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When a GMP variable is used as a function parameter, it’s effectively a
call-by-reference, meaning if the function stores a value there it will change
the original in the caller. Parameters which are input-only can be designated
const
to provoke a compiler error or warning on attempting to modify
them.
When a function is going to return a GMP result, it should designate a
parameter that it sets, like the library functions do. More than one value
can be returned by having more than one output parameter, again like the
library functions. A return
of an mpz_t
etc doesn’t return the
object, only a pointer, and this is almost certainly not what’s wanted.
Here’s an example accepting an mpz_t
parameter, doing a calculation,
and storing the result to the indicated parameter.
void foo (mpz_t result, const mpz_t param, unsigned long n) { unsigned long i; mpz_mul_ui (result, param, n); for (i = 1; i < n; i++) mpz_add_ui (result, result, i*7); } int main (void) { mpz_t r, n; mpz_init (r); mpz_init_set_str (n, "123456", 0); foo (r, n, 20L); gmp_printf ("%Zd\n", r); return 0; }
foo
works even if the mainline passes the same variable for
param
and result
, just like the library functions. But
sometimes it’s tricky to make that work, and an application might not want to
bother supporting that sort of thing.
For interest, the GMP types mpz_t
etc are implemented as one-element
arrays of certain structures. This is why declaring a variable creates an
object with the fields GMP needs, but then using it as a parameter passes a
pointer to the object. Note that the actual fields in each mpz_t
etc
are for internal use only and should not be accessed directly by code that
expects to be compatible with future GMP releases.
Next: Memory Management, Previous: Variable Conventions, Up: GMP Basics [Index]