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;
}

 

结论：

int型数组a：在内存中不是连续存放的，每个元素占用4个字节单元；char型数组b：在内存中是连续存放的，每个元素占用1个字节单元。

数组名a对应的值，和&a[0]是一样的，数组名b对应的值，和&b[0]是不一样的。

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: %c\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;
}

结论：

int型二位数组a，在内存中是“按行连续存放”的，每个元素占用4个字节单元，数组名a的值和&a[0][0]的值，在数字字面值上是一样的。

char型二维数组b，在内存中是“按行连续存放”的，每个元素占用1个内存字节单元，数组名b的值和&b0][0]的值，在数字字面值上是一样的。

a[0],a[1]和b[0],b[1]的值：a[][]一排有三个元素，一个元素占4个字节，因此a[0]与a[1]差12个字节；b[][]一排也有三个元素，而字符型数组一个元素占1个字节，因此b[0]与b[1]差3个字节。

task2:

#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);
    
}
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);
}

结论：

test1的是两个不同的数组，引用时不用加[]

test2的是一个数组里的两行句子，引用时要写[]

task3_1:

#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; 
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;
}

task3_2：

#include <stdio.h>
#define N 1000
int main() {
char line[N];
int word_len; 
int max_len; 
int end; 
int i;
while(gets(line) != NULL) {
word_len = 0;
max_len = 0;
end = 0;
i = 0;
while(1) {

while(line[i] == ' ') {
word_len = 0;
i++;
}

while(line[i] != '\0' && line[i] != ' ') {
word_len++;
i++;
}

if(max_len < word_len) {
max_len = word_len;
end = i; 
}

if(line[i] == '\0')
break;
}

printf("最长单词: ");
for(i = end - max_len; i < end; ++i)
printf("%c", line[i]);
printf("\n\n");
}
return 0;
}

task4:

#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输出课程分数: \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输出课程分数(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){
    double ave,sum = 0;
    int i;
    for(i=0;i<n;i++){
        sum = sum+x[i];
    }
    ave=sum/n;
    return ave;
}
void bubble_sort(int x[], int n)
{int i,j;
for(i = 0;i<n-1;i++){
    for(j = 0;j<n-1-i;j++)
    if(x[j]<x[j+1]){
        int temp = x[j];
        x[j] = x[j+1];
        x[j+1] = temp;
    }
}
    
}

 

 task5:

#include <stdio.h>
#define N 100
void dec2n(int x, int n); // 函数声明
int main() 
{
    int x;
    printf("输入一个十进制整数: ");
    while(scanf("%d", &x) != EOF) 
    {
        dec2n(x, 2); // 函数调用: 把x转换成二进制输出
        dec2n(x, 8); // 函数调用: 把x转换成八进制输出
        dec2n(x, 16); // 函数调用: 把x转换成十六进制输出
        printf("\n输入一个十进制整数: ");
    }
    return 0;
}
void dec2n(int x, int n)
{
    if(n==2)
        {
            int a[33]={0},i=0;
            while(x>0)
            {
                a[i]=x%2;
                i++;
                x/=2;
            }
            for(i=i-1;i>=0;i--)
                printf("%d",a[i]);
            printf("\n");
        }
    if(n==8)
        printf("%o\n",x);
    if(n==16)
        printf("%x\n",x);
}

task6：

#include <stdio.h>
#define N 100
#define M 4
void output(int x[][N], int n); 
void rotate_to_right(int x[][N], int n); 
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;
    
    for(i = 0;i < n; i++) {
        int 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;
}

结论：

replace的功能：用new_char代替old_char

'\0'：表示句子结束

 task7_2:

#include <stdio.h>

#define N 80

void delet(char x[], char ch);

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

    printf("输入一个字符串：");
    gets(text);
    
    printf("输入一个字符："); 
    ch = getchar();
    
    printf("原始文本: \n");
    printf("%s\n", text);
    
    printf("截断处理...\n");
    delet(text, ch);
    
    printf("处理后文本: \n");
    printf("%s\n", text);
    
    return 0;
}

void delet(char x[], char ch) {
    int i, t = 0;
    
    for(i = 0; x[i] != '\0'; i++)
        if(x[i] != ch)
            x[t++] = x[i];
        else break ;
    
    x[t] = '\0';
}

task8:

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