C Primer Plus第十二章编程练习答案

学完C语言之后，我就去阅读《C Primer Plus》这本经典的C语言书籍，对每一章的编程练习题都做了相关的解答，仅仅代表着我个人的解答思路，如有错误，请各位大佬帮忙点出！

1.不使用全局变量，重写程序清单12.4。

#include <stdio.h>
void critic(int* units)
{
    printf("No luck, my friend. Try again.\n");
    scanf("%d", units);
}
int main(void)
{
    int units;

    printf("How many pounds to a firkin of butter?\n");
    scanf("%d", &units);

    while (units != 56)
    {
        critic(&units);
    }
    printf("You must have looked it up!\n");

    return 0;
}

2.在美国，通常以英里/加仑来计算油耗；在欧洲，以升/100 公里来计算。下面是程序的一部分，提示用户选择计算模式（美制或公制），然后接收数据并计算油耗。

// pe12-2b.c

// 与 pe12-2a.c 一起编译

#include <stdio.h>

#include "pe12-2a.h"

int main(void)

{

int mode;

printf("Enter 0 for metric mode, 1 for US mode: ");

scanf("%d", &mode);

while (mode >= 0)

{ set_mode(mode);

get_info(); show_info();

printf("Enter 0 for metric mode, 1 for US mode");

printf(" (-1 to quit): "); scanf("%d", &mode); }

printf("Done.\n"); return 0; }

下面是是一些输出示例：

Enter 0 for metric mode, 1 for US mode: 0

Enter distance traveled in kilometers: 600

Enter fuel consumed in liters: 78.8

Fuel consumption is 13.13 liters per 100 km.

Enter 0 for metric mode, 1 for US mode (-1 to quit): 1

Enter distance traveled in miles: 434

Enter fuel consumed in gallons: 12.7

Fuel consumption is 34.2 miles per gallon.

Enter 0 for metric mode, 1 for US mode (-1 to quit): 3

Invalid mode specified. Mode 1(US) used.

Enter distance traveled in miles: 388

Enter fuel consumed in gallons: 15.3

Fuel consumption is 25.4 miles per gallon.

Enter 0 for metric mode, 1 for US mode (-1 to quit): -1

Done.

如果用户输入了不正确的模式，程序向用户给出提示消息并使用上一次输入的正确模式。请提供pe12-2a.h头文件和pe12-2a.c源文件。源代码文件应定义3个具有文件作用域、内部链接的变量。一个表示模式、一个表示距离、一个表示消耗的燃料。get_info()函数根据用户输入的模式提示用户输入相应数据，并将其储存到文件作用域变量中。show_info()函数根据设置的模式计算并显示油耗。可以假设用户输入的都是数值数据。

#include <stdio.h>
#include "pe12-2a.h"
int main(void)
{
    int mode;

    printf("Enter 0 for metric mode, 1 for US mode: ");
    scanf("%d", &mode);
    while (mode >= 0)
    {
        set_mode(mode);
        get_info();
        show_info();
        printf("Enter 0 for metric mode, 1 for US mode");
        printf(" (-1 to quit): ");
        scanf("%d", &mode);
    }
    printf("Done.\n");

    return 0;
}

3.重新设计编程练习2，要求只使用自动变量。该程序提供的用户界面不变，即提示用户输入模式等。但是，函数调用要作相应变化。

#include <stdio.h>
#include "pe12-2a.h"
int main(void)
{
    int temp, mode;
    double range, fuel;

    printf("Enter 0 for metric mode, 1 for US mode: ");
    scanf("%d", &mode);
    temp = mode;
    while (mode >= 0)
    {
        set_mode(&mode, &temp);
        get_info(temp, &range, &fuel);
        show_info(temp, range, fuel);
        printf("Enter 0 for metric mode, 1 for US mode");
        printf(" (-1 to quit): ");
        scanf("%d", &mode);
    }
    printf("Done.\n");

    return 0;
}

4.在一个循环中编写并测试一个函数，该函数返回它被调用的次数。

#include <stdio.h>
static int count = 0;
int counter(void)
{
    return ++count;
}
int main(void)
{
    int i, j;
    printf("Please enter a integer: ");
    scanf("%d", &i);

    for (j = 1; j <= i; j++)
    {
        printf("count = %d\n", counter());
    }
    printf("The function called %d times.\n", count);

    return 0;
}

5.编写一个程序，生成100个1～10范围内的随机数，并以降序排列（可以把第11章的排序算法稍加改动，便可用于整数排序，这里仅对整数排序）。

#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#define LEN 100
void sort(int a[], int n)
{
    int i, j, t;

    for (i = 0; i < n - 1; i++)
    {
        for (j = i + 1; j < n; j++)
        {
            if (a[i] < a[j])
            {
                t = a[j];
                a[j] = a[i];
                a[i] = t;
            }
        }
    }
    return;
}
void show_array(const int a[], int n)
{
    int i;

    for (i = 0; i < n; i++)
    {
        printf("%-3d ", a[i]);
        if (0 == (i + 1) % 10)
        {
            printf("\n");
        }
    }
    printf("\n");
}
int main(void)
{
    int i, a[LEN];

    srand((unsigned int)time(0));
    for (i = 0; i < LEN; i++)
    {
        a[i] = rand() % 10 + 1;
    }
    printf("Array:\n");
    show_array(a, LEN);
    sort(a, LEN);
    printf("After sorting:\n");
    show_array(a, LEN);

    return 0;
}

6.编写一个程序，生成1000个1～10范围内的随机数。不用保存或打印这些数字，仅打印每个数出现的次数。用 10 个不同的种子值运行，生成的数字出现的次数是否相同？可以使用本章自定义的函数或ANSI C的rand()和 srand()函数，它们的格式相同。这是一个测试特定随机数生成器随机性的方法。

#include <stdio.h>
#include <stdlib.h>
#define N 10
#define LEN 1000
int main(void)
{
    int i, temp, a[N + 1];
    unsigned int seeds;

    for (seeds = 1; seeds <= N; seeds++)
    {
        printf("Time #%d:\n", seeds);
        srand(seeds);
        for (i = 0; i < N + 1; i++)
        {
            a[i] = 0;
        }
        for (i = 0; i < LEN; i++)
        {
            temp = rand() % 10 + 1;
            a[temp]++;
        }
        for (i = 1; i < N + 1; i++)
        {
            printf("%-3d appeared %d times.\n", i, a[i]);
        }
        printf("Total random numbers: %d\n\n", LEN);
    }

    return 0;
}

7.编写一个程序，按照程序清单12.13输出示例后面讨论的内容，修改该程序。使其输出类似： Enter the number of sets;

enter q to stop : 18

How many sides and how many dice? 6 3

Here are 18 sets of 3 6-sided throws.

12 10 6 9 8 14 8 15 9 14 12 17 11 7 10

13 8 14 H

ow many sets?

Enter q to stop: q

#include <stdio.h>
#include <time.h>
#include <stdlib.h>
int rollem(int sides)
{
    return rand() % sides + 1;
}
int main(void)
{
    int dice, count, roll;
    int sides;
    int set, sets;

    srand((unsigned int)time(0));
    printf("Enter the number of sets.\nEnter q to stop: ");
    while (scanf("%d", &sets) == 1)
    {
        printf("How many sides and how many dice?\n");
        printf("Please two integers: ");
        if (scanf("%d %d", &sides, &dice) != 2)
        {
            puts("Not integers -- terminating input loop.");
            break;
        }
        printf("Here are %d sets of %d %d-sided throws.\n", sets, dice, sides);
        for (set = 0; set < sets; set++)
        {
            for (roll = 0, count = 0; count < dice; count++)
            {
                roll += rollem(sides);
            }
            printf("%-3d", roll);
            if (0 == (set + 1) % 8)
            {
                putchar('\n');
            }
        }
        printf("\nHow many sets? Enter q to stop: ");
    }
    puts("GOOD FORTUNE TO YOU!\n");

    return 0;
}

8.下面是程序的一部分：

// pe12-8.c

#include <stdio.h>

int * make_array(int elem, int val);

void show_array(const int ar [], int n);

int main(void)

{ int * pa;

int size;

int value;

printf("Enter the number of elements: ");

while (scanf("%d", &size) == 1 && size > 0)

{ printf("Enter the initialization value: ");

scanf("%d", &value);

pa = make_array(size, value);

if (pa) { show_array(pa, size); free(pa); }

printf("Enter the number of elements (<1 to quit): "); }

printf("Done.\n");

return 0; }

提供make_array()和show_array()函数的定义，完成该程序。make_array() 函数接受两个参数，第1个参数是int类型数组的元素个数，第2个参数是要赋给每个元素的值。该函数调用malloc()创建一个大小合适的数组，将其每个元素设置为指定的值，并返回一个指向该数组的指针。show_array()函数显示数组的内容，一行显示8个数。

#include <stdio.h>
#include <stdlib.h>
int* make_array(int elem, int val)
{
    int i;
    int* pt;
    pt = (int*)malloc(elem * sizeof(int));
    if (NULL == pt)
    {
        printf("Memory allocation failed!\n");
        exit(0);
    }
    printf("Output %d numbers:\n", val);
    for (i = 0; i < elem; i++)
    {
        pt[i] = val;
    }
    return pt;
}
void show_array(const int ar[], int n)
{
    int i;

    for (i = 0; i < n; i++)
    {
        printf("%d ", ar[i]);
        if (0 == (i + 1) % 8)
        {
            printf("\n");
        }
    }
    printf("\n");
    return;
}
int main(void)
{
    int* pa;
    int size;
    int value;

    printf("Enter the number of elements: ");
    while (scanf("%d", &size) == 1 && size > 0)
    {
        printf("Enter the initialization value: ");
        scanf("%d", &value);
        pa = make_array(size, value);
        if (pa)
        {
            show_array(pa, size);
            free(pa);
        }
        printf("Enter the number of elements (<1 to quit): ");
    }
    printf("Done.\n");

    return 0;
}

9.编写一个符合以下描述的函数。首先，询问用户需要输入多少个单词。然后，接收用户输入的单词，并显示出来，使用malloc()并回答第1个问题（即要输入多少个单词），创建一个动态数组，该数组内含相应的指向 char的指针（注意，由于数组的每个元素都是指向char的指针，所以用于储存malloc()返回值的指针应该是一个指向指针的指针，且它所指向的指针指向char）。在读取字符串时，该程序应该把单词读入一个临时的char数组，使用malloc()分配足够的存储空间来储存单词，并把地址存入该指针数组（该数组中每个元素都是指向 char 的指针）。然后，从临时数组中把单词拷贝到动态分配的存储空间中。因此，有一个字符指针数组，每个指针都指向一个对象，该对象的大小正好能容纳被储存的特定单词。下面是该程序的一个运行示例：

How many words do you wish to enter? 5

Enter 5 words now:

I enjoyed doing this exerise

Here are your words: I

enjoyed

doing

this

exercise

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define LEN 256
int main(void)
{
    char** pt;
    int i, n, length;
    static char temp[LEN];

    printf("How many words do you wish to enter? ");
    scanf("%d", &n);
    if ((pt = (char**)malloc(n * sizeof(char*))) != NULL)
    {
        printf("Enter %d words now:\n", n);
        for (i = 0; i < n; i++)
        {
            scanf("%255s", temp);
            length = strlen(temp) + 1;
            pt[i] = (char*)malloc(length * sizeof(char));
            if (NULL == pt[i])
            {
                printf("Memory allocation failed!\n");
                exit(EXIT_FAILURE);
            }
            strcpy(pt[i], temp);
        }
        printf("Here are your words:\n");
        for (i = 0; i < n; i++)
        {
            puts(pt[i]);
            free(pt[i]);
            pt[i] = NULL;
        }
        free(pt);
        pt = NULL;
    }
    else
    {
        printf("Memory allocation failed!\n");
        exit(EXIT_FAILURE);
    }

    return 0;
}