C语言程序设计笔记04函数

C语言程序设计-笔记04-函数

例5-1  计算圆柱体的体积。输入圆柱的高和半径,求圆柱体积volume=πxr^2xh。要求定义和调用函数cylinder(r,h)计算圆柱体的体积。

#include<stdio.h>

double cylinder(double r,double h);

int main(void)

{

      double height,radius,volume;

     

      printf("Enter radius and height:");

      scanf("%lf%lf",&radius,&height);

      volume=cylinder(radius,height);

      printf("volume=%.3f\n",volume);

     

      return 0;

}

double cylinder(double r,double h)

{

      double result;

     

      result=3.1415926*r*r*h;

     

      return result;

}

a2

 

a1

 

例5-2  计算五边形的面积。将一个五边形分割成3个三角形,输入这些三角形的7条边长,计算该五边形的面积。要求定义和调用函数area(x,y,z)计算边长为x,y,z的三角形面积。

               
   

a7

 
   
 
   

a3

 
 
   

a4

 
 

 

 

#include<stdio.h>

#include<math.h>

int main(void)

{

      double a1,a2,a3,a4,a5,a6,a7,s;

      double area(double x,double y,double z);

      printf("Please input 7 side lengths in the order a1 to a7:\n");

      scanf("%lf%lf%lf%lf%lf%lf%lf",&a1,&a2,&a3,&a4,&a5,&a6,&a7);

      s=area(a1,a5,a6)+area(a6,a7,a4)+area(a7,a2,a3);

      printf("The area of the pentagon is %.2f\n",s);

     

      return 0;

}

double area(double x,double y,double z)

{

      double p=(x+y+z)/2;

     

      return sqrt(p*(p-x)*(p-y)*(p-z));//海伦-秦九韶公式

}

例5-3  使用函数判断完全平方数。定义一个判断完全平方数的函数IsSquare(n),当n为完全平方数时返回1,否则返回0,不允许调用数学库函数。

#include<stdio.h>

int IsSquare(n)

{

      int i;

     

      for(i=1;n>0;i=i+2)

      {

           n=n-i;

      }

      if(n==0)

      {

           return 1;

      }

      else

      {

           return 0;

      }

}

int main(void)

{

      int n;

      printf("Enter n:");

      scanf("%d",&n);

      if(IsSquare(n)==1)

      {

           printf("%d is a square number.\n",n);

      }

      else

      {

           printf("%d is not a square number.\n",n);     

      }

}

例5-4  使用函数求最大公约数。定义函数gcd(int m,int n),计算m和n的最大公约数。

#include<stdio.h>

int gcd(int m,int n)

{

      int r,temp;

     

      if(m<n)

      {

           temp=m;

           m=n;

           n=temp;

      }

      r=m%n;

      while(r!=0)

      {

           m=n;

           n=r;

           r=m%n;

      }

     

      return n;

}

int main(void)

{

      int m,n;

     

      printf("Enter m,n:");

      scanf("%d%d",&m,&n);

      printf("%d\n",gcd(m,n));

     

      return 0;

}

例5-5  使用函数判断素数。定义函数prime(m)判断m是否为素数,当m为素数时返回1,否则返回0.

#include<stdio.h>

#include<math.h>

int prime(int m)

{

      int i,limit;

     

      if(m<=1)

      {

           return 0;

      }

      else if(m==2)

      {

           return 1;

      }

      else

      {

           limit=sqrt(m)+1;

           for(i=2;i<=limit;i++)

           {

                 if(m%i==0)

                 {

                      return 0;

                 }

           }

          

           return 1;

      }

}

int main(void)

{

      int n;

      printf("Enter n:");

      scanf("%d",&n);

      if(prime(n)==1)

      {

           printf("%d is prime.\n",n);

      }

      else

      {

           printf("%d is not prime.\n",n);

      }

     

      return 0;

}

例5-6  数字金字塔。输入一个正整数n,输出n行数字金字塔。

#include<stdio.h>

int main(void)

{

      int n;

      void pyramid(int n);

     

      printf("Enter n:");

      scanf("%d",&n);

      pyramid(n);

     

      return 0;

}

void pyramid(int n)

{

      int i,j;

     

      for(i=1;i<=n;i++)

      {

           for(j=1;j<=n-i;j++)//输出每行数字前的空格

           {

                 printf(" ");

           }

           for(j=1;j<=i;j++)//输出每行的数字和数字后的空格

           {

                 printf("%d",i);

                 printf(" ");

           }

           putchar('\n');

      }

}

例5-7  计算2个复数之和与之积。分别输入2个复数的实部与虚部,用函数实现计算2个复数之和与之积。

#include<stdio.h>

double result_real,result_imag;    //全局变量

int main(void)

{

      double imag1,imag2,real1,real2;

      void complex_prod(double real1,double real2,double imag1,double imag2);

      void complex_add(double real1,double real2,double imag1,double imag2);

     

      printf("Enter 1st complex number(real and imaginary):");

      scanf("%lf%lf",&real1,&imag1);

      printf("Enter 2nd complex number(real and imaginary):");

      scanf("%lf%lf",&real2,&imag2);

      complex_add(real1,imag1,real2,imag2);

      printf("addition of complex is %f+%fi\n",result_real,result_imag);

      complex_prod(real1,imag1,real2,imag2);

      printf("product of complex is %f+%fi\n",result_real,result_imag);

     

      return 0;

}

void complex_add(double real1,double imag1,double real2,double imag2)

{

      result_real=real1+real2;

      result_imag=imag1+imag2;

}

void complex_prod(double real1,double imag1,double real2,double imag2)

{

      result_real=real1*real2-imag1*imag2;

      result_imag=real1*imag2+real2*imag1;

}

例5-8  用函数实现财务现金记账。先输入操作类型(1 收入,2 支出,0 结束),再输入操作金额,计算现金剩余金额,经多吃操作直到输入操作类型为0时结束。要求定义并调用函数,其中现金收入与现金支出分别用不同函数实现。

#include<stdio.h>

double cash;

int main(void)

{

      int choice;

      double value;

      void income(double number),expend(double number);

     

      cash=0;

      printf("Enter operate choice(0--end,1--income,2--expend):");

      scanf("%d",&choice);

      while(choice!=0)

      {

           if(choice==1||choice==2)

           {

                 printf("Enter cash value:");

                 scanf("%lf",&value);

                 if(choice==1)

                 {

                      income(value);

                 }

                 else

                 {

                      expend(value);

                 }

                 printf("current cash %.2f\n",cash);

           }

      printf("Enter operate choice(0--end,1--income,2--expend):");

      scanf("%d",&choice);    

      }

     

      return 0;

}

void income(double number)

{

      cash=cash+number;

}

void expend(double number)

{

      cash=cash-number;

}

例5-9  输入正整数n,输出1!-n!的值。要求定义并调用含静态变量的函数fact_s(n)计算n!.

#include<stdio.h>

double fact_s(int n);

int main(void)

{

      int i,n;

      printf("Input n:");

      scanf("%d",&n);

      for(i=1;i<=n;i++)

      {

           printf("%3d!=%.0f\n",i,fact_s(i));

      }

     

      return 0;

}

double fact_s(int n)

{

      static double f=1;

      f=f*n;

     

      return (f);

}

参考资料

C语言程序设计/何钦铭,颜晖主编.---4版.---北京:高等教育出版社,2020.9

原文地址:https://www.cnblogs.com/halflife/p/15532471.html