4 ms·
You don't need any specific instructions. The hex encoded double contains the final binary representation of the floating point number. So no conversion is requ
by firebacon 6y ago
You don't need any specific instructions. The hex encoded double contains the final binary representation of the floating point number. So no conversion is required to load it, except maybe for swapping around bytes on some architectures. Conceptually:
double v;
memcpy(&v, "\x00\x00\x00\x00\x00\xe4\x94\x40", sizeof(v)); // LE