Chapter 7 of C language programming (fifth edition of Tan Haoqiang) analysis and answer of modular programming exercises with functions

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subject 1： Write two functions , Find the maximum common divisor and the minimum common multiple of two integers respectively , Call these two functions with the main function , And output the result . Two integers are entered by the keyboard .

Explain ： Let two integers be u and v, The algorithm for finding the maximum common divisor by rolling division is as follows ∶

ifv>u
 Put the variable  u And v The value of the swap 		（ Make the big one u For divisor ）
  while（u/v The remainder of  r≠0）
{
     u=v								（ Divisor v Become a divisor  u）
  v=r								（ Make the remainder r Become a divisor  v）
}

Output the greatest common divisor r

Minimum common multiple 1= u* v/ greatest common divisor r

You can use the following two methods ∶

Method 1 ： Use two functions hcf and lcd Find the greatest common divisor and the least common multiple respectively . Enter two integers in the main function u and v, And pass it to the function hcf, The maximum common divisor is returned to the integer variable assigned by the main function h, Then take it. h And two integers u ,v Pass it to the function as an argument lcd, Find the least common multiple , Return to the main function and assign it to the integer variable l. Output the maximum common divisor and the minimum common multiple .

Write the program accordingly ：

#include <stdio.h>
int main()
{
    

	int hef(int, int);		// Function declaration 
	int lcd(int, int, int); // Function declaration 
	int u, v, h, l;
	scanf("%d,%d", &u, &v);
	h = hef(u, v);
	printf("H.C.F=%d\n", h);
	l = lcd(u, v, h);
	printf("L.C.D=%d\n", l);
	return 0;
}
int hef(int u, int v)
{
    
	int t, r;
	if (v > u)
	{
    
		t = u;
		u = v;
		v = t;
	}
	while ((r = u % v) != 0)

	{
    
		u = v;
		v = r;
	}
	return (v);
}
int lcd(int u, int v, int h)
{
    
	return (u * v / h);
}

Running results ：

Input 24 and 16 Two Numbers , The maximum common divisor of program output is 8, The least common multiple is 48.

Method 2 ： Using global variables . Global variables Hcf and Lcd Represent the greatest common divisor and the least common multiple respectively . Use two functions to find the maximum common divisor and the minimum common multiple , The value is not brought back by the function , It's assigned to global variables Hcf and Lcd. Output their values in the main function .

The procedure is as follows ：

#include <stdio.h>
int Hcf, Lcd; // Hcf  and  Lcd Global variable 
int main()
{
    
	void hcf(int, int);
	void lcd(int, int);
	int u, v;
	scanf("%d,%d", &u, &v);
	hcf(u, v); // call hcf  function 
	lcd(u, v); // call lcd  function 
	printf("H.C.F=%d\n", Hcf);
	printf("L.C.D=%d\n", Lcd);
	return 0;
}
void hcf(int u, int v)
{
    
	int t, r;
	if (v > u)
	{
    
		t = u;
		u = v;
		v = t;
	}
	while ((r = u % v) != 0)
	{
    
		u = v;
		v = r;
	}
	Hcf = v; // Assign the maximum common divisor to the global variable  Hcf
}
void lcd(int u, int v)
{
    
	Lcd = u * v / Hcf; // Assign the least common multiple to the global variable Lcd
}

The running result is the same as method 1 .

Hcf Global variable ,hcf Is the function name , The case of the two names is different , There is no confusion . stay hcf Find the maximum common divisor in the function and assign it to the global variable Hcf, stay lcd The global variable... Is referenced in the function Hcf Value , The least common multiple obtained is assigned to the global variable Lcd. Output... In the main function Hcf and Lcd Value .

subject 2： Find the equation $ ax^2+bx+c=0 $ The root of the , use 3 When two functions are solved separately ：$ b^2-4ac $ Greater than 0、 be equal to 0 And less than 0 And output the result . Enter... From the main function a,b,c Value .

Explain ：

Answer code ：

#include <stdio.h>
#include <math.h>
float x1, x2, disc, p, q;
int main()
{
    
	void greater_than_zero(float, float);
	void equal_to_zero(float, float);
	void smaller_than_zero(float, float);
	float a, b, c;
	printf("input a,b,c:");
	scanf("%f,%f,%f", &a, &b, &c);
	printf("equation:%5.2f*x*x+%5.2f*x+%5.2f=0\n", a, b, c);
	disc = b * b - 4 * a * c;
	printf("root:\n");
	if (disc > 0)
	{
    
		greater_than_zero(a, b);
		printf("x1=%f\t\tx2=%f\n", x1, x2);
	}
	else if (disc == 0)
	{
    
		equal_to_zero(a, b);
		printf("xl=%f\t\tx2=%f\n", x1, x2);
	}
	else
	{
    
		smaller_than_zero(a, b);
		printf("x1=%f+%fi\tx2=%f-%fi\n", p, q, p, q);
	}
	return 0;
}
void greater_than_zero(float a, float b)
{
    
	x1 = (-b + sqrt(disc)) / (2 * a);
	x2 = (-b - sqrt(disc)) / (2 * a);
}
void equal_to_zero(float a, float b)
{
    
	x1 = x2 = (-b) / (2 * a);
}
void smaller_than_zero(float a, float b)
{
    
	p = -b / (2 * a);
	q = sqrt(-disc) / (2 * a);
}

Running results ：

① Two unequal real roots .

② Two equal real roots .

③ The complex roots of two conjugates .

subject 3： Write a function that judges prime numbers , Enter an integer in the main function , Output information about whether it is a prime number .

Explain ：

#include <stdio.h>
int main()
{
    
	int prime(int);
	int n;
	printf("input an integer:");
	scanf("%d", &n);
	if (prime(n))
		printf("%d is a prime.\n", n);
	else
		printf("%d is not a prime.\n", n);
	return 0;
}
int prime(int n)
{
    
	int flag = 1, i;
	for (i = 2; i < n / 2 && flag == 1; i++)
		if (n % i == 0)
			flag = 0;
	return (flag);
}

Running results ：

①

②

subject 4： Write a function , Make a given 3×3 Transpose a two-dimensional integer array , That's row and column interchange .

Explain ：

Answer code ：

#include <stdio.h>
#define N 3
int array[N][N];
int main()
{
    
	void convert(int array[][3]);
	int i, j;
	printf("input array:\n");
	for (i = 0; i < N; i++)
		for (j = 0; j < N; j++)
			scanf("%d", &array[i][j]);
	printf("\noriginal array:\n");
	for (i = 0; i < N; i++)
	{
    
		for (j = 0; j < N; j++)
			printf("%5d", array[i][j]);
		printf("\n");
	}
	convert(array);
	printf("convert array:\n");
	for (i = 0; i < N; i++)
	{
    
		for (j = 0; j < N; j++)
			printf("%5d", array[i][j]);
		printf("\n");
	}
	return 0;
}
void convert(int array[][3]) // Functions that define transpose arrays 
{
    
	int i, j, t;
	for (i = 0; i < N; i++)
		for (j = i + 1; j < N; j++)
		{
    
			t = array[i][j];
			array[i][j] = array[j][i];
			array[j][i] = t;
		}
}

Running results ：

subject 5： Write a function , Causes an input string to be stored in reverse order , Enter and output strings in the main function .

Explain ：

Answer code ：

#include <stdio.h>
#include <string.h>
int main()
{
    
	void inverse(char str[]);
	char str[100];
	printf("input string:");
	scanf("%s", str);
	inverse(str);
	printf("inverse string:%s\n", str);
	return 0;
}
void inverse(char str[])
{
    
	char t;
	int i, j;
	for (i = 0, j = strlen(str); i < (strlen(str) / 2); i++, j--)
	{
    
		t = str[i];
		str[i] = str[j - 1];
		str[j - 1] = t;
	}
}

Running results ：

Input string ∶abcdefg, Output ∶gfedcba.

subject 6： Write a function , Connect two strings .

Explain ：

Answer code ：

#include <stdio.h>
#include <string.h>
int main()
{
    
	void concatenate(char stringl[], char string2[], char string[]);
	char s1[100], s2[100], s[100];
	printf("input string1:");
	scanf("%s", s1);
	printf("input string2:");
	scanf("%s", s2);
	concatenate(s1, s2, s);
	printf("\nThe new string is %s\n", s);
	return 0;
}
void concatenate(char string1[], char string2[], char string[])
{
    
	int i, j;
	for (i = 0; string1[i] != '\0'; i++)
		string[i] = string1[i];
	for (j = 0; string2[j] != '\0'; j++)
		string[i + j] = string2[j];
	string[i + j] = '\0';
}

Running results ：

Enter two strings ∶country and side, The program connects two strings into a string and outputs ;countryside.

subject 7： Write a function , Copy vowels from one string to another , Then the output .

Explain ：

#include <stdio.h>
int main()
{
    
	void cpy(char[], char[]);
	char str[80], c[80];
	printf("input string:");
	gets(str);
	cpy(str, c);
	printf("The vowel letters are:%s\n", c);
	return 0;
}

void cpy(char s[], char c[])
{
    
	int i, j;
	for (i = 0, j = 0; s[i] != '\0'; i++)
		if (s[i] == 'a' || s[i] == 'A' || s[i] == 'e' || s[i] == 'E' || s[i] == 'i' || s[i] == 'I' || s[i] == 'o' || s[i] == 'O' || s[i] == 'u' || s[i] == 'U')
		{
    
			c[j] = s[i];
			j++;
		}
	c[j] = '\0';
}

Running results ：

take abcdefghijklm Vowel output in .

subject 8： Write a function , Enter a 4 Digit number , This is required to be output 4 Number characters , But there is a space between every two numbers . Such as the input 1990, The output should be "1 9 9 0".

Explain ：

Answer code ：

#include <stdio.h>
#include <string.h>
int main()
{
    
	void insert(char[]);
	char str[80];
	printf("input four digits:");
	scanf("%s", str);
	insert(str);
	return 0;
}
void insert(char str[])
{
    
	int i;
	for (i = strlen(str); i > 0; i--)
	{
    
		str[2 * i] = str[i];
		str[2 * i - 1] = ' ';
	}
	printf("output:\n%s\n", str);
}

Running results ：

subject 9： Write a function , Pass a string from the argument , Count the letters in this string 、 Numbers 、 Number of spaces and other characters , Enter the string in the main function and output the above results .

Explain ：

Answer code ：

#include <stdio.h>
int letter, digit, space, others;
int main()
{
    
	void count(char[]);
	char text[80];
	printf("input string:\n");
	gets(text);
	printf("string:");
	puts(text);
	letter = 0;
	digit = 0;
	space = 0;
	others = 0;
	count(text);
	printf("\nletter:%d\ndigit:%d\nspace:%d\nothers:%d\n", letter, digit, space, others);
	return 0;
}
void count(char str[])
{
    
	int i;
	for (i = 0; str[i] != '\0'; i++)
		if ((str[i] >= 'a' && str[i] <= 'z') || (str[i] >= 'A' && str[i] <= 'Z'))
			letter++;
		else if (str[i] >= '0' && str[i] <= '9')
			digit++;
		else if (str[i] == 32)
			space++;
		else
			others++;
}

Running results ：

subject 10： Write a function , Enter a line of characters , Output the longest word in this string .

Explain ： Think of a word as a string of all letters , Program setting longest The function is to find the position of the longest word . The return value of this function is the starting position of the longest word in the line of characters .longest Functional N-S Pictured graph 7.1 Shown .

chart 7.1 of use flag Indicates whether the word has started ,flag=0 Indicates that... Has not started ,flag=1 Indicates the beginning of the word ; len Represents the accumulated number of letters of the current word ; length Represents the length of the longest word in the previous word ; point Represents the starting position of the current word （ Use subscripts to indicate ）; place Represents the starting position of the longest word . function alphabetic The function of is to judge whether the current character is a letter , If so, return to 1 , Otherwise return to 0 .

Answer code ：

#include <stdio.h>
#include <string.h>
int main()
{
    
	int alphabetic(char);
	int longest(char[]);
	int i;
	char line[100];
	printf("input one line:\n");
	gets(line);
	printf("The longest word is :");
	for (i = longest(line); alphabetic(line[i]); i++)
		printf("%c", line[i]);
	printf("\n");
	return 0;
}
int alphabetic(char c)
{
    
	if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'z'))
		return (1);
	else
		return (0);
}
int longest(char string[])
{
    
	int len = 0, i, length = 0, flag = 1, place = 0, point;
	for (i = 0; i <= strlen(string); i++)
		if (alphabetic(string[i]))
			if (flag)
			{
    
				point = i;
				flag = 0;
			}
			else
				len++;
		else
		{
    
			flag = 1;
			if (len >= length)
			{
    
				length = len;
				place = point;
				len = 0;
			}
		}
	return (place);
}

Running results ：

subject 11： Write a function , use " Foaming method " For the input 10 The characters are arranged in descending order .

Explain ： Of the main function N-S Pictured graph 7.2 Shown .sort The function is used to sort , Its N-S Pictured graph 7.3 Shown .

Answer code ：

#include <stdio.h>
#include <string.h>
#define N 10
char str[N];
int main()
{
    
	void sort(char[]);
	int i, flag;
	for (flag = 1; flag == 1;)
	{
    
		printf("input string:\n");
		scanf("%s", &str);
		if (strlen(str) > N)
			printf("string too long,input again!");
		else
			flag = 0;
	}
	sort(str);
	printf("string sorted:\n");
	for (i = 0; i < N; i++)
		printf("%c", str[i]);
	printf("\n");
	return 0;
}
void sort(char str[])
{
    
	int i, j;
	char t;
	for (j = 1; j < N; j++)
		for (i = 0; (i < N - j) && (str[i] != '0'); i++)
			if (str[i] > str[i + 1])
			{
    
				t = str[i];
				str[i] = str[i + 1];
				str[i + 1] = t;
			}
}

Running results ：

subject 12： Find the root by Newton iterative method . The equation is $ ax^3+bx2+cx+d=0 $ , coefficient a,b,c,d The values of are 1,2,3,4, Input by the main function . seek x stay 1 A real root nearby . After finding the root, the main function outputs .

Explain ：

Newton's iterative formula is $ x=x_0-\frac{f(x_0)}{f’(x_0)} $

among ,$ x_0 $ Is the approximate root obtained last time , At the beginning, according to the questions $ x_0=1 $（ I hope to ask x stay 1 A real root nearby , So the first 1 The approximate value of times can be set to 1 ）. today $ f(x)=ax^3+bx2+cx+d $ , Plug in a,b,c,d Value , obtain $ f(x) =x^3+2x2+3x+4 $ . $ f’(x) $ yes $ f(x) $ The derivative of , today $ f’(x)=3x^2+6x+3 $.The\; first1TimeOverlappinggeneration,$ x=1-\frac{f(1)}{f’(1)}=1-\frac{1+2+3+4}{3+6+3}=1-\frac{10}{12}=0.1666666 $ . The first 2 Iterations are performed with 0.1666666 As 
$ x_0 $ Substitute into the iterative formula , Find out x The next approximation of . And so on , Every iteration starts with x Find the next one closer to the true value x. Iterate until $ |x-x_0|\le10^{-3} $ End of time .

Find the function of the root of the equation by Newton iterative method solut Of N-S chart , Pictured 7.4 Shown .

The procedure is as follows ：

#include <stdio.h>
#include <math.h>

int main()
{
    
	float solut(float a, float b, float c, float d);
	float a, b, c, d;
	printf("input a,b,c,d:");
	scanf("%f,%f,%f,%f", &a, &b, &c, &d);
	printf("x=%10.7f\n", solut(a, b, c, d));
	return 0;
}
float solut(float a, float b, float c, float d)
{
    
	float x = 1, x0, f, f1;
	do
	{
    
		x0 = x;
		f = ((a * x0 + b) * x0 + c) * x0 + d;
		f1 = (3 * a * x0 + 2 * b) * x0 + c;
		x = x0 - f / f1;
	} while (fabs(x - x0) >= 1e-3);
	return (x);
}

Running results

Input coefficient 1,2,3,4, Find the approximate root is -1.6506292.

subject 13： Find... Recursively n The value of Legendre polynomials of order , The recursive formula is
$$p_n(x)=\{\begin{array}{ll}1& (n=0)\\ x& (n=1)\\ ((2n-1)\times x-p_{n-1}(x)-(n-1)\times p_{n-2}(x))/n& (n\ge 1)\end{array}$$

Explain ： Find a recursive function p Of N-S chart , Pictured 7.5 Shown .

Answer code ：

#include <stdio.h>

int main()
{
    
	int x, n;
	float p(int, int);
	printf("\ninput n & x:");
	scanf("%d,%d", &n, &x);
	printf("n=%d,x=%d\n", n, x);
	printf("P%d(%d)=%6.2f\n", n, x, p(n, x));
	return 0;
}
float p(int n, int x)
{
    
	if (n == 0)
		return (1);
	else if (n == 1)
		return (x);
	else
		return (2 * n - 1) * x * p((n - 1), x) - (n - 1) * p((n - 2), x) / n;
}

Running results ：

①：

②：

③：

subject 14： Input 10 A student 5 Results of courses , Use functions to realize the following functions ：

① Calculate the average score of each student ;

② Calculate the average score of each course ;

③ Find out all 50 The student and course with the highest score among the scores ;

④ Calculate the mean score variance ∶
$$\sigma =\frac{1}{n}\sum x_i^2-(\frac{\sum x_i}{n}{)}^2$$

among ,x; Is the average score of a student .

Explain ： Of the main function N-S Pictured graph 7.6 Shown .

function input_stu The execution result is to give the whole process variable student score array score Input the initial value of each element .

function aver_stu The function of is to calculate the average score of each student , And assign the result to the whole variable array a_stu Elements in . function aver cour The function of is to calculate the average score of each course , The calculation results are stored in the whole variable array a_Cur.

function highest The return value of is the highest score ,r,c It's two global variables , Each represents the row with the highest score 、 Column number . Of this function N-S See Figure 7.7.

function s_var The return value of is the variance of the average score .

Answer code ：

#include <stdio.h>
#define N 10
#define M 5
float score[N][M];		   // Global array 
float a_stu[N], a_cour[M]; // Global array 
int r, c;				   // Global variables 
int main()
{
    
	int i, j;
	float h;			  // Function declaration 
	float s_var(void);	  // Function declaration 
	float highest();	  // Function declaration 
	void input_stu(void); // Function declaration 
	void aver_stu(void);  // Function declaration 
	void aver_cour(void);
	input_stu(); // Function call , Input  10 Students' grades 
	aver_stu();	 // Function call , Calculation  10 Student GPA 
	aver_cour();
	printf("\n NO. cour1 cour2 cour3 cour4 cour5 aver\n");
	for (i = 0; i < N; i++)
	{
    
		printf("\n NO %2d", i + 1); // Output a student number 
		for (j = 0; j < M; j++)
			printf("%8.2f", score[i][j]); //  Output a student's grades in each course 
		printf("%8.2f\n", a_stu[i]);	  // Output the average score of a student 
	}
	printf("\naverage:"); // Output 5  Average grade of each course 
	for (j = 0; j < M; j++)
		printf("%8.2f", a_cour[j]);
	printf("\n");
	h = highest();											 // Call function , Get the highest score and which student it belongs to 、 Which course 
	printf("highest:%7.2f NO. %2d course %2d\n", h, r, c); // Output the highest score and student number 、 Course no. 
	printf("variance %8.2f\n", s_var());					 // Call function , Calculate and output variance 
	return 0;
}
void input_stu(void) // Input  10 A function of student achievement 
{
    
	int i, j;
	for (i = 0; i < N; i++)
	{
    
		printf("\ninput score of student%2d:\n", i + 1); // Student number from 1 Start 
		for (j = 0; j < M; j++)
			scanf("%f", &score[i][j]);
	}
}
void aver_stu(void) // Calculation  10  A function of a student's average grade 
{
    
	int i, j;
	float s;
	for (i = 0; i < N; i++)
	{
    
		for (j = 0, s = 0; j < M; j++)
			s += score[i][j];
		a_stu[i] = s / 5.0;
	}
}

void aver_cour(void) // Calculation  5  A function of the average grade of a course 
{
    
	int i, j;
	float s;
	for (j = 0; j < M; j++)
	{
    
		s = 0;
		for (i = 0; i < N; i++)
			s += score[i][j];
		a_cour[j] = s / (float)N;
	}
}
float highest() // Get the highest score and which student it belongs to 、 Which course's function 
{
    
	float high;
	int i, j;
	high = score[0][0];
	for (i = 0; i < N; i++)
		for (j = 0; j < M; j++)
			if (score[i][j] > high)
			{
    
				high = score[i][j];
				r = i + 1; // Array line number i from 0 Start , Student number r from 1 Start , so r=i+1
				c = j + 1; // Array column number j from 0 Start , Course no. c from 1 Start , so c=j+1
			}
	return (high);
}
float s_var(void) // Find the function of variance 
{
    
	int i;
	float sumx, sumxn;
	sumx = 0.0;
	sumxn = 0.0;
	for (i = 0; i < N; i++)
	{
    
		sumx += a_stu[i] * a_stu[i];
		sumxn += a_stu[i];
	}
	return (sumx / N - (sumxn / N) * (sumxn / N));
}

Running results

The above is the input 10 Of a student 5 Results of courses , Here is the output ：

subject 15： Write several functions ∶

① Input 10 The name and number of an employee ;

② Sort by employee number from small to large , The order of names is also adjusted ;

③ An employee number is required , Find out the name of the employee by half search method ; Enter the employee number to be searched from the main function , Output the employee's name .

Explain ：

input The function is to complete 10 Data entry of employees .sort The function is to sort by selection , The process is similar to the problem solving section in this book 6 Chapter one 2 topic .

search The employee name() function is used to find the employee name of the specified employee number by half searching , The search algorithm is shown in the problem solving section of this book 6 Chapter one 9 topic .

Answer code ：

#include <stdio.h>
#include <string.h>
#define N 10
int main()
{
    
	void input(int[], char name[][8]);
	void sort(int[], char name[][8]);
	void search(int, int[], char name[][8]);
	int num[N], number, flag = 1, c;
	char name[N][8];
	input(num, name);
	sort(num, name);
	while (flag == 1)
	{
    
		printf("\ninput number to look for:");
		scanf("%d", &number);
		search(number, num, name);
		printf("continue ot not(Y/N)?");
		getchar();
		c = getchar();
		if (c == 'N' || c == 'n')
			flag = 0;
	}
	return 0;
}

void input(int num[], char name[N][8]) // Function of input data 
{
    
	int i;
	for (i = 0; i < N; i++)
	{
    
		printf("input NO.:");
		scanf("%d", &num[i]);
		printf("input name:");
		getchar();
		gets(name[i]);
	}
}

void sort(int num[], char name[N][8]) // Sort function 
{
    
	int i, j, min, temp1;
	char temp2[8];
	for (i = 0; i < N - 1; i++)
	{
    
		min = i;
		for (j = i; j < N; j++)
			if (num[min] > num[j])
				min = j;
		temp1 = num[i];
		strcpy(temp2, name[i]);
		num[i] = num[min];
		strcpy(name[i], name[min]);

		num[min] = temp1;
		strcpy(name[min], temp2);
	}
	printf("\n result:\n");
	for (i = 0; i < N; i++)
		printf("\n %5d%10s", num[i], name[i]);
}

void search(int n, int num[], char name[N][8]) // Half search function 
{
    
	int top, bott, mid, loca, sign;
	top = 0;
	bott = N - 1;
	loca = 0;
	sign = 1;
	if ((n < num[0]) || (n > num[N - 1]))
		loca = -1;
	while ((sign == 1) && (top <= bott))
	{
    
		mid = (bott + top) / 2;
		if (n == num[mid])
		{
    
			loca = mid;
			printf("NO.%d,his name is %s.\n", n, name[loca]);
			sign = -1;
		}
		else if (n < num[mid])
			bott = mid - 1;
		else
			top = mid + 1;
	}
	if (sign == 1 || loca == -1)
		printf("%d not been found.\n", n);
}

Running results ：

First enter 10 The number and name of an employee , Then sort according to the employee number from small to large . The inquiry employee number is 3 and 4 The name of .

subject 16： Write a function , Enter a hexadecimal number , Output the corresponding decimal number .

Explain ：

The main function main Of N-S Pictured graph 7.8 Shown .

A function for finding decimal numbers htoi Of N-S chart , Pictured 7.9 Shown .

Answer code ：

#include <stdio.h>
#define MAX 1000
int main()
{
    
	int htoi(char s[]);
	int c, i, flag, flag1;
	char t[MAX];
	i = 0;
	flag = 0;
	flag1 = 1;
	printf("input a HEX number:");
	while ((c = getchar()) != '\0' && i < MAX && flag1)
	{
    
		if (c >= '0' && c <= '9' || c >= 'a' && c <= 'f' || c >= 'A' && c <= 'F')
		{
    
			flag = 1;
			t[i++] = c;
		}
		else if (flag)
		{
    
			t[i] = '\0';
			printf("decimal number %d\n", htoi(t));
			printf("continue or not?");
			c = getchar();
			if (c == 'N' || c == 'n')
				flag1 = 0;
			else
			{
    
				flag = 0;
				i = -0;
				printf("\ninput a HEX number:");
			}
		}
	}
	return 0;
}
int htoi(char s[])
{
    
	int i, n;
	n = 0;
	for (i = 0; s[i] != '\0'; i++)
	{
    
		if (s[i] >= '0' && s[i] <= '9')
			n = n * 16 + s[i] - '0';
		if (s[i] >= 'a' && s[i] <= 'f')
			n = n * 16 + s[i] - 'a' + 10;
		if (s[i] >= 'A' && s[i] <= 'F')
			n = n * 16 + s[i] - 'A' + 10;
	}
	return (n);
}

Running results ：

subject 17： Recursively convert an integer n Convert to string . for example , Input 483, Expected output string "483".n The number of digits is uncertain , It can be an integer with any number of digits .

Explain ： Of the main function N-S Pictured graph 7.10 Shown .

Answer code ：

#include <stdio.h>
int main()
{
    
	void convert(int n);
	int number;
	printf("input an integer;");
	scanf("%d", &number);
	printf("output:");
	if (number < 0)
	{
    
		putchar('-'); // Output one first " One " Sign and space 
		putchar(' ');
		number = -number;
	}
	convert(number);
	printf("\n");
	return 0;
}

void convert(int n) // Recursive function 
{
    
	int i;
	if ((i = n / 10) != 0)
		convert(i);
	putchar(n % 10 + '0');
	putchar(32);
}

Running results ：

①:

②:

explain ： If it's a negative number , To convert it to a positive number , At the same time, manually output a "-" Number .convert The function only deals with positive numbers . If number The value of is 345, call convert Function, put 345 Pass to n. Execute function body ,
n/10 Value （ It's also i Value ） by 34, It's not equal to 0. Call again convert function , At this time parameter n The value of is 34. Then execute the function body ,n/10 Value （ It's also i Value ） by 3, It's not equal to 0. Call again convert function , At this time parameter n The value of is 3. Then execute the function body ,n/10 Value （ It's also i Value ） be equal to 0. No more calls convert function , And perform putchar （n%10+‘0’）, here n The value of is 3, so n%10 The value of is 3（% Is the complement operator ）, character ‘0’ Of ASCII The code is 48,3 Add 4 be equal to 51,51 Is the character ’3’ Of ASCII Code , therefore putchar（n%10＋’0’） The output characters ‘3’. next putchar（32） Output a space , So that two characters are separated by spaces .

then , The process returns to the last call convert At function , It should be followed by putchar（n%10＋’0’）, Pay attention to the n It's the last call convert Function n, Its value is 34, therefore n%10 The value of is 4, add ‘0’ be equal to 52,52 Is the character ‘4’ Of ASCII Code , therefore putchar（n%10＋’0’） The output characters ‘4’ , next putchar（32） Output a space .

The process returns to the last call convert At function , It should be followed by putchar（n%10＋’0’）, Pay attention to the n It's No 1 Secondary call convert Function n , Its value is 345 , therefore n%10 The value of is 5, add ‘0’ be equal to 53,53 Is the character ‘5’ Of ASCII Code , therefore putchar（n%10＋’0’） The output characters ‘5’ , next putchar（32） Output a space .

thus , Yes convert The recursive call to the function ends , Return the main function , Output a newline , Program end .

putchar（n%10＋’0’） It can also be rewritten as putchar（n%10＋48）, because 48 Is the character ‘0’ Of ASCII Code .

subject 18： Given year 、 month 、 Japan , Calculate the day of the year .

Explain ： The main function receives the date entered from the keyboard , And call sum_day and leap The function calculates the number of days . Its N-S See Figure 7.11.sum_day Number of days to calculate the input date .leap Function returns whether it is a leap year .

Answer code ：

#include <stdio.h>
int main()
{
    
	int sum_day(int month, int day);
	int leap(int year);
	int year, month, day, days;
	printf("input date(year,month,day):");
	scanf("%d,%d,%d", &year, &month, &day);
	printf("%d/%d/%d ", year, month, day); // Call function  sum_day
	days = sum_day(month, day);			   // Call function  leap
	if (leap(year) && month >= 3)
		days = days + 1;
	printf("is the %dth day in this year.\n", days);
	return 0;
}

int sum_day(int month, int day) // function  sum_day∶ Calculate the date 
{
    
	int day_tab[13] = {
    0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
	int i;
	for (i = 1; i < month; i++)
		day += day_tab[i]; // Accumulate the number of days before the current month 
	return (day);
}

int leap(int year) // function leap∶ Determine if it's a leap year 
{
    
	int leap;
	leap = year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
	return (leap);
}

Running results ：