任务1
task1.1
#include <stdio.h>
#define N 5void input (int x[],int n);
void output(int x[],int n);
void find_min_max(int x[],int n,int *pmin,int *pmax);int main()
{int a[N];int min,max;printf("录入%d个数据：\n",N);input(a,N);printf("数据是：\n");output(a,N);printf("数据处理...\n");find_min_max(a,N,&min,&max);printf("输出结果：\n");printf("min=%d,max=%d\n",min,max);return 0;
}void input (int x[],int n)
{int i;for(i=0;i<n;++i)scanf("%d",&x[i]);}void output(int x[],int n)
{int i;for(i=0;i<n;++i)printf("%d",x[i]);printf("\n");}void find_min_max(int x[],int n,int *pmin,int *pmax)
{int i;*pmin=*pmax=x[0];for(i=0;i<n;++i)if(x[i]<*pmin)*pmin=x[i];else if(x[i]>*pmax)*pmax=x[i];
}
问题1
功能是找到输入数据中的最小值和最大值
问题2
pmin指向变量min的地址，pmax指向变量max的地址
task1.2
#include <stdio.h>
#define N 5void input(int x[], int n);
void output(int x[], int n);
int *find_max(int x[], int n);int main()
{int a[N];int *pmax;printf("录入%d个数据:\n", N);input(a, N);printf("数据是: \n");output(a, N);printf("数据处理...\n");pmax=find_max(a, N);printf("输出结果:\n");printf("max = %d\n",*pmax);return 0;}void input(int x[], int n)
{int i;for(i = 0; i < n; ++i)scanf("%d", &x[i]);
}void output(int x[], int n)
{int i;for(i = 0; i < n; ++i)printf("%d", x[i]);printf("\n");
}int *find_max(int x[], int n)
{int max_index=0;int i;for(i = 0; i < n; ++i)if(x[i] > x[max_index])max_index=i;return &x[max_index];
}
问题1
功能是找到输入数据中最大数所存储的地址，返回最大值的地址
问题2
可以
任务2
task2.1
#include<stdio.h>
#include<string.h>
#define N 80int main()
{char s1[N]="Learning makes me happy";char s2[N]="Learning makes me sleepy";char tmp[N];printf("sizeof(s1) vs. strlen(s1):\n");printf("sizeof(s1)=%d\n",sizeof(s1));printf("strlen(s1)=%d\n",strlen(s1));printf("\nbefore swap:\n");printf("s1:%s\n",s1);printf("s2:%s\n",s2);printf("\nswapping...\n");strcpy(tmp,s1);strcpy(s1,s2);strcpy(s2,tmp);printf("\nafter swap:\n");printf("s1;%s\n",s1);printf("s2:%s\n",s2);return 0;
}
问题1
大小为80，sizeof计算的是数组s1占用的字节数，strlen统计的是字符串s1的有效字符数
问题2
不能，s1是一个地址常量，不能被赋值
问题3
是
task2.2
#include<stdio.h>
#include<string.h>
#define N 80int main()
{char *s1="Learning makes me happy";char *s2="Learning makes me sleepy";char *tmp;printf("sizeof(s1) vs. strlen(s1):\n");printf("sizeof(s1)=%d\n",sizeof(s1));printf("strlen(s1)=%d\n",strlen(s1));printf("\nbefore swap:\n");printf("s1:%s\n",s1);printf("s2:%s\n",s2);printf("\nswapping...\n");tmp=s1;s1=s2;s2=tmp;printf("\nafter swap:\n");printf("s1:%s\n",s1);printf("s2:%s\n",s2);return 0;}
问题1
存放的是字符串”Learning makes me happy“的首地址，sizeof（s1）计算的是s1指向的字符串所占用字节数，strlen（s1）统计的是s1所指向字符串中的有效字符数
问题2
可以，task2.1中的line6意思为将字符串存入字符数组s1中，task2.2中的line6意思为将字符串的首地址纯存入指针变量s1中
问题3
交换的是s1和s2所存储的地址，没有交换
任务3
#include<stdio.h>int main()
{int x[2][4]={{1,9,8,4},{2,0,4,9}};int i,j;int *ptr1;int (*ptr2)[4];printf("输出1：使用数组名、下标直接访问二维数组元素\n");for(i=0;i<2;++i){for(j=0;j<4;++j)printf("%d ",x[i][j]);printf("\n");}printf("\n输出2：使用指针变量ptr1（指向元素）间接访问\n");for(ptr1=&x[0][0],i=0;ptr1<&x[0][0]+8;++ptr1,++i){printf("%d ",*ptr1);if((i+1)%4==0)printf("\n");}printf("\n输出3：使用指针变量ptr2（指向一维数组）间接访问\n");for(ptr2=x;ptr2<x+2;++ptr2){for(j=0;j<4;++j)printf("%d ",*(*ptr2+j));printf("\n");}return 0;
}
任务4
#include<stdio.h>
#define N 80void replace(char *str,char old_char,char new_char);int main()
{char text[N]="Programming is difficult or not,it is a question.";printf("原始文本：\n");printf("%s\n",text);replace(text,'i','*');printf("处理后文本：\n");printf("%s\n",text);return 0;
}void replace(char *str,char old_char,char new_char)
{int i;while(*str){if(*str==old_char)*str=new_char;str++;}
}
问题1
功能是将字符串中的字母i替换成*
问题2
可以
任务5
#include<stdio.h>
#define N 80char *str_trunc(char *str,char x);int main()
{char str[N];char ch;while(printf("输入字符串："),gets(str)!=NULL){printf("输入一个字符：");ch=getchar();printf("截断处理...\n");str_trunc(str,ch);printf("截断处理后的字符串：%s\n\n",str);getchar();}return 0;
}char *str_trunc(char *str,char x)
{char *p=str;while(*str){if(*str==x){*str='\0';break;}elsestr++;}return p;
}
问题
第二次要输入字符串时会把上次输入时按的回车当作字符串，直接进行输入字符的命令
作用是储存输入完成时按的回车，不让其被当作输入字符存入字符串执行命令
任务6
#include <stdio.h>
#include <string.h>
#define N 5
int check_id(char *str);int main()
{char *pid[N]={"31010120000721656X","3301061996X0203301","53010220051126571","510104199211197977","53010220051126133Y"};int i;for(i=0;i<N;++i){if(check_id(pid[i]))printf("%s\tTrue\n",pid[i]);elseprintf("%s\tFalse\n",pid[i]);}return 0;
}int check_id(char *str)
{if(strlen(str)!=18)return 0;while(*str){if(*str>'9'&&*str!='X'||*str<'0')return 0;if(*str=='X'&&*(str+1)!='\0')return 0;str++;}return 1;
}
任务7
#include<stdio.h>
#define N 80
void encoder(char *str,int n);
void decoder(char *str,int n);int main()
{char words[N];int n;printf("输入英文文本：");gets(words);printf("输入n:");scanf("%d",&n);printf("编码后的英文文本：");encoder(words,n);printf("%s\n",words);printf("对编码后的英文文本解码：");decoder(words,n);printf("%s\n",words);return 0;}void encoder(char *str,int n)
{while(*str){if(*str>='A'&&*str<='Z')*str=(*str-'A'+n)%26+'A';else if(*str>='a'&&*str<='z')*str=(*str-'a'+n)%26+'a';else ;str++;}
}void decoder(char *str,int n)
{while(*str){if(*str>='A'&&*str<='Z')*str=(*str-'A'+26-n)%26+'A';else if(*str>='a'&&*str<='z')*str=(*str-'a'+26-n)%26+'a';else ;str++;}
}
任务8
#include <stdio.h>
#include <string.h>
void sort(int n,char *s[]);int main(int argc, char *argv[]) {int i;sort(argc-1,argv+1);for(i = 1; i < argc; ++i)printf("hello, %s\n", argv[i]);return 0;
}
void sort(int n,char *s[])
{int i,j;char *tmp;for(i=0;i<n-1;++i)for(j=0;j<n-1-i;++j)if(strcmp(s[j],s[j+i])>0){tmp=s[j];s[j]=s[j+1];s[j+1]=tmp;}}
