习题1: 定义一个函数，给函数传递任意多个浮点数，计算出这些数的平均值，从键盘输入任意个值，并输出平均值，以说明这个函数的执行过程。
- /* Exercise 8.1 A function to calculate an average */
- #include <stdio.h>
- #include <stdlib.h>
- #include <ctype.h>
- double average(double data[], int count)
- {
- double sum = 0.0;
- for(int i = 0 ; i<count ; sum += data[i++])
- ;
- return sum/count;
- }
- #define CAPACITY_INCREMENT 6 /* Increment in the capacity for data values */
- int main(void)
- {
- double *data = NULL; /* Pointer to array of data values */
- double *newdata = NULL; /* Pointer to new array of data values */
- double *averages = NULL; /* Pointer to array of averages */
- int count = 0; /* Number of data values */
- int capacity = 0; /* Number of data values that can be stored */
- char answer = 'n';
- do
- {
- if(count == capacity)
- {
- capacity += CAPACITY_INCREMENT;
- /* Create new array of pointers */
- newdata = (double*)malloc(capacity*sizeof(double));
- /* Create an array of values of type double for each new day and store the address */
- if(data)
- {
- /* Copy the existing values to the new array */
- for(int i = 0 ; i<count ; i++)
- newdata[i] = data[i];
- free(data); /* Free memory for the old array of pointers */
- }
- data = newdata; /* copy the address of the new array of pointers */
- newdata = NULL; /* Reset the pointer */
- }
- printf("Enter a data value: ");
- scanf(" %lf", data+count++);
- printf("Do you want to enter another (y or n)? ");
- scanf(" %c", &answer);
- } while(tolower(answer) != 'n');
- printf("\nThe average of thew values you entered is %10.2lf\n", average(data, count));
- free(data);
- return 0;
- }
习题2: 定义一个函数，返回其整数参数的字符串表示。例如，如果这个参数是 25，函数就返回"25"。如果参数是-98，函数就返回"-98"。用于适当的main()版本说明函数的执行过程。
- /* Exercise 8.2 A function to return a string representation of an integer */
- #include <stdio.h>
- #include <stdbool.h>
- /* Convert an integer to a string */
- /* Caller must allocate string array */
- /* Function returns the string to allow */
- /* Use of the function in an expression. */
- char* itoa(int n, char str[])
- {
- int i = 0; /* Loop counter */
- bool negative = 0; /* Indicate negative integer */
- int length = 0; /* Length of string */
- int temp = 0; /* Temporary storage */
- if(negative = (n<0)) /* Is it negative? */
- n = -n; /* make it positive */
- /* Generate digit characters in reverse order */
- do
- {
- str[i++] = '0'+n%10; /* Create a rightmost digit */
- n /= 10; /* Remove the digit */
- }while(n>0); /* Go again if there's more digits */
- if(negative) /* If it was negative */
- str[i++] = '-'; /* Append minus */
- str[i] = '\0'; /* Append terminator */
- length = i; /* Save the length */
- /* Now reverse the string in place */
- /* by switching first and last, */
- /* second and last but one, etc */
- for(i = 0 ; i<length/2 ;i++)
- {
- temp = str[i];
- str[i] = str[length-i-1];
- str[length-i-1] = temp;
- }
- return str; /* Return the string */
- }
- int main(void)
- {
- char str[15]; /* Stores string representation of integer */
- long testdata[] = { 30L, -98L, 0L, -1L, 999L, -12345L};
- for (int i = 0 ; i< sizeof testdata/sizeof(long) ; i++)
- printf("Integer value is %d, string is %s\n", testdata[i], itoa(testdata[i],str));
- return 0;
- }
习题3: 扩展为上一题定义的函数，使函数接受两个参数，以指定结果的字段宽度，使返回的字符串表示右对齐。例如，如果第一个变元的值是-98，字段宽度变元是5，返回的字符串就应是" -98"。用适当的main()版本说明函数的执行过程。
- /* Exercise 8.3 A function to return a string representation of an integer with a given width */
- #include <stdio.h>
- /* Convert an integer to a string with a fixed width. */
- /* if the widh is too small, the minimum width is assumed. */
- char* itoa(int n, char str[], int width)
- {
- int i = 0; /* Loop counter */
- int j = 0; /* Loop counter */
- int negative = 0; /* Indicate negative integer */
- int length = 0; /* Length of string */
- int temp = 0; /* Temporary storage */
- if(negative = (n<0)) /* Is it negative? */
- n = -n; /* make it positive */
- /* Generate digit characters in reverse order */
- do
- {
- str[i++] = '0'+n%10; /* Create a rightmost digit */
- n /= 10; /* Remove the digit */
- }while(n>0); /* Go again if there's more digits */
- if(negative) /* If it was negative */
- str[i++] = '-'; /* Append minus */
- str[i] = '\0'; /* Append terminator */
- length = i; /* Save the length */
- /* Now reverse the string in place */
- /* by switching first and last, */
- /* second and last but one, etc */
- for(i = 0 ; i<length/2 ;i++)
- {
- temp = str[i];
- str[i] = str[length-i-1];
- str[length-i-1] = temp;
- }
- /* Shift the string to the right and insert blanks */
- if(width>length)
- {
- for(i=length, j = width ; i>= 0 ; i--, j--)
- str[j] = str[i];
- for(i = 0 ; i<width-length ; i++)
- str[i] = ' ';
- }
- return str; /* Return the string */
- }
- int main(void)
- {
- char str[15]; /* Stores string representation of integer */
- long testdata[] = { 30L, -98L, 0L, -1L, 999L, -12345L};
- for (int i = 0 ; i< sizeof testdata/sizeof(long) ; i++)
- printf("Integer value is %10d, string is %s\n", testdata[i], itoa(testdata[i],str, 14));
- return 0;
- }
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