《Linux C编程一站式学习》习题解答

programming
Author

Oscar Tuo

Published

January 29, 2025

学这本书主要是为了满足ysyx的要求,故会先以一个较快的速度更新完ysyx要求的部分(第1-9章, 第11-16章, 第21章, 第23-25章, 以及第26章第1节),未经特殊标记的程序默认采用C17标准编译通过。

编译环境:AMD Ryzen 7 7840HS,EndeavourOS x86-64,Linux 6.12.10,gcc 14.2.1(标了C23的用的是15.0)；跑不通的建议先想想是不是Windows的问题,再想想是不是英特尔的问题,再想想是不是Clang的问题

你可以在这里读到这本书: https://akaedu.github.io/book/index.html

第二章

习题2-2

#include <stdio.h>
int main(void)
{
    printf("%%\n");
}

习题2-5

#include <stdio.h>
#include <math.h>
int main(void)
{
    printf("%d\n",(int)ceil(17.0/4));
}

第三章

习题3-3

1.

不能increment里的x是形参，不会影响传进去的实参

2.

printf在stdio.h里，不把stdio.h给include进去的话printf没有定义

第四章

习题4-1

错误:if的条件满足后会执行一个空语句，下面那行并没有被if的判定所限制
编译能通过:没有语法错误

习题4-2

1.

x%10;(int) (x%100)/10

2.

void printdigit(int x)
{
    printf("%d,%d\n",x%10,(int) (x%100)/10)
}

习题4-3

1.

if (x>=10 || x<=0)
    printf("x is out of range\n");

2.

if (x<=0 && y<=0)
    printf("test failed!\n");
else
    printf("test ok!\n");

3.

y=1 || x=1

4.

总之是第二个

第五章

习题5-1

1.

请用C23编译


bool is_leap_year(int year)
{
    if(year%4!=0 || (year%100==0 && year%400!=0))
        return false;
    else
        return true;
}

2.

#include <math.h>
double myround(double x)
{
  if(x<0)
    return ceil(x-0.5);
  else 
    return floor(x+0.5);
}

用gcc编译的话请加入-lm参数,我也不知道这个大哥为什么默认不会去找math.h

习题5-3

1.

int euclid(int a, int b)
{
    if (a%b==0)
        return b;
    else
        return(euclid(b,a%b));
}

证明交给屏幕前的各位未来的yau们自己完成(

2.

int fibo(int n)
{
    if (n==0||n==1)
        return 1;
    else
        return(fibo(n-1)+fibo(n-2));
}

第六章

习题6-1

1.

用循环写递归本质上就是让你的脑子递归

(1).
int euclid(int a,int b)
{int remain = 1;
while (remain!=0)
{
    remain = a%b;
    a = b;
    b = remain;
}
return a;}
(2).
int fib(int n){
int fibo[100];
fibo[0]=1;
fibo[1]=1;
for(int i = 2;i<=n;i++)
{
    fibo[n]=fibo[n-2]+fibo[n-1];
}}

2.

int nine(void){
int nines=0;
for(int i = 1;i<101;i++)
{
    if ((i%10)==9||((int) i/10) == 9)
    nines++;
}
return nines;

习题6-4

1.

#include <stdio.h>
int is_prime(int n)
{
    int i=2;
    while(i < n && (n % i) !=0)
        i++;
    if (i == n)
        return 1;
    else
        return 0;
}

int main(void)
{
    int i;
    for (i = 1; i <= 100; i++) {
        if (is_prime(i))
        printf("%d\n", i);
    }
    return 0;
}

2.

可能会有某几次循环表达式3没有被执行

习题6-5

1.

#include <stdio.h>

int main(void)
{
    int i, j;
    for (i=1; i<=9; i++) {
        for (j=1; j<=9; j++)
            if(j<=i)    
            printf("%d\t", i*j);
        printf("\n");
    }
    return 0;
}

2.

void diamond(int n,char s)
{
  if(n%2==0)return;
  for(int i = 1;i<=n;i+=2)
  {
    for(int t = 0;t<(n-i)/2;t++)printf(" ");
    for(int t = 0;t<i;t++)
      printf("%c",s);
    printf("\n");
  }
  for(int i = n-2;i>0;i-=2)
  {
    for(int t = 0;t<(n-i)/2;t++)printf(" ");
    for(int t = 0;t<i;t++)
      printf("%c",s);
    printf("\n");
  }
}

第七章

习题7-2

1.

#include <stdio.h>
struct complex_struct{
    double x,y;
};
void output_complex(struct complex_struct a){
    if(a.x!=0)
  {
    printf("%.2lf",a.x);
    if(a.y!=0)printf("+");
    else printf("\n");
  }

  if(a.y!=0)
    printf("%.2lfi\n",a.y);
}
int main(void)
{
    struct complex_struct a = {1,2};
    struct complex_struct b = {0,1};
    struct complex_struct c = {1,0};
    output_complex(a);
    output_complex(b);
    output_complex(c);
}

2.

#include <stdio.h>

struct rational{
  int num,deno;
};


int euclid(int a, int b)
{
    if (a%b==0)
        return b;
    else
        return(euclid(b,a%b));
}
struct rational elliminate(struct rational c)
{
  int factor = euclid(c.num,c.deno);
  c.num = c.num/factor;
  c.deno = c.deno/factor;
  return c;
}
struct rational add_rational(struct rational a, struct rational b)
{
  struct rational sum;
  sum.num = a.num*b.deno + b.num*a.deno;
  sum.deno = a.deno*b.deno;
  sum = elliminate(sum);
  return sum;
}
struct rational sub_rational(struct rational a, struct rational b)
{
  struct rational differ;
  differ.num = a.num*b.deno - b.num*a.deno;
  differ.deno = a.deno*b.deno;
  differ = elliminate(differ);
  return differ;
}
struct rational mul_rational(struct rational a, struct rational b)
{
  struct rational product;
  product.num = a.num*b.num;
  product.deno = a.deno*b.deno;
  product = elliminate(product);
  return product;
}
struct rational div_rational(struct rational a, struct rational b)
{
  struct rational quotient;
  quotient.num = a.num*b.deno;
  quotient.deno = a.deno*b.num;
  quotient = elliminate(quotient);
  return quotient;
}
struct rational make_rational(int a, int b)
{
  struct rational c;
  c.num = a;
  c.deno = b;
  return c;
}
void print_rational(struct rational a)
{
  if(a.deno!=1)
    printf("%d/%d\n",a.num,a.deno);
  else
    printf("%d\n",a.num);
}
int main(void)
{
    struct rational a = make_rational(1, 8); /* a=1/8 */
    struct rational b = make_rational(-1, 8); /* b=-1/8 */
    print_rational(add_rational(a, b));
    print_rational(sub_rational(a, b));
    print_rational(mul_rational(a, b));
    print_rational(div_rational(a, b));

    return 0;
}

习题7-3

1.


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

enum coordinate_type { RECTANGULAR, POLAR };
struct complex_struct {
    enum coordinate_type t;
    double a, b;
};
struct complex_struct make_from_real_img(double x, double y)
{
    struct complex_struct z;
    z.t = RECTANGULAR;
    z.a = x;
    z.b = y;
    return z;
}
struct complex_struct make_from_mag_ang(double r, double A)
{
    struct complex_struct z;
    z.t = POLAR;
    z.a = r;
    z.b = A;
    return z;
}

double real_part(struct complex_struct in)
{
  if (in.t == RECTANGULAR)
    return in.a;
  else 
    return (in.a*cos(in.b));
}

double img_part(struct complex_struct in)
{
  if (in.t == RECTANGULAR)
    return in.b;
  else 
    return (in.a*sin(in.b));
}

double magnitude(struct complex_struct in)
{
  if (in.t == POLAR)
    return in.a;
  else 
    return sqrt(in.a*in.a+in.b*in.b);
}

double angle(struct complex_struct in)
{
  if (in.t == POLAR)
    return in.b;
  else 
    return atan(in.b/in.a);
}

int main(void)
{
  struct complex_struct a = make_from_real_img(2,3);
  struct complex_struct b = make_from_mag_ang(1,2);
  printf("%lf %lf %lf %lf %lf %lf %lf %lf\n",real_part(a),real_part(b),img_part(a),img_part(b),magnitude(a),magnitude(b),angle(a),angle(b));
  return 0;
}

使用gcc编译时请加入-lm指令，main函数主要是用来测试

2.

enumtest.cast
cols: 80
rows: 7
autoPlay: true
speed: 4
theme: nord
poster: "npt:1:23"

原因:enum里的RECTANGULAR被后边的初始化语句在main里给覆盖了，而POLAR则还是定义的2

第八章

习题8-1

#include <stdio.h>
#include <string.h>
int main(void)
{
  int a[5]={5,4,3,2,1};
  int b[5];
  memcpy(b,a,sizeof(a));
}

习题8-2

rand()%11+10

第十章

暂缺…

第十一章

习题11-4

#include <stdio.h>
int a[10]={1,2,3,4,5,6,7,8,9,0};
int partition(int start, int end)
{
    /*从a[start..end]中选取一个pivot元素（比如选a[start]为pivot）;
    在一个循环中移动a[start..end]的数据，将a[start..end]分成两半，
    使a[start..mid-1]比pivot元素小，a[mid+1..end]比pivot元素大，而a[mid]就是pivot元素;*/
    int mid = (int) (start+end)/2;
  for (int i = start;i<mid;i++)
  {
    if(a[i]>a[mid])
    {
      int temp=a[i];
      for(int t = i;t<mid;t++)
      {
        a[t]=a[t+1];
      }
      a[mid]=temp;
      mid--;
    }
  }
  for (int i = end;i>mid;i--)
  {
    if(a[i]<a[mid])
    {
      int temp=a[i];
      for(int t = i;t>mid;t--)
      {
        a[t]=a[t-1];
      }
      a[mid]=temp;
      mid++;
    }
  }

  return mid;
}

void quicksort(int start, int end)
{
    int mid;
    if (end > start) {
        mid = partition(start, end);
        quicksort(start, mid-1);
        quicksort(mid+1, end);
    }
}

int main(void)
{
  quicksort(0,9);
  for(int i = 0;i<10;i++)printf("%d\n",a[i]);
}

我估计这个应该不是最优算法,在原书内含和partition函数以外的部分都是用于测试

习题11-5

1.

没有，至少得把每个元素都遍历一遍才能确定是最小的

2.

#include <stdio.h>

int main(void)
{
  int data[5] = {1,2,3,4,5};
  int min=1,second=2;
  for(int i = 1;i<5;i++)
  {
    if (data[i]<=second && data[i] >= min)
      second = data[i];
    else if (data[i]<=min)
      min = data[i];
  }
  printf("%d\n",second);
}

我不是很确定这个的复杂度算不算是n

3.

#include <stdio.h>


int a[5] = {1,2,3,4,5};

int partition(int start, int end)
{
    int mid = (int) (start+end)/2;
  for (int i = start;i<mid;i++)
  {
    if(a[i]>a[mid])
    {
      int temp=a[i];
      for(int t = i;t<mid;t++)
      {
        a[t]=a[t+1];
      }
      a[mid]=temp;
      mid--;
    }
  }
  for (int i = end;i>mid;i--)
  {
    if(a[i]<a[mid])
    {
      int temp=a[i];
      for(int t = i;t>mid;t--)
      {
        a[t]=a[t-1];
      }
      a[mid]=temp;
      mid++;
    }
  }


  return mid;
}

void quicksort(int start, int end)
{
    int mid;
    if (end > start) {
        mid = partition(start, end);
        quicksort(start, mid-1);
        quicksort(mid+1, end);
    }
}

int order_statistic(int start, int end, int k)
{
    int i = partition(start,end);
    if (k-1 == i)
      return a[i];
    else if (k >= i)
  {
    quicksort(i+1,4);
    return a[k-1];
  }
    else
  { 
    quicksort(0,i-1);
    return a[k-1];
  }
}

int main(void)
{
  printf("%d\n",order_statistic(0,4,1));
}

ds说这个的平均复杂度接近\[n^2\],我不太相信但是这个的平均复杂度肯定比n大得多，我还需要再钻研一下。