Number of bytes to be allocated by mpz_export()
kgoldman at us.ibm.com
Tue Nov 16 21:38:51 CET 2010
I have to determine in advance the number of bytes that mpz_export() will
I see in the manual:
numb = 8*size - nail;
count = (mpz_sizeinbase (z, 2) + numb-1) / numb;
p = malloc (count * size);
but I'm not sure of the parameters. If my eventual exported array is
- size is 1?
- I don't know what nail is
I think I want 'nails' to be 0 in mpz_export() so nothing gets discarded.
Could 'nail' be the same? This would yield
byteCount = (mpz_sizeinbase (z, 2) + 7) / 8
*** I think I have to do this because I have some (always positive) fixed
size byte arrays. Thus, I plan to get the byteCount, zero pad my array,
and then mpz_export() to the index where non-zero data will start.
Does this sound reasonable?
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