结构作为函数参数，将结构地址传递给函数

#include <stdio.h>
struct Xxx
{
	int aa;
	double bb[2];
	char *cc;
	char *dd[3];
};
void function1(struct Xxx *qq,struct Xxx *rr);
int main()
{
	struct Xxx mm,nn;
	function1(&mm,&nn);
	
	printf("%d %lf %lf %s%s%s%s\n",mm.aa,mm.bb[0],mm.bb[1],
	mm.cc,mm.dd[0],mm.dd[1],mm.dd[2]);
	printf("%d %lf %lf %s%s%s%s\n",nn.aa,nn.bb[0],nn.bb[1],
	nn.cc,nn.dd[0],nn.dd[1],nn.dd[2]);

}
void function1(struct Xxx *qq,struct Xxx *rr)
{
	(*qq).aa=12;
	(*qq).bb[0]=23.4;
	(*qq).bb[1]=34.5;
	(*qq).cc = "Structure ";
	(*qq).dd[0]="passed ";
	(*qq).dd[1]="to ";
	(*qq).dd[2]="function.";
	
	rr->aa = 15;
	rr->bb[0]=45.6;
	rr->bb[1]=67.8;
	rr->cc="Pointer ";
	rr->dd[0]="operators ";
	rr->dd[1]="can ";
	rr->dd[2]="be used.";
	}