实验4

TASK1_1

#include<stdio.h>
#define N 4

int main(){
    int a[N]={2,0,2,3};
    char b[N]={'2','0','2','3'};
    int i;
    printf("sizeof(int) = %d\n",sizeof(int));
    printf("sizeof(char)=%d\n",sizeof(char));
    printf("\n");
    
    for(i=0;i<N;i++)
    printf("%p:%d\n",&a[i],a[i]);
    printf("\n");
    
    for(i=0;i<N;++i)
    printf("%p:%c\n",&b[i],b[i]);
    
    printf("\n");
    
    printf("a=%p\n",a);
    printf("b=%p\n",b);
    
    return 0;
}

1.int型数组连续存放，每个元素占用4个字节；

2.char型数组连续存放，每个元素占用1个字节；

3.一样

task1_2

#include<stdio.h>
#define N 2
#define M 3

int main(){
    int a[N][M]={{1,2,3},{4,5,6}};
    char b[N][M]={{'1','2','3'},{'4','5','6'}};
    int i,j;
    
    for (i=0;i<N;++i)
    for(j=0;j<M;++j)
    printf("%p:%d\n",&a[i][j],a[i][j]);
    
    printf("\n");
    
    printf("a=%p\n",a);
    printf("a[0]=%p\n",a[0]);
    printf("a[1]=%p\n",a[1]);
    printf("\n");
    
    for(i=0;i<N;++i)
    for(j=0;j<M;++j)
    printf("%p:%d\n",&b[i][j],b[i][j]);
    printf("\n");
    
    printf("b=%p\n",b);
    printf("b[0]=%p\n",b[0]);
    printf("b[1]=%p\n",b[1]);
    printf("\n");
    
    return 0;
    
}

1.是。4个

2.a与a[0]相同，与a[1]不同

3.是。1个

4.b与b【0】在数值上是一样的，与b[1]不同

5.他们表示的都是本列数组的第一个字符的地址

 

TESA 2

#include<stdio.h>

#include<string.h>
#define N 80

void swap_str(char s1[N],char s2[N]);
void test1();
void test2();

int main(){
 printf("测试1：用两个一维数组，实现两个字符串交换\n");
 test1();
 
 printf("\n测试：用二维数组，实现两个字符串交换\n");
 test2();
 
 return 0;
}

void test1(){
 char views1[N] = "hey,C,I hate u.";
 char views2[N] = "hey,C,I love u.";
 
 printf("交换前：\n");
 puts(views1);
 puts(views2);
 
 swap_str(views1,views2);
 
 printf("交换后：\n");
 puts(views1);
 puts(views2);
}

void test2(){
 char views[2][N] = {"hey,C, I hate u","hey,C,I love u."};
 
 printf("交换前：\n");
 puts(views[0]);
 puts(views[1]);
 
 swap_str(views[0],views[1]);
 
 printf("交换后：\n");
 puts(views[0]);
 puts(views[1]);
}

void swap_str(char s1[N],char s2[N]){
 char tmp[N];
 
 strcpy(tmp,s1);
 strcpy(s1,s2);
 strcpy(s2,tmp);
}

因为一维数组实参不加[]表示该一维数组地址，而二维数组加中括号表示第一行字符串的地址，两者在数值上相同

TASK 3_2

#include<stdio.h>
#define N 80

int count(char x[]);

int main(){
 char words[N+1];
 int n;
 
 while(gets(words)!=NULL){
 n=count(words);
 printf("单词数：%d\n\n",n);
}
return 0;
}
int count(char x[]){
 int i;
 int word_flag=0;//标记单词开始为1，单词结束为0
 int number = 0;//标记单词个数
 
 for(i=0;x[i]!='\0';i++){
  if(x[i]==' ')
  word_flag=0;
  else if(word_flag==0){
  word_flag=1;
  number++;}
 }

return number;
}

TASK3.2

#include<stdio.h>

#define N 100
int count(char x[]);

int main(){
 char words[N+1];
 int n;
 
 while(gets(words)!=NULL){
  n=count(words);
     printf("单词数:%d\n\n",n);
 }
 
 return 0;
 
}

int count(char x[]){
 int i;
 int word_flag=0;
 int number =0;
 
 for(i=0;x[i]!='\0';i++){
  if(x[i]==' ')
  word_flag=0;
  else if(word_flag==0){
   word_flag=1;
   number++;
  }
 }
 return number;
}

TASK 4

#include<stdio.h>
#define N 5

void input(int x[],int n);
void output(int x[],int n);
double average(int x[],int n);
void bubble_sort(int x[],int n);

int main(){
int scores[N];
double ave;

printf("录入%d个分数:\n",N);
input(scores,N);

printf("\n输出课程数据:\n");
output(scores,N);

printf("\n课程分数处理:计算均分、排序...\n");
ave = average(scores,N);
bubble_sort(scores,N);

printf("\n输出课程均分:%.2f\n",ave);
printf("\n输出课程均分（高-->低）：\n");
output(scores,N);

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");
}

double average(int x[],int n){
int i;
double sum,ave;
sum=0;

for (i=0;i<n;i++)
sum+=x[i];

ave=sum/n;
return ave;
}

void bubble_sort(int x[],int n){
int i,t;
int m;

for(t=n;t>0;t--)
{for(i=0;i<t;i++)
if(x[i]<x[i+1])
{m=x[i+1];
x[i+1]=x[i];
x[i]=m;
}
}
}

 

TASK 5

#include<stdio.h>
#define N 100
void dec2n(int x,int n);

int main(){
int x;

printf("输入一个十进制整数:");
while(scanf("%d",&x)!=EOF){
dec2n(x,2);
dec2n(x,8);
dec2n(x,16);

printf("输入一个十进制整数:");
}
return 0;
}

void dec2n(int x, int n){
char a[]="0123456789ABCDEF",b[N];
int i,t=0,y;

for(i=0;x!=0;i++){
y=x%n;
b[i]=a[y];
x=x/n;
t++;
}

for(i=t-1;i>=0;i--)
printf("%c",b[i]);

printf("\n");
}

TASK 6

 

#include <stdio.h>
#define N 100
#define M 4

void output(int x[][N], int n); // 函数声明
void rotate_to_right(int x[][N], int n); // 函数声明

int main() {
int t[][N] = {{21, 12, 13, 24},
{25, 16, 47, 38},
{29, 11, 32, 54},
{42, 21, 33, 10}};

printf("原始矩阵:\n");
output(t, M);

rotate_to_right(t, M);
printf("变换后矩阵:\n");
output(t, M);

return 0;
}

void output(int x[][N], int n) {
int i, j;
for (i = 0; i < n; ++i) {
for (j = 0; j < n; ++j)
printf("%4d", x[i][j]);
printf("\n");
}
}
void rotate_to_right(int x[][N],int n){
int i,j,t;
for(i=0;i<n;i++){
t=x[i][n-1];
for(j=n-1;j>0;j--)
x[i][j]=x[i][j-1];
x[i][0]=t;
}
}

TASK7.1

#include<stdio.h>
#define N 80

void replace(char x[], char old_char, char new_char);

int main() {
char text[N] = "c 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 x[], char old_char, char new_char) {
int i;

for (i = 0; x[i] != '\0'; ++i)
if (x[i] == old_char)
x[i] = new_char;
}

1.replace功能为把所有的i换成*

2.\0是字符串结束的标志。当字符串结束，循环结束

TASK7.2

#include <stdio.h>
#define N 80

int main() {
char str[N], ch;
int i;

printf("输入字符串: ");
gets(str);

printf("输入一个字符: ");
ch = getchar();

printf("截断处理......");

i = 0;
while (str[i] != '\0') {
if (str[i] == ch)
str[i]='\0';
else
i++;

}

printf("\n截断处理后字符串: %s\n", str);

return 0;
}

TASK 8

 

#include <stdio.h>
#include <string.h>

#define N 5
#define M 20

void bubble_sort(char str[][M], int n);

int main() {
char name[][M] = {"Bob", "Bill", "Joseph", "Taylor", "George"};
int i;

printf("输出初始名单:\n");
for (i = 0; i < N; i++)
printf("%s\n", name[i]);

printf("\n排序中...\n");
bubble_sort(name, N);

printf("\n按字典序输出名单:\n");
for (i = 0; i < N; i++)
printf("%s\n", name[i]);

return 0;
}

void bubble_sort(char str[][M],int n){
int i,j;
char t[20];

for(j=0;j<n;j++)
{for(i=0;i<n-j-1;i++)
if(strcmp(str[i],str[i+1])>0)
{strcpy(t,str[i]);
strcpy(str[i],str[i+1]);
strcpy(str[i+1],t);}
}
}