当前位置 : 主页 > 编程语言 > 其它开发 >

实验四

来源:互联网 收集:自由互联 发布时间:2022-05-30
task 1-1 #include stdio.h#define N 4int 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) pri
task 1-1
#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;
}

image
①int型数组a在内存中连续存放,每个元素占用4个内存字节单元

②char型数组b在内存中连续存放,每个元素占用一个内存字节单元。

③数组名a对应的值和&a[0]一样,数组b对应的值和&b[0]是一样的

task 1-2
#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");

 for (i = 0; i < N; ++i)
     for (j = 0; j < M; ++j)
         printf("%p: %c\n", &b[i][j], b[i][j]);

return 0;
}

image
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。

task 2
#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) != 0) 

{
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;
}

image

task 3
#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 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输入课程分数:\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;
}

void input(int x[], int n)
{
int i;

for (i = 0; i < n; ++i)
	scanf_s("%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 sum = 0, i;
double ave;

for (i = 0; i < n; ++i)
{
	sum += x[i];
}
ave = 1.0*sum / N;
return ave;
}

void sort(int x[], int n)
{
int i, j, temp;
for (i = 1; i < n; i++)
{
	for (j = 0; j < n-1; j++)
	{
		if (x[j] < x[j + 1])
		{
			temp = x[j];
			x[j] = x[j + 1];
			x[j + 1] = temp;
		}
	}
}
}

image

task 4
#include<stdio.h>
void dec2n(int x, int n);

int main()
{
int x;

printf("输入一个十进制整数:");
scanf("%d", &x);

dec2n(x, 2);
dec2n(x, 8);
dec2n(x, 16);

return 0;
}

void dec2n(int x, int n)
{
if (x == 0)
	printf("\n");
else
	{
		dec2n(x / n, n);

		if (n >= 10)
		{
			if (x % n >= 10)
			printf("%c\n", (char)((x % n - 10) + 'A'));
			else
			printf("%d", x % n);
		}
		else
		printf("%d", x % n);
}
}

image

task 5
#include<stdio.h>
#define N 100

int main()
{
int n, i, j, x[N][N];
printf("Enter n:");
while (scanf("%d", &n) != 0)
{
	for (i = 0; i <= n - 1; i++)
	{
		for (j = 0; j <= n - 1; j++)
		{
			if (i <= j)
				x[i][j] = i + 1;
			else
				x[i][j] = j + 1;
				printf("%2d", x[i][j]);
		}
		printf("\n");
	}
	printf("\nEnter n:");
}
return 0;
}

image

task 6
#include<stdio.h>
void output(char x[]);
void sort(char x[], char y[]);
#define N 80

int main()
{
char views1[N] = "hey,c,i hate u.";
char views2[N] = "hey,c,i love u.";

printf("original views:");

printf("\nviews1:");
output(views1);
printf("\nviews2:");
output(views2);

printf("\n\nswapping...\n");

sort(views1, views2);

printf("\nviews1:");
output(views1);
printf("\nviews2:");
output(views2);

return 0;
}

void output(char x[])
{
for (int i=0; x[i] != '\0'; i++)
	printf("%c", x[i]);
}

void sort(char x[], char y[])
{
int temp;
for (int i = 0; x[i] != '\0'; i++)
{
	temp = x[i];
	x[i] = y[i];
	y[i] = temp;
}
}

image

task 7
#define _CRT_SECURE_NO_WARNINGS
#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)
{
char temp[M];
for (int i = 0; i < n - 1; i++)
{
	for (int j = 0; j < n - 1; j++)
	{
		if (strcmp(str[j], str[j + 1]) > 0)
		{
			strcpy(temp, str[j]);
			strcpy(str[j], str[j + 1]);
			strcpy(str[j + 1], temp);
		}
	}
}
}

image

上一篇:MySQL索引相关
下一篇:没有了
网友评论