实验四
实验四
task1.1
① int型数组a数组a在内存中是连续存放的,每个元素占用4个内存字节单
② char型数组b数组b在内存中是连续存放的,每个元素占用1个内存字节单元。
③对应的值一样。
#include <stdio.h>
#define N 4
int main()
{
int a[N] = {2, 0, 2, 2};
char b[N] = {'2', '0', '2', '2'};
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;
}
task1.2
①按行连续存放,每个元素占4个内存字节单位
②按行连续存放,每个元素占用1个内存字节单元。
#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");
for (i = 0; i < N; ++i)
for (j = 0; j < M; ++j)
printf("%p: %c\n", &b[i][j], b[i][j]);
return 0;
}
task2
#include<stdio.h>
int days_of_year(int year, int month, int day);
int main()
{
int year, month, day;
int days;
while (scanf("%d%d%d", &year, &month, &day) != EOF)
{
days = days_of_year(year, month, day);
printf("%4d-%02d-%02d是这一年中的第%d天.\n\n", year, month, day, days);
}
return 0;
}
int days_of_year(int year, int month, int day)
{
int n=0,i;
int x[13]={0,31,28,31,30,31,30,31,31,30,31,30,31 };
if ((year%4==0&&year%100!=0)||(year%400==0))
x[2]=29;
for (i=1;i<month;i++)
n=n+x[i];
return n+day;
}
task3
#include <stdio.h>
#define N 5
// 函数声明
void input(int x[], int n);
void output(int x[], int n);
double average(int x[], int n);
void 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);
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;
}
// 函数定义
// 输入n个整数保存到整型数组x中
void input(int x[], int n)
{
int i;
for(i=0; i<n; ++i)
scanf("%d", &x[i]);
}
// 输出整型数组x中n个元素
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 s=0;
for(i=0;i<n;i++)
s+=x[i];
return s/n;
}
void sort(int x[],int n)
{
int i,j,t;
for(i=0;i<n-1;i++)
{
for(j=0;j<n-1-i;j++)
{
if(x[j]<x[j+1])
{
t =x[j];
x[j]=x[j+1];
x[j+1]=t;
}
}
}
}
task4
#include <stdio.h>
void dec2n(int x, int n); // 函数声明
int main()
{
int x;
printf("输入一个十进制整数: ");
scanf("%d", &x);
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
return 0;
}
void dec2n(int x,int n)
{
int a[100];
int i,k,j=0;
while(x!=0)
{
i=x%n;
a[j++]=i;
x=x/n;
}
for(k=j-1;k>=0;k--)
{
if(a[k]>9&&a[k]<16)
printf("%c",a[k]-10+'A');
else
printf("%d",a[k]);
}
printf("\n");
}
task5
#include <stdio.h>
int main (void)
{
int n,i,j;
while( printf("Enter n:"),scanf("%d",&n)!=EOF)
{
int a[n][n];
for (i=0;i<n;i++)
{
for (j=i;j<n;j++)
{
a[i][j]=i+1;
}
}
for (j=0;j<n;j++)
{
for (i=j;i<n;i++)
{
a[i][j]=j+1;
}
}
for (i=0;i<n;i++)
{
for (j=0;j<n;j++)
{
printf("%d ",a[i][j]);
}
printf("\n");
}
printf("\n");
}
return 0;
}
task6
#include<stdio.h>
#define N 80
int main()
{
char views1[N]="hey,c,i hate u.";
char views2[N]="hey,c,i love u.";
int i,j;
char t;
printf("original views:\n");
printf("views1:\n");
for(j=0;j<N;j++)
printf("%c",views1[j]);
printf("\n");
printf("views2:\n");
for(j=0;j<N;j++)
printf("%c",views2[j]);
printf("\n");
printf("swapping...\n");
for(i=0;i<N;i++)
{
t=views1[i];
views1[i]=views2[i];
views2[i]=t;
}
printf("views1:");
for(j=0;j<N;j++)
printf("%c",views1[j]);
printf("\n");
printf("views2:");
for(j=0;j<N;j++)
printf("%c",views2[j]);
}
task 7
#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 a[M];
for(i=0;i<N;i++)
{
for(j=0;j<N-1;j++)
{
if (strcmp(str[j], str[j+1]) > 0)
{
strcpy(a, str[j]);
strcpy(str[j], str[j+1]);
strcpy(str[j+1], a);
}
}
}
}