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实例：
设计一个模拟社会关系的数据结构，每个人的信息用结构表示，包含名字、性别和指向父亲、母亲、配偶、子女的指针（只限两个子女）。要求编写以下函数：
（1）增加一个新人的函数
（2）建立人与人之间关系的函数：父-子、母-子、配偶等。
（3）检查两人之间是否为堂兄妹
思路解析：
能够充分的联系指针的应用。书中的代码在增加一个新人时，只为新人提供名字和性别，关于新人的其他信息通过调用其他函数建立。
书中代码如下：
1 #include <stdio.h>
2 #include <stdlib.h>
3 #include <string.h>
4
5 #define CHILDREN 2
6
7 struct person{
8 char *name; /*名字符串指针*/
9 char sex; /*性别：男用字符'M'；女用字符'F'*/
10 struct person *father; /*指向父亲*/
11 struct person *mother; /*指向母亲*/
12 struct person *mate; /*指向配偶*/
13 struct person *childern[CHILDREN];/*指向子女*/
14 };
15
16 /* [函数]newperson增加新人 */
17 struct person *newperson(char *name, char sex)
18 {
19 struct person *p;
20 int index;
21
22 p = (struct person *)malloc(sizeof(struct person));
23 p->name = (char *)malloc(strlen(name)+1);
24 strcpy(p->name, name);
25 p->sex = sex;
26 p->father = NULL;
27 p->mother = NULL;
28 p->mate = NULL;
29 for(index=0; index<CHILDREN; index++)
30 p->childern[index] = NULL;
31 return p;
32 }
33
34 /* [函数]father_child建立父－子关系 */
35 void father_child(struct person *father, struct person *child)
36 {
37 int index;
38
39 for(index=0; index<CHILDREN-1; index++)/*寻找一个空缺的子女指针*/
40 if(father->childern[index]==NULL) /*若没有空缺，则填在最后*/
41 break;
42 father->childern[index] = child; /*建立父－子关系*/
43 child->father = father;
44 }
45
46 void mother_child(struct person *mother, struct person *child)
47 {
48 int index;
49 for(index=0; index<CHILDREN-1; index++)/*寻找一个空缺的子女指针*/
50 if(mother->childern[index]==NULL) /*若没有空缺，则填在最后*/
51 break;
52 mother->childern[index] = child; /*建立母－子关系*/
53 child->mother = mother;
54 }
55
56 /* [函数]mate 建立配偶关系 */
57 void mate(struct person *h, struct person *w)
58 {
59 h->mate = w;
60 w->mate = h;
61 }
62
63 /* [函数]brotherinlow 检查两人是否是堂兄妹 */
64 int brothersinlaw(struct person *p1, struct person *p2)
65 {
66 struct person *f1, *f2;
67 if(p1==NULL||p2==NULL||p1==p2)
68 return 0;
69 if(p1->sex==p2->sex) return 0;/*不可能是堂兄妹*/
70 f1 = p1->father;
71 f2 = p2->father;
72 if(f1!=NULL && f1==f2) return 0;/*是兄妹，不是堂兄妹*/
73 while(f1!=NULL&&f2!=NULL&&f1!=f2)/*考虑远房情况*/
74 {
75 f1 = f1->father;
76 f2 = f2->father;
77 if(f1!=NULL && f2!=NULL && f1==f2) return 1;
78 }
79 return 0;
80 }
81
82 /* 函数print_relate用于输出人物p的姓名，性别和各种关系 */
83 void print_relate(struct person *p)
84 {
85 int index,i;
86 if(p->name == NULL)
87 return;
88 if(p->sex == 'M')
89 printf(" %s is male.", p->name);
90 else
91 printf(" %s is female.",p->name);
92 if(p->father != NULL)
93 printf(" %s's father is %s.",p->name,p->father->name);
94 if(p->mother != NULL)
95 printf(" %s's mother is %s.",p->name,p->mother->name);
96 printf("\n");
97 if(p->mate != NULL)
98 if(p->sex == 'M')
99 printf(" His wife is %s.", p->mate->name);
100 else
101 printf(" Her husband is %s.",p->mate->name);
102 if(p->childern != NULL)
103 {
104 for(index=0; index<CHILDREN-1; index++)
105 if(p->childern[index]==NULL)
106 break;
107 if(index>0)
108 printf(" Children are:");
109 for(i=0; i<index; i++)
110 printf(" %s",p->childern[i]->name);
111 }
112 printf("\n");
113 }
114
115 int main()
116 {
117 char *name[8] = {"John","Kate","Maggie","Herry","Jason","Peter","Marry","Jenny"};
118 char male='M', female='F';
119 struct person *pGrandfather, *pFather1, *pFather2, *pMother1, *pMother2;
120 struct person *pSon, *pDaughter, *pCousin;
121
122 pGrandfather = newperson(name[0],male);
123 pFather1 = newperson(name[3],male);
124 pFather2 = newperson(name[4],male);
125 pMother1 = newperson(name[1],female);
126 pMother2 = newperson(name[2],female);
127 pSon = newperson(name[5],male);
128 pDaughter = newperson(name[6],female);
129 pCousin = newperson(name[7],female);
130
131 father_child(pGrandfather,pFather1);
132 father_child(pGrandfather,pFather2);
133 father_child(pFather1,pSon);
134 father_child(pFather1,pDaughter);
135 father_child(pFather2,pCousin);
136
137 mate(pFather1,pMother1);
138 mate(pFather2,pMother2);
139 mother_child(pMother1,pSon);
140 mother_child(pMother1,pDaughter);
141 mother_child(pMother2,pCousin);
142
143 print_relate(pGrandfather);
144 print_relate(pFather1);
145 print_relate(pFather2);
146 print_relate(pMother1);
147 print_relate(pMother2);
148 print_relate(pSon);
149 print_relate(pDaughter);
150 print_relate(pCousin);
151
152
153 if(!brothersinlaw(pDaughter,pCousin))
154 printf("%s and %s are not brothers (sisters) in law.\n",pDaughter->name,pCousin->name);
155 else
156 printf("%s and %s are brothers (sisters) in law.\n",pDaughter->name,pCousin->name);
157 if(!brothersinlaw(pSon,pCousin))
158 printf("%s and %s are not brothers (sisters) in law.\n",pSon->name,pCousin->name);
159 else
160 printf("%s and %s are brothers (sisters) in law.\n",pSon->name,pCousin->name);
161 if(!brothersinlaw(pSon,pDaughter))
162 printf("%s and %s are not brothers (sisters) in law.\n",pSon->name,pDaughter->name);
163 else
164 printf("%s and %s are brothers (sisters) in law.\n",pSon->name,pDaughter->name);
165
166 //printf("Hello world!\n");
167 return 0;
168 }
分类:
C语言实例解析精粹学习笔记
实例：
设计一个模拟社会关系的数据结构，每个人的信息用结构表示，包含名字、性别和指向父亲、母亲、配偶、子女的指针（只限两个子女）。要求编写以下函数：
（1）增加一个新人的函数
（2）建立人与人之间关系的函数：父-子、母-子、配偶等。
（3）检查两人之间是否为堂兄妹
思路解析：
能够充分的联系指针的应用。书中的代码在增加一个新人时，只为新人提供名字和性别，关于新人的其他信息通过调用其他函数建立。
书中代码如下：
1 #include <stdio.h>
2 #include <stdlib.h>
3 #include <string.h>
4
5 #define CHILDREN 2
6
7 struct person{
8 char *name; /*名字符串指针*/
9 char sex; /*性别：男用字符'M'；女用字符'F'*/
10 struct person *father; /*指向父亲*/
11 struct person *mother; /*指向母亲*/
12 struct person *mate; /*指向配偶*/
13 struct person *childern[CHILDREN];/*指向子女*/
14 };
15
16 /* [函数]newperson增加新人 */
17 struct person *newperson(char *name, char sex)
18 {
19 struct person *p;
20 int index;
21
22 p = (struct person *)malloc(sizeof(struct person));
23 p->name = (char *)malloc(strlen(name)+1);
24 strcpy(p->name, name);
25 p->sex = sex;
26 p->father = NULL;
27 p->mother = NULL;
28 p->mate = NULL;
29 for(index=0; index<CHILDREN; index++)
30 p->childern[index] = NULL;
31 return p;
32 }
33
34 /* [函数]father_child建立父－子关系 */
35 void father_child(struct person *father, struct person *child)
36 {
37 int index;
38
39 for(index=0; index<CHILDREN-1; index++)/*寻找一个空缺的子女指针*/
40 if(father->childern[index]==NULL) /*若没有空缺，则填在最后*/
41 break;
42 father->childern[index] = child; /*建立父－子关系*/
43 child->father = father;
44 }
45
46 void mother_child(struct person *mother, struct person *child)
47 {
48 int index;
49 for(index=0; index<CHILDREN-1; index++)/*寻找一个空缺的子女指针*/
50 if(mother->childern[index]==NULL) /*若没有空缺，则填在最后*/
51 break;
52 mother->childern[index] = child; /*建立母－子关系*/
53 child->mother = mother;
54 }
55
56 /* [函数]mate 建立配偶关系 */
57 void mate(struct person *h, struct person *w)
58 {
59 h->mate = w;
60 w->mate = h;
61 }
62
63 /* [函数]brotherinlow 检查两人是否是堂兄妹 */
64 int brothersinlaw(struct person *p1, struct person *p2)
65 {
66 struct person *f1, *f2;
67 if(p1==NULL||p2==NULL||p1==p2)
68 return 0;
69 if(p1->sex==p2->sex) return 0;/*不可能是堂兄妹*/
70 f1 = p1->father;
71 f2 = p2->father;
72 if(f1!=NULL && f1==f2) return 0;/*是兄妹，不是堂兄妹*/
73 while(f1!=NULL&&f2!=NULL&&f1!=f2)/*考虑远房情况*/
74 {
75 f1 = f1->father;
76 f2 = f2->father;
77 if(f1!=NULL && f2!=NULL && f1==f2) return 1;
78 }
79 return 0;
80 }
81
82 /* 函数print_relate用于输出人物p的姓名，性别和各种关系 */
83 void print_relate(struct person *p)
84 {
85 int index,i;
86 if(p->name == NULL)
87 return;
88 if(p->sex == 'M')
89 printf(" %s is male.", p->name);
90 else
91 printf(" %s is female.",p->name);
92 if(p->father != NULL)
93 printf(" %s's father is %s.",p->name,p->father->name);
94 if(p->mother != NULL)
95 printf(" %s's mother is %s.",p->name,p->mother->name);
96 printf("\n");
97 if(p->mate != NULL)
98 if(p->sex == 'M')
99 printf(" His wife is %s.", p->mate->name);
100 else
101 printf(" Her husband is %s.",p->mate->name);
102 if(p->childern != NULL)
103 {
104 for(index=0; index<CHILDREN-1; index++)
105 if(p->childern[index]==NULL)
106 break;
107 if(index>0)
108 printf(" Children are:");
109 for(i=0; i<index; i++)
110 printf(" %s",p->childern[i]->name);
111 }
112 printf("\n");
113 }
114
115 int main()
116 {
117 char *name[8] = {"John","Kate","Maggie","Herry","Jason","Peter","Marry","Jenny"};
118 char male='M', female='F';
119 struct person *pGrandfather, *pFather1, *pFather2, *pMother1, *pMother2;
120 struct person *pSon, *pDaughter, *pCousin;
121
122 pGrandfather = newperson(name[0],male);
123 pFather1 = newperson(name[3],male);
124 pFather2 = newperson(name[4],male);
125 pMother1 = newperson(name[1],female);
126 pMother2 = newperson(name[2],female);
127 pSon = newperson(name[5],male);
128 pDaughter = newperson(name[6],female);
129 pCousin = newperson(name[7],female);
130
131 father_child(pGrandfather,pFather1);
132 father_child(pGrandfather,pFather2);
133 father_child(pFather1,pSon);
134 father_child(pFather1,pDaughter);
135 father_child(pFather2,pCousin);
136
137 mate(pFather1,pMother1);
138 mate(pFather2,pMother2);
139 mother_child(pMother1,pSon);
140 mother_child(pMother1,pDaughter);
141 mother_child(pMother2,pCousin);
142
143 print_relate(pGrandfather);
144 print_relate(pFather1);
145 print_relate(pFather2);
146 print_relate(pMother1);
147 print_relate(pMother2);
148 print_relate(pSon);
149 print_relate(pDaughter);
150 print_relate(pCousin);
151
152
153 if(!brothersinlaw(pDaughter,pCousin))
154 printf("%s and %s are not brothers (sisters) in law.\n",pDaughter->name,pCousin->name);
155 else
156 printf("%s and %s are brothers (sisters) in law.\n",pDaughter->name,pCousin->name);
157 if(!brothersinlaw(pSon,pCousin))
158 printf("%s and %s are not brothers (sisters) in law.\n",pSon->name,pCousin->name);
159 else
160 printf("%s and %s are brothers (sisters) in law.\n",pSon->name,pCousin->name);
161 if(!brothersinlaw(pSon,pDaughter))
162 printf("%s and %s are not brothers (sisters) in law.\n",pSon->name,pDaughter->name);
163 else
164 printf("%s and %s are brothers (sisters) in law.\n",pSon->name,pDaughter->name);
165
166 //printf("Hello world!\n");
167 return 0;
168 }