Decimal floating point to binary floating point

[Dan]

Thanks Thomas and Albert,

Perhaps a look-up table, storing m * 2^n, where m,n satisfy m * 2^n = 1.25^N for N element [-99,99]. Then

8.25 * 10^84 = 8.25 * (2^3 * 1.25)^84
= 8.25 * 2^252 * 1.25^84
= 8.25 * 2^252 * m * 2^n from look-up table, so fast
= 8.25 * m * 2^(252+n)

Then call assembly subroutine (I'm doing mixed C/assembly) converting 8.25 * m to double precision floating point, giving sign bit and fraction and intermediate exponent and add 252+n to it to obtain final exponent.