So if I could take the 52 numerical bits into a long long int 64bit (or should I just work with the bytes in a second array), then read the 11 exponent bits into an int, sutract 1023 from it for the actual exponent. I would need to determine how many decimal places the exponent indicates shifting and add 4 more shifts to the right. This should be the trunication point of the real int number. Then I need to use those bits 37 places left for my interger and add the sign bit.
OK that said I've got my head confused. If I work with it all at the byte/bit level, would the sign bit be the leftmost bit in the lowest addressed byte in the array of 8 bytes making up the original float?