C Primer Plus 第六版(中文版)第十四章(代码注释版)编程练习答案（14.18.4）

14.18.4（a）

#include 
#include 
#include 
#include 

#define N 20
#define LEN 30
void show(struct person *ptr);

struct fullname
{
	char lastname[N];
	char middname[N];
	char firstname[N];

};

struct  person
{
	int num; //社保号
	struct fullname name; //姓名 包含三个结构体
};



int main()
{
	//定义该结构体类型数组5个
	struct person record[5] = {

		{1234567,{"xia", "long" ,"Han" }},
		{2234567,{"qiang", "" ,	"Li" }},//无中间字的
		{3234567,{"yi", "zun" ,	"Li" }},
		{4234567,{"ze", "ling" ,"Li" }},
		{5234567,{"zhu", "" ,"Qing" }},

	};

	show(record);//record数组名代表首地址

	return 0;
}

void show(struct person *ptr)//结构体指针，指向结构体数组
{
	int i;
	printf("All numbers messages:\n");
	for (i = 0; i<5; i++)
	{
		//显示数组分情况也需要middname是否存在

		if (strlen (ptr[i].name.middname) )//统计middname中字符数>0
		{
			printf("%s, %s %c - %d\n",ptr[i].name.firstname, ptr[i].name.lastname,ptr[i].name.middname[0],ptr[i].num );
	     }
		else //middname无字符
		{
			printf("%s, %s  - %d\n", ptr[i].name.firstname, ptr[i].name.lastname,  ptr[i].num);
		}
	
	}


}

14.18.4(b)

#include 
#include 
#include 
#include 

#define N 20
#define LEN 30
void show(struct person );

struct fullname
{
	char lastname[N];
	char middname[N];
	char firstname[N];

};

struct  person
{
	int num; //社保号
	struct fullname name; //姓名 包含三个结构体
};



int main()
{
	int i;
	//定义该结构体类型数组5个
	struct person record[5] = {

		{1234567,{"xia", "long" ,"Han" }},
		{2234567,{"qiang", "" ,	"Li" }},//无中间字的
		{3234567,{"yi", "zun" ,	"Li" }},
		{4234567,{"ze", "ling" ,"Li" }},
		{5234567,{"zhu", "" ,"Qing" }},

	};

	for(i = 0; i < 5; i++)//循环去输出值
	show(record[i]);

	return 0;
}

void show(struct person pers)//结构体
{
	int i;
	printf("All numbers messages:\n");
	    
	//外层有套循环去输出数组值了，这里就不在需要循环
	//显示数组分情况也需要middname是否存在

		if (strlen (pers.name.middname) )//统计middname中字符数>0
		{
			printf("%s, %s %c - %d\n", pers.name.firstname, pers.name.lastname, pers.name.middname[0],pers.num );
	     }
		else //middname无字符
		{
			printf("%s, %s  - %d\n", pers.name.firstname, pers.name.lastname, pers.num);
		}
	
	


}