C程序设计(第5版)谭浩强习题解答 第8章 善于利用指针

C程序设计(第5版)谭浩强习题解答

第8章 善于利用指针

1. 输入3个整数，要求按由小到大的顺序输出。

#include 
int main()
{
	void swap(int *p1, int *p2);
	int n1, n2, n3;
	int *p1, *p2, *p3;
	printf("input three integer n1,n2,n3:");
	scanf("%d,%d,%d", &n1, &n2, &n3);
	p1 = &n1;
	p2 = &n2;
	p3 = &n3;
	if (n1 > n2) swap(p1, p2);
	if (n1 > n3) swap(p1, p3);
	if (n2 > n3) swap(p2, p3);
	printf("Now,the order is:%d,%d,%d\n", n1, n2, n3);
	return 0;
}

void swap(int *p1, int *p2)
{
	int p;
	p = *p1; *p1 = *p2; *p2 = p;
}

2. 输入3个字符串，要求按由小到大的顺序输出。

#include 
#include 
int main()
{
	void swap(char *, char *);
	char str1[20], str2[20], str3[20];
	printf("input three line:\n");
	gets(str1);
	gets(str2);
	gets(str3);
	if (strcmp(str1, str2) > 0)  swap(str1, str2);
	if (strcmp(str1, str3) > 0)  swap(str1, str3);
	if (strcmp(str2, str3) > 0)  swap(str2, str3);
	printf("Now,the order is:\n");
	printf("%s\n%s\n%s\n", str1, str2, str3);
	return 0;
}

void swap(char *p1, char *p2)
{
	char p[20];
	strcpy(p, p1); strcpy(p1, p2); strcpy(p2, p);
}

3. 输入10个整数,将其中最小的数与第一个数对换, 把最大的数与最后一个数对换。写3个函数:①输人10个数;②进行处理;③输出10个数。

#include 
int main()
{
	void input(int *);
	void max_min_value(int *);
	void output(int *);
	int number[10];
	input(number);
	max_min_value(number);
	output(number);
	return 0;
}

void input(int *number)
{
	int i;
	printf("input 10 numbers:");
	for (i = 0; i < 10; i++)
		scanf("%d", &number[i]);
}

void max_min_value(int *number)
{
	int *max, *min, *p, temp;
	max = min = number;
	for (p = number + 1; p < number + 10; p++)
		if (*p > *max) max = p;
		else if (*p < *min) min = p;
	temp = number[0]; number[0] = *min; *min = temp;
	if (max == number) max = min;
	temp = number[9]; number[9] = *max; *max = temp;
}

void output(int *number)
{
	int *p;
	printf("Now,they are:    ");
	for (p = number; p < number + 10; p++)
		printf("%d ", *p);
	printf("\n");
}

4. 有n个整数,使前面各数顺序向后移m个位置,最后m个数变成最前面m个数,见图8.43。 写一函数实现以上功能,在主函数中输人n个整数和输出调整后的n个数。

#include 
int main()
{
	void move(int[20], int, int);
	int number[20], n, m, i;
	printf("how many numbers?");
	scanf("%d", &n);
	printf("input %d numbers:\n", n);
	for (i = 0; i < n; i++)
		scanf("%d", &number[i]);
	printf("how many place you want move?");
	scanf("%d", &m);
	move(number, n, m);
	printf("Now,they are:\n");
	for (i = 0; i < n; i++)
		printf("%d  ", number[i]);
	printf("\n");
	return 0;
}

void move(int array[20], int n, int m)
{
	int *p, array_end;
	array_end = *(array + n - 1);
	for (p = array + n - 1; p > array; p--)
		*p = *(p - 1);
	*array = array_end;
	m--;
	if (m > 0) move(array, n, m);
}

5. 有n个人围成一圈，顺序排号。从第1个人开始报数(从1到3报数),凡报到3的人退出圈子,问最后留下的是原来第几号的那位。

#include 
int main()
{
	int i, k, m, n, num[50], *p;
	printf("\ninput number of person: n=");
	scanf("%d", &n);
	p = num;
	for (i = 0; i < n; i++)
		*(p + i) = i + 1;
	i = 0;
	k = 0;
	m = 0;
	while (m < n - 1)
	{
		if (*(p + i) != 0)  k++;
		if (k == 3)
		{
			*(p + i) = 0;
			k = 0;
			m++;
		}
		i++;
		if (i == n) i = 0;
	}
	while (*p == 0) p++;
	printf("The last one is NO.%d\n", *p);
	return 0;
}

6. 写一函数,求一个字符串的长度。在main函数中输入字符串,并输出其长度。

#include 
int main()
{
	int length(char *p);
	int len;
	char str[20];
	printf("input string:  ");
	scanf("%s", str);
	len = length(str);
	printf("The length of string is %d.\n", len);
	return 0;
}

int length(char *p)
{
	int n;
	n = 0;
	while (*p != '\0')
	{
		n++;
		p++;
	}
	return(n);
}

7. 有一字符串,包含n个字符。写一函数,将此字符串中从第m个字符开始的全部字符复制成为另一个字符串。

#include 
#include 
int main()
{
	void copystr(char *, char *, int);
	int m;
	char str1[20], str2[20];
	printf("input string:");
	gets(str1);
	printf("which character that begin to copy?");
	scanf("%d", &m);
	if (strlen(str1) < m)
		printf("input error!");
	else
	{
		copystr(str1, str2, m);
		printf("result:%s\n", str2);
	}
	return 0;
}

void copystr(char *p1, char *p2, int m)
{
	int n;
	n = 0;
	while (n < m - 1)
	{
		n++;
		p1++;
	}
	while (*p1 != '\0')
	{
		*p2 = *p1;
		p1++;
		p2++;
	}
	*p2 = '\0';
}

8. 输入一行文字,找出其中大写字母、小写字母、空格、数字以及其他字符各有多少。

#include 
int main()
{
	int upper = 0, lower = 0, digit = 0, space = 0, other = 0, i = 0;
	char *p, s[20];
	printf("input string:  ");
	while ((s[i] = getchar()) != '\n') i++;
	p = &s[0];
	while (*p != '\n')
	{
		if (('A' <= *p) && (*p <= 'Z'))
			++upper;
		else if (('a' <= *p) && (*p <= 'z'))
			++lower;
		else if (*p == ' ')
			++space;
		else if ((*p <= '9') && (*p >= '0'))
			++digit;
		else
			++other;
		p++;
	}
	printf("upper case:%d     lower case:%d", upper, lower);
	printf("     space:%d     digit:%d      other:%d\n", space, digit, other);
	return 0;
}

9. 写一函数,将一个3x3的整型矩阵转置。

#include 
int main()
{
	void move(int *pointer);
	int a[3][3], *p, i;
	printf("input matrix:\n");
	for (i = 0; i < 3; i++)
		scanf("%d %d %d", &a[i][0], &a[i][1], &a[i][2]);
	p = &a[0][0];
	move(p);
	printf("Now,matrix:\n");
	for (i = 0; i < 3; i++)
		printf("%d %d %d\n", a[i][0], a[i][1], a[i][2]);
	return 0;
}

void move(int *pointer)
{
	int i, j, t;
	for (i = 0; i < 3; i++)
		for (j = i; j < 3; j++)
		{
			t = *(pointer + 3 * i + j);
			*(pointer + 3 * i + j) = *(pointer + 3 * j + i);
			*(pointer + 3 * j + i) = t;
		}
}

10. 将一个5x5的矩阵中最大的元素放在中心,4个角分别放4个最小的元素(顺序为从左到右,从上到下依次从小到大存放),写一函数实现之。用main函数调用。

//8.10.1
#include 
int main()
{
	void change(int *p);
	int a[5][5], *p, i, j;
	printf("input matrix:\n");
	for (i = 0; i < 5; i++)
		for (j = 0; j < 5; j++)
			scanf("%d", &a[i][j]);
	p = &a[0][0];
	change(p);
	printf("Now,matrix:\n");
	for (i = 0; i < 5; i++)
	{
		for (j = 0; j < 5; j++)
			printf("%d ", a[i][j]);
		printf("\n");
	}
	return 0;
}

void change(int *p)
{
	int i, j, temp;
	int *pmax, *pmin;
	pmax = p;
	pmin = p;
	for (i = 0; i < 5; i++)
		for (j = i; j < 5; j++)
		{
			if (*pmax < *(p + 5 * i + j)) pmax = p + 5 * i + j;
			if (*pmin > *(p + 5 * i + j)) pmin = p + 5 * i + j;
		}
	temp = *(p + 12);
	*(p + 12) = *pmax;
	*pmax = temp;
	temp = *p;
	*p = *pmin;
	*pmin = temp;
	pmin = p + 1;
	for (i = 0; i < 5; i++)
		for (j = 0; j < 5; j++)
			if (((p + 5 * i + j) != p) && (*pmin > *(p + 5 * i + j))) pmin = p + 5 * i + j;
	temp = *pmin;
	*pmin = *(p + 4);
	*(p + 4) = temp;
	pmin = p + 1;
	for (i = 0; i < 5; i++)
		for (j = 0; j < 5; j++)
			if (((p + 5 * i + j) != (p + 4)) && ((p + 5 * i + j) != p) && (*pmin > *(p + 5 * i + j)))pmin = p + 5 * i + j;
	temp = *pmin;
	*pmin = *(p + 20);
	*(p + 20) = temp;
	pmin = p + 1;
	for (i = 0; i < 5; i++)
		for (j = 0; j < 5; j++)
			if (((p + 5 * i + j) != p) && ((p + 5 * i + j) != (p + 4)) && ((p + 5 * i + j) != (p + 20)) && (*pmin > *(p + 5 * i + j)))
				pmin = p + 5 * i + j;
	temp = *pmin;
	*pmin = *(p + 24);
	*(p + 24) = temp;
}


//8.10.2
#include 
int main()
{
	void change(int *p);
	int a[5][5], *p, i, j;
	printf("input matrix:\n");
	for (i = 0; i < 5; i++)
		for (j = 0; j < 5; j++)
			scanf("%d", &a[i][j]);
	p = &a[0][0];
	change(p);
	printf("Now,matrix:\n");
	for (i = 0; i < 5; i++)
	{
		for (j = 0; j < 5; j++)
			printf("%d ", a[i][j]);
		printf("\n");
	}
	return 0;
}

void change(int *p)          //交换函数
{
	int i, j, temp;
	int *pmax, *pmin;
	pmax = p;
	pmin = p;
	for (i = 0; i < 5; i++)          //找最大值和最小值的地址,并赋给 pmax,pmin
		for (j = i; j < 5; j++)
		{
			if (*pmax < *(p + 5 * i + j)) pmax = p + 5 * i + j;
			if (*pmin > *(p + 5 * i + j)) pmin = p + 5 * i + j;
		}
	temp = *(p + 12);              //将最大值与中心元素互换
	*(p + 12) = *pmax;
	*pmax = temp;

	temp = *p;                   //将最小值与左上角元素互换
	*p = *pmin;
	*pmin = temp;

	pmin = p + 1;
	//将a[0][1]的地址赋给pmin，从该位置开始找最小的元素
	for (i = 0; i < 5; i++)         //找第二最小值的地址赋给 pmin 
		for (j = 0; j < 5; j++)
		{
			if (i == 0 && j == 0) continue;
			if (*pmin > *(p + 5 * i + j)) pmin = p + 5 * i + j;
		}
	temp = *pmin;               //将第二最小值与右上角元素互换 
	*pmin = *(p + 4);
	*(p + 4) = temp;

	pmin = p + 1;
	for (i = 0; i < 5; i++)        //找第三最小值的地址赋给pmin 
		for (j = 0; j < 5; j++)
		{
			if ((i == 0 && j == 0) || (i == 0 && j == 4)) continue;
			if (*pmin > *(p + 5 * i + j)) pmin = p + 5 * i + j;
		}
	temp = *pmin;              //将第三最小值与左下角元素互换
	*pmin = *(p + 20);
	*(p + 20) = temp;

	pmin = p + 1;
	for (i = 0; i < 5; i++)       //找第四最小值的地址赋给pmin 
		for (j = 0; j < 5; j++)
		{
			if ((i == 0 && j == 0) || (i == 0 && j == 4) || (i == 4 && j == 0)) continue;
			if (*pmin > *(p + 5 * i + j)) pmin = p + 5 * i + j;
		}
	temp = *pmin;             //将第四最小值与右下角元素互换
	*pmin = *(p + 24);
	*(p + 24) = temp;
}

11. 在主函数中输入10个等长的字符串。用另一函数对它们排序。然后在主函数输出这10个已排好序的字符串。

//8.11.1
#include 
#include 
int main()
{
	void sort(char s[][6]);
	int i;
	char str[10][6];
	printf("input 10 strings:\n");
	for (i = 0; i < 10; i++)
		scanf("%s", str[i]);
	sort(str);
	printf("Now,the sequence is:\n");
	for (i = 0; i < 10; i++)
		printf("%s\n", str[i]);
	return 0;
}

void sort(char s[10][6])
{
	int i, j;
	char *p, temp[10];
	p = temp;
	for (i = 0; i < 9; i++)
		for (j = 0; j < 9 - i; j++)
			if (strcmp(s[j], s[j + 1]) > 0)
			{
				strcpy(p, s[j]);
				strcpy(s[j], s[+j + 1]);
				strcpy(s[j + 1], p);
			}
}


//8.11.2
#include 
#include 
int main()
{
	void sort(char(*p)[6]);
	int i;
	char str[10][6];
	char(*p)[6];
	printf("input 10 strings:\n");
	for (i = 0; i < 10; i++)
		scanf("%s", str[i]);
	p = str;
	sort(p);
	printf("Now,the sequence is:\n");
	for (i = 0; i < 10; i++)
		printf("%s\n", str[i]);
	return 0;
}

void sort(char(*s)[6])
{
	int i, j;
	char temp[6], *t = temp;
	for (i = 0; i < 9; i++)
		for (j = 0; j < 9 - i; j++)
			if (strcmp(s[j], s[j + 1]) > 0)
			{
				strcpy(t, s[j]);
				strcpy(s[j], s[+j + 1]);
				strcpy(s[j + 1], t);
			}
}

12. 用指针数组处理上一题目,字符串不等长。

#include 
#include 
int main()
{
	void sort(char *[]);
	int i;
	char *p[10], str[10][20];
	for (i = 0; i < 10; i++)
		p[i] = str[i];
	printf("input 10 strings:\n");
	for (i = 0; i < 10; i++)
		scanf("%s", p[i]);
	sort(p);
	printf("Now,the sequence is:\n");
	for (i = 0; i < 10; i++)
		printf("%s\n", p[i]);
	return 0;
}

void sort(char *s[])
{
	int i, j;
	char *temp;
	for (i = 0; i < 9; i++)
		for (j = 0; j < 9 - i; j++)
			if (strcmp(*(s + j), *(s + j + 1)) > 0)
			{
				temp = *(s + j);
				*(s + j) = *(s + j + 1);
				*(s + j + 1) = temp;
			}
}

13. 写一个用矩形法求定积分的通用函数,分别求∫01sinxdx,∫01cosxdx,∫01exdx

#include
#include
int main()
{
	float integral(float(*)(float), float, float, int);//对integarl函数的声明
	float fsin(float);          //对fsin函数的声明
	float fcos(float);          //对fcos函数的声明
	float fexp(float);          //对fexp函数的声明
	float a1, b1, a2, b2, a3, b3, c, (*p)(float);
	int n = 20;
	printf("input a1,b1:");
	scanf("%f,%f", &a1, &b1);
	printf("input a2,b2:");
	scanf("%f,%f", &a2, &b2);
	printf("input a3,b3:");
	scanf("%f,%f", &a3, &b3);
	p = fsin;
	c = integral(p, a1, b1, n);
	printf("The integral of sin(x) is:%f\n", c);
	p = fcos;
	c = integral(p, a2, b2, n);
	printf("The integral of cos(x) is:%f\n", c);
	p = fexp;
	c = integral(p, a3, b3, n);
	printf("The integral of exp(x) is:%f\n", c);
	return 0;
}

float integral(float(*p)(float), float a, float b, int n)
{
	int i;
	float x, h, s;
	h = (b - a) / n;
	x = a;
	s = 0;
	for (i = 1; i <= n; i++)
	{
		x = x + h;
		s = s + (*p)(x)*h;
	}
	return(s);
}
float fsin(float x)
{
	return sin(x);
}
float fcos(float x)
{
	return cos(x);
}
float fexp(float x)
{
	return exp(x);
}

14. 将n个数按输入时顺序的逆序排列,用函数实现。

#include 
int main()
{
	void sort(char *p, int m);
	int i, n;
	char *p, num[20];
	printf("input n:");
	scanf("%d", &n);
	printf("please input these numbers:\n");
	for (i = 0; i < n; i++)
		scanf("%d", &num[i]);
	p = &num[0];
	sort(p, n);
	printf("Now,the sequence is:\n");
	for (i = 0; i < n; i++)
		printf("%d ", num[i]);
	printf("\n");
	return 0;
}

void sort(char *p, int m) //将n个数逆序排列函数  
{
	int i;
	char temp, *p1, *p2;
	for (i = 0; i < m / 2; i++)
	{
		p1 = p + i;
		p2 = p + (m - 1 - i);
		temp = *p1;
		*p1 = *p2;
		*p2 = temp;
	}
}

15. 有一个班4个学生,5门课程。①求第1门课程的平均分;②找出有两门以上课程不及格的学生,输出他们的学号和全部课程成绩及平均成绩;③找出平均成绩在90分以上或全部课程成绩在85分以上的学生。分别编3个函数实现以上3个要求。

#include 
int main()
{
	void avsco(float *, float *);
	void avcour1(char(*)[10], float *);
	void fali2(char course[5][10], int num[], float *pscore, float aver[4]);
	void good(char course[5][10], int num[4], float *pscore, float aver[4]);
	int i, j, *pnum, num[4];
	float score[4][5], aver[4], *pscore, *paver;
	char course[5][10], (*pcourse)[10];
	printf("input course:\n");
	pcourse = course;
	for (i = 0; i < 5; i++)
		scanf("%s", course[i]);
	printf("input NO. and scores:\n");
	printf("NO.");
	for (i = 0; i < 5; i++)
		printf(",%s", course[i]);
	printf("\n");
	pscore = &score[0][0];
	pnum = &num[0];
	for (i = 0; i < 4; i++)
	{
		scanf("%d", pnum + i);
		for (j = 0; j < 5; j++)
			scanf("%f", pscore + 5 * i + j);
	}
	paver = &aver[0];
	printf("\n\n");
	avsco(pscore, paver);                  //求出每个学生的平均成绩 
	avcour1(pcourse, pscore);                //求出第一门课的平均成绩 
	printf("\n\n");
	fali2(pcourse, pnum, pscore, paver);       //找出2门课不及格的学生 
	printf("\n\n");
	good(pcourse, pnum, pscore, paver);        //找出成绩好的学生 
	return 0;
}

void avsco(float *pscore, float *paver)  //求每个学生的平均成绩的函数
{
	int i, j;
	float sum, average;
	for (i = 0; i < 4; i++)
	{
		sum = 0.0;
		for (j = 0; j < 5; j++)
			sum = sum + (*(pscore + 5 * i + j));       //累计每个学生的各科成绩
		average = sum / 5;                   //计算平均成绩
		*(paver + i) = average;
	}
}

void avcour1(char(*pcourse)[10], float *pscore)      //求第一课程的平均成绩的函数 
{
	int i;
	float sum, average1;
	sum = 0.0;
	for (i = 0; i < 4; i++)
		sum = sum + (*(pscore + 5 * i));               //累计每个学生的得分
	average1 = sum / 4;                        //计算平均成绩
	printf("course 1:%s average score:%7.2f\n", *pcourse, average1);
}

void fali2(char course[5][10], int num[], float *pscore, float aver[4])
//找两门以上课程不及格的学生的函数 
{
	int i, j, k, labe1;
	printf("        ==========Student who is fail in two courses=======  \n");
	printf("NO. ");
	for (i = 0; i < 5; i++)
		printf("%11s", course[i]);
	printf("    average\n");
	for (i = 0; i < 4; i++)
	{
		labe1 = 0;
		for (j = 0; j < 5; j++)
			if (*(pscore + 5 * i + j) < 60.0) labe1++;
		if (labe1 >= 2)
		{
			printf("%d", num[i]);
			for (k = 0; k < 5; k++)
				printf("%11.2f", *(pscore + 5 * i + k));
			printf("%11.2f\n", aver[i]);
		}
	}
}

void good(char course[5][10], int num[4], float *pscore, float aver[4])
//找成绩优秀学生(各门85以上或平均90分以上)的函数 
{
	int i, j, k, n;
	printf("         ======Students whose score is good======\n");
	printf("NO. ");
	for (i = 0; i < 5; i++)
		printf("%11s", course[i]);
	printf("    average\n");
	for (i = 0; i < 4; i++)
	{
		n = 0;
		for (j = 0; j < 5; j++)
			if (*(pscore + 5 * i + j) > 85.0) n++;
		if ((n == 5) || (aver[i] >= 90))
		{
			printf("%d", num[i]);
			for (k = 0; k < 5; k++)
				printf("%11.2f", *(pscore + 5 * i + k));
			printf("%11.2f\n", aver[i]);
		}
	}
}

16. 输入一个字符串,内有数字和非数字字符,例如：A123x456 17960? ,302tab5876，将其中连续的数字作为一个整数,依次存放到一数组a中。例如,123放在a[0],456放在a1[1]…统计共有多少个整数,并输出这些数。

#include 
int main()
{
	char str[50], *pstr;
	int i, j, k, m, e10, digit, ndigit, a[10], *pa;
	printf("input a string:\n");
	gets(str);
	pstr = &str[0];    //字符指针pstr置于数组str 首地址
	pa = &a[0];        //指针pa置于a数组首地址
	ndigit = 0;        //ndigit代表有多少个整数
	i = 0;             //代表字符串中的第几个字符
	j = 0;
	while (*(pstr + i) != '\0')
	{
		if ((*(pstr + i) >= '0') && (*(pstr + i) <= '9'))
			j++;
		else
		{
			if (j > 0)
			{
				digit = *(pstr + i - 1) - 48;          //将个数位赋予digit
				k = 1;
				while (k < j)     //将含有两位以上数的其它位的数值累计于digit
				{
					e10 = 1;
					for (m = 1; m <= k; m++)
						e10 = e10 * 10;                  //e10代表该位数所应乘的因子
					digit = digit + (*(pstr + i - 1 - k) - 48)*e10;  //将该位数的数值\累加于digit
					k++;                   //位数K自增
				}
				*pa = digit;               //将数值赋予数组a
				ndigit++;
				pa++;                    //指针pa指向a数组下一元素
				j = 0;
			}
		}
		i++;
	}
	if (j > 0)                         //以数字结尾字符串的最后一个数据
	{
		digit = *(pstr + i - 1) - 48;          //将个数位赋予digit
		k = 1;
		while (k < j)          //将含有两位以上数的其它位的数值累加于digit
		{
			e10 = 1;
			for (m = 1; m <= k; m++)
				e10 = e10 * 10;            //e10代表位数所应乘的因子
			digit = digit + (*(pstr + i - 1 - k) - 48)*e10;  //将该位数的数值累加于digit
			k++;  //位数K自增
		}
		*pa = digit;                 //将数值赋予数组a
		ndigit++;
		j = 0;
	}
	printf("There are %d numbers in this line, they are:\n", ndigit);
	j = 0;
	pa = &a[0];
	for (j = 0; j < ndigit; j++)            //打印数据
		printf("%d ", *(pa + j));
	printf("\n");
	return 0;
}

17. 写一函数，实现两个字符串的比较。即自己写一个strcmp函数,函数原型为int strcmp(char * p1 ,char * p2); 设p1指向字符串s1, p2指向字符串s2。要求当s1=s2时,返回值为0;若s1≠s2,返回它们二者第1个不同字符的ASCII码差值(如"BOY"与"BAD" ,第2个字母不同,0与A之差为79- 65=14)。如果s1>s2,则输出正值;如果s1

#include
int main()
{
	int strcmp(char *p1, char *p2);
	int m;
	char str1[20], str2[20], *p1, *p2;
	printf("input two strings:\n");
	scanf("%s", str1);
	scanf("%s", str2);
	p1 = &str1[0];
	p2 = &str2[0];
	m = strcmp(p1, p2);
	printf("result:%d,\n", m);
	return 0;
}

int strcmp(char *p1, char *p2)           //两个字符串比较函数
{
	int i;
	i = 0;
	while (*(p1 + i) == *(p2 + i))
		if (*(p1 + i++) == '\0') return(0);     //相等时返回结果0 
	return(*(p1 + i) - *(p2 + i));              //不等时返回结果为第一个不等字符ASCII码的差值
}

18. 编一程序,输入月份号,输出该月的英文月名。例如，输人3,则输出"March" ,要求用指针数组处理。

#include 
int main()
{
	char *month_name[13] = { "illegal month","January","February","March","April",
	  "May","June","july","August","September","October", "November","December" };
	int n;
	printf("input month:\n");
	scanf("%d", &n);
	if ((n <= 12) && (n >= 1))
		printf("It is %s.\n", *(month_name + n));
	else
		printf("It is wrong.\n");
	return 0;
}

19.(1) 编写一个函数new,对n个字符开辟连续的存储空间,此函数应返回一个指针(地址),指向字符串开始的空间。new(n)表示分配n个字节的内存空间。(2)写一函数free,将前面用new函数占用的空间释放。free§表示将p(地址)指向的单元以后的内存段释放。

//8.19.1
#include 
#define NEWSIZE 1000                    //指定开辟存区的最大容量
char newbuf[NEWSIZE];                   //定义字符数组newbuf 
char *newp = newbuf;                      //定义指针变量newp，指向可存区的始端             

char *new(int n)                        //定义开辟存区的函数new,开辟存储区后返回指针
{ 
	if (newp + n <= newbuf + NEWSIZE)         //开辟区未超过newbuf数组的大小
	{
		newp += n;                          //newp指向存储区的末尾
		return(newp - n);                  //返回一个指针,它指向存区的开始位置
	}
	else
		return(NULL);                    //当存区不够分配时,返回一个空指针
}


//8.19.2
#include 
#define NEWSIZE 1000
char newbuf[NEWSIZE];
char *newp = newbuf;

void free(char *p)                             //释放存区函数
{
	if (p >= newbuf && p < newbuf + NEWSIZE)
		newp = p;
}

20. 用指向指针的指针的方法对5个字符串排序并输出。

#define LINEMAX 20              //定义字符串的最大长度
int main()
{
	int i;
	char **p, *pstr[5], str[5][LINEMAX];
	for (i = 0; i < 5; i++)
		pstr[i] = str[i];   //将第i个字符串的首地址赋予指针数组 pstr 的第i个元素
	printf("input 5 strings:\n");
	for (i = 0; i < 5; i++)
		scanf("%s", pstr[i]);
	p = pstr;
	sort(p);
	printf("strings sorted:\n");
	for (i = 0; i < 5; i++)
		printf("%s\n", pstr[i]);
}

sort(char **p)            //冒泡法对5个字符串排序函数
{
	int i, j;
	char *temp;
	for (i = 0; i < 5; i++)
	{
		for (j = i + 1; j < 5; j++)
		{
			if (strcmp(*(p + i), *(p + j)) > 0)      //比较后交换字符串地址
			{
				temp = *(p + i);
				*(p + i) = *(p + j);
				*(p + j) = temp;
			}
		}
	}
	return 0;
}

21. 用指向指针的指针的方法对n个整数排序并输出。要求将排序单独写成一个函数。n个整数在主函数中输入,最后在主函数中输出。

#include
int main()
{
	void sort(int **p, int n);
	int i, n, data[20], **p, *pstr[20];
	printf("input n:\n");
	scanf("%d", &n);
	for (i = 0; i < n; i++)
		pstr[i] = &data[i];   //将第i个整数的地址赋予指针数组 pstr 的第i个元素  
	printf("input %d integer numbers:", n);
	for (i = 0; i < n; i++)
		scanf("%d", pstr[i]);
	p = pstr;
	sort(p, n);
	printf("Now,the sequence is:\n");
	for (i = 0; i < n; i++)
		printf("%d  ", *pstr[i]);
	printf("\n");
	return 0;
}
void sort(int **p, int n)
{
	int i, j, *temp;
	for (i = 0; i < n - 1; i++)
	{
		for (j = i + 1; j < n; j++)
		{
			if (**(p + i) > **(p + j))  //比较后交换整数地址 
			{
				temp = *(p + i);
				*(p + i) = *(p + j);
				*(p + j) = temp;
			}
		}
	}
}