任务1
代码1.
1 #include<stdio.h>
2 #define N 5
3 void input(int x[],int n);
4 void output(int x[],int n);
5 void fin_min_max(int x[],int n,int *pmin,int *pmax);
6
7 int main(){
8 int a[N];
9 int min,max;
10
11 printf("录入%d个数据：\n",N);
12 input(a,N);
13
14 printf("数据是:\n");
15 output(a,N);
16
17 printf("数据处理...\n");
18 find_min_max(a,N,&min,&max);
19
20 printf("输出结果：\n");
21 printf("min=%d,max=%d\n",min,max);
22
23 return 0;
24 }
25
26 void input(int x[],int n){
27 int i;
28 for(i=0;i<n;++i)
29 scanf("%d",&x[i]);
30 }
31
32 void output(int x[],int n){
33 int i;
34 for(i=0;i<n;++i)
35 printf("%d",x[i]);
36 printf("\n");
37 }
38
39 void find_min_max(int x[],int n,int *pmin, int *pmax){
40 int i;
41 *pmin=*pmax=x[0];
42
43 for(i=0;i<n;++i)
44 if(x[i]<*pmin)
45 *pmin=x[i];
46 else if(x[i]>*pmax)
47 *pmax=x[i];
48
49 }
View Code
运行结果
1.find_min_max功能是找到五个数据中的最大值和最小值
2,pmin,pmax指向x[0]的地址
代码1.2
1 #include<stdio.h>
2 #define N 5
3 void input(int x[],int n);
4 void output(int x[],int n);
5 int *find_max(int x[],int n);
6
7 int main(){
8 int a[N];
9 int *pmax;
10
11 printf("录入%d个数据：\n",N);
12 input(a,N);
13
14 printf("数据是:\n");
15 output(a,N);
16
17 printf("数据处理...\n");
18 pmax=find_max(a,N);
19
20 printf("输出结果：\n");
21 printf("max=%d\n",*pmax);
22
23 return 0;
24 }
25
26 void input(int x[],int n){
27 int i;
28 for(i=0;i<n;++i)
29 scanf("%d",&x[i]);
30 }
31
32 void output(int x[],int n){
33 int i;
34 for(i=0;i<n;++i)
35 printf("%d",x[i]);
36 printf("\n");
37 }
38
39 int *find_max(int x[],int n){
40 int max_index=0;
41 int i;
42
43 for(i=0;i<n;++i)
44 if(x[i]>x[max_index])
45 max_index=i;
46
47 return &x[max_index];
48 }
View Code
运行结果
1.find_max的功能返回最大值的地址；
2.可以实现,ptr为指针可以储存最大值的地址；
任务2
代码2.1
1 #include<stdio.h>
2 #include<string.h>
3 #include<stdlib.h>
4 #define N 80
5
6 int main(){
7 char s1[N] = "Learning makes me happy";
8 char s2[N] = "Learning nakes me sleep";
9 char tmp[N];
10
11 printf("sizeof(s1) vs. strlen(s1): \n");
12 printf("sizeof(s1) = %d\n",sizeof(s1));
13 printf("strlen(s1) = %d\n",strlen(s1));
14
15 printf("\nbefore swap: \n");
16 printf("s1: %s\n", s1);
17 printf("s2: %s\n", s2);
18
19 printf("\nswapping...\n");
20 strcpy(tmp, s1);
21 strcpy(s1, s2);
22 strcpy(s2, tmp);
23
24 printf("\nafter swap: \n");
25 printf("s1: %s\n", s1);
26 printf("s2: %s\n", s2);
27
28 system("pause");
29 return 0;
30 }
View Code
结果
1.sizeof(s1)计算的是预处理命令中N决定的，为80字节，sizeof(s1)计算的s1占据的内存空间，strlen(s1)统计的是字符串的长度；
2.s1是数组名常量，不能修改；
3.用中间量tmp,将字符数组s1,s2中的内容进行了交换；
代码2.2
1 #include<stdio.h>
2 #include<string.h>
3 #include<stdlib.h>
4 #define N 80
5
6 int main(){
7 char *s1 = "Learning makes me happy";
8 char *s2= "Learning nakes me sleep";
9 char *tmp;
10
11 printf("sizeof(s1) vs. strlen(s1): \n");
12 printf("sizeof(s1) = %d\n",sizeof(s1));
13 printf("strlen(s1) = %d\n",strlen(s1));
14
15 printf("\nbefore swap: \n");
16 printf("s1: %s\n", s1);
17 printf("s2: %s\n", s2);
18
19 printf("\nswapping...\n");
20 tmp = s1;
21 s1 = s2;
22 s2 = tmp;
23
24 printf("\nafter swap: \n");
25 printf("s1: %s\n", s1);
26 printf("s2: %s\n", s2);
27
28 system("pause");
29 return 0;
30 }
View Code
1.s1是指针变量，存放的是首地址，sizeof(s1)计算的s1占据的内存空间，strlen(s1)统计的是字符串的长度;
2.可以修改，指针变量可以赋值；
3.交换的是地址；
task3.
代码3
1 #include<stdio.h>
2
3 int main(){
4 int x[2][4] = {{1,9,8,4},{2,0,4,9}};
5 int i,j;
6 int *ptr1;
7 int(*ptr2)[4];
8
9 printf("输出1：使用数组名、下标直接访问二维数组元素\n");
10 for(i=0;i<2;++i){
11 for(j=0;j<4;++j)
12 printf("%d",x[i][j]);
13 printf("\n");
14 }
15
16 printf("\n输出2：使用指针变量ptr1(指向元素）间接访问\n");
17 for(ptr1=&x[0][0],i=0;ptr1<&x[0][0]+8;++ptr1,++i){
18 printf("%d",*ptr1);
19
20 if((i+1)%4==0)
21 printf("\n");
22 }
23
24 printf("\n输出3：使用指针变量ptr2(指向一维数组）间接访问\n");
25 for(ptr2=x;ptr2<x+2;++ptr2){
26 for(j=0;j<4;++j)
27 printf("%d",*(*ptr2+j));
28 printf("\n");
29 }
30
31 return 0;
32 }
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任务4
1 #include<stdio.h>
2 #define N 80
3
4 void replace(char *str,char old_char,char new_char);
5
6 int main(){
7 char text[N]="Programming is difficult or not, it is a question.";
8
9 printf("原始文本：\n");
10 printf("%s\n",text);
11
12 replace(text,'i','*');
13
14 printf("处理后文本：\n");
15 printf("%s\n",text);
16
17 return 0;
18 }
19 void replace(char *str,char old_char,char new_char){
20 int i;
21
22 while(*str){
23 if(*str==old_char)
24 *str=new_char;
25 str++;
26 }
27 }
图片
1.replace作用将old_char全部转化为new_char;
2.可以，两者是等价的
task5
1 #include <stdio.h>
2 #define N 80
3
4 char *str_trunc(char *str,char x);
5
6 int main(){
7 char str[N];
8 char ch;
9
10 while(printf("输入字符串："),gets(str)!=NULL){
11 printf("输入一个字符：");
12 ch =getchar();
13
14 printf("截断处理...\n");
15 str_trunc(str,ch);
16
17 printf("截断处理后的字符串：%s\n\n",str);
18 getchar();
19
20 }
21 return 0;
22 }
23
24 char *str_trunc(char *str,char x){
25 int i,j;
26 for (i=0;i<N;i++){
27 if(str[i]==x){
28 str[i]='\0';
29 break;
30 }
31 }
32 return str;
33 }
图片
删去line18行后
读取确认的回车键‘\n’，不然‘\n’会被下一次循环读取；
task6
1 #include<stdio.h>
2 #include<string.h>
3 #define N 5
4
5 int check_id(char *str);
6
7 int main()
8 {
9 char *pid[N] ={"31010120000721656X",
10 "3301061996X0203301",
11 "53010220051126571",
12 "510104199211197977",
13 "53010220051126133Y"};
14 int i;
15
16 for(i=0;i<N;++i)
17 if(check_id(pid[i]))
18 printf("%s\tTrue\n",pid[i]);
19 else
20 printf("%s\tFalse\n",pid[i]);
21
22 return 0;
23
24 }
25
26 int check_id(char *str) {
27 int len = strlen(str);
28 if (len != 18) {
29 return 0;
30 }
31 int i;
32 for ( i = 0; i < 17; ++i) {
33 if (!isdigit(str[i])) {
34 return 0;
35 }
36 }
37 if (isdigit(str[17]) || (toupper(str[17]) == 'X')) {
38 return 1;
39 } else {
40 return 0;
41 }
42 }
图片
task7
1 #include <stdio.h>
2 #define N 80
3 void encoder(char *str, int n);
4 void decoder(char *str, int n);
5
6 int main() {
7 char words[N];
8 int n;
9
10 printf("输入英文文本: ");
11 gets(words);
12
13 printf("输入n: ");
14 scanf("%d", &n);
15
16 printf("编码后的英文文本: ");
17 encoder(words, n);
18 printf("%s\n", words);
19
20 printf("对编码后的英文文本解码: ");
21 decoder(words, n);
22 printf("%s\n", words);
23
24 return 0;
25 }
26
27
28 void encoder(char *str, int n) {
29 while(*str!='\0'){
30 if(*str >= 'a'&&*str <= 'z')
31 {
32 *str = (*str-'a'+n)%26+'a';
33 }
34 if(*str >= 'A'&&*str <= 'Z')
35 {
36 *str = (*str-'A'+n)%26+'A';
37 }
38 str++;
39 }
40 }
41
42
43 void decoder(char *str, int n) {
44 while(*str!='\0'){
45 if(*str >= 'a'&&*str <= 'z')
46 {
47 *str = (*str-'a'-n+26)%26+'a';
48 }
49 if(*str >= 'A'&&*str <= 'Z')
50 {
51 *str = (*str-'A'-n+26)%26+'A';
52 }
53 str++;
54 }
55 }
图片
task8
1 #include<stdio.h>
2
3 int main(int argc, char *argv[]){
4 int i;
5
6 for(i=1;i<argc;++i)
7 printf("hello,%s\n",argv[i]);
8
9 return 0;
10 }
task8.1
1 #include<stdio.h>
2 #include<string.h>
3
4 void sort(int argc, char *argv[]);
5
6 int main(int argc, char *argv[]){
7 int i;
8
9 sort(argc,argv);
10
11 for(i=1;i<argc;++i)
12 printf("hello,%s\n",argv[i]);
13
14 return 0;
15 }
16
17 void sort(int argc, char *argv[]){
18 int i,j;
19 char *temp;
20 for(i = 0; i < argc-1 ; i++ ){
21 for(j = 0; j< argc-1-i; j++){
22 if(strcmp(argv[j],argv[j+1])>0){
23 temp = argv[j];
24 argv[j] = argv[j+1];
25 argv[j+1] = temp;
26 }
27 }
28 }
29 }
图片
标签：int,s1,char,实验,str,printf,include From： https://www.cnblogs.com/YamadaRyo1314/p/18580980