The limit is in the mpz type. It uses an int for the number of limbs,
i.e. limit is 2^32 limbs = 2^38 bits. So your number must be < 2^38.
There are no plans to remove this restriction at the moment. But you
can use the mpn functions, which are not so limited.
Bill.
On 10 June 2012 09:03, Jason Moxham <ja...@thecodecavern.co.uk> wrote:
> Hi
>
> Sorry for the delay.
>
> I seem to remember that 64bit builds are limited to 2^37 bits (16GB) , I read
> it somewhere on the GMP website ,I have no idea which part of MPIR has this
> restriction.
>
> Jason
>
>
> -----------------------------------------------------------
> MPIR-2.5.1 bug
>
> #include <stdio.h>
> #include <stdlib.h>
> #include <gmp.h>
>
> int main(int argc, char *argv[])
> {
> // Demonstrates a MPIR bug probably with mpz_mul
> //
> // Need a computer with at least 60G of memory
> //
> // % foo
> // 3^(2^36) mod 3 equals 0
> // 3^(2^37) mod 3 does NOT equal 0 - Error
>
> unsigned long p;
> unsigned long e=68719476736; // 2^36
> mpz_t pp, b;
>
> p = 3; // bug does not happen with p = 2
> mpz_init(pp);
> mpz_init(b);
> // p^(2^36)
> mpz_ui_pow_ui(pp,p,e);
> // p^(2^36) mod p
> mpz_mod_ui(b,pp,p);
> printf("%d^(2^36) mod %d ", p, p);
> if (mpz_cmp_ui(b,0) == 0) {
> printf("equals 0\n");
> } else {
> printf("does NOT equal 0 - Error\n");
> exit(0);
> }
> fflush(stdout);
> // square - (p^(2^36))^2 = p^(2^37)
> mpz_mul(pp,pp,pp);
> // p^(2^37) mod p
> mpz_mod_ui(b,pp,p);
> printf("%d^(2^37) mod %d ", p, p);
> if (mpz_cmp_ui(b,0) == 0) {
> printf("equals 0\n");
> } else {
> printf("does NOT equal 0 - Error\n");
> exit(0);
> }
> fflush(stdout);
> }
>
> --
> Kate Minola
>
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