissa: 0x3851EB851EB85 (=0011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101)

Converting to binary, it is: \(1.0011\ 1000\ 0101\ 0001\ 1110\ 1011\ 1000\ 0101\ 0001\ 1110\ 1011\ 1000\ 0101 \times 2^{-1}\) and it matches the mantissa of 0.61: