动态内存分配重载函数和内联函数的试验

1、实验二、动态内存分配、重载函数和内联函数的试验试验目的通过本试验掌握使用new和delete的运算符创建动态数组的方法.。并熟悉重载、内联函数的编写与调用，了解面向对象程序设计的多态性原理。试验要求要求独立完成。以WORD文档形式，在webcc上提交源程序和运行结果（图）。编程特别强调程序可读性，注意要有足够的注释；使用代码缩进格式；对齐及有合适的空行。试验内容1. 动态数组能实现按需定义数组的大小，节省内存。试参照书中的程序2-9编写一程序将一行任意个数以空格分隔的数值数据读到一个数组中，并输出数据的个数、最大值、最小值、平均值及过零的次数(相邻两个数据符号相反)。#include <

2、iostream>using namespace std;double max(double *data,int number ) /求最大值double i=0.0;i=*data;while(-number)data+;if(i<*data)i=*data;return i;double min(double *data,int number ) /求最小值double i=0.0;i=*data;while(-number)data+;if(i>*data)i=*data;return i;double average(double *data,int number)

3、/求平均值double sum=0.0;int i=number;while(number-)sum+=*data+;return sum/i;int postive(double *data,int number) /求正数的个数int i=0;while(number-)if(*data+>0)i+;return i;int main() double data, *p_data, *p_temp=NULL; int num=0,i; cout << "Type a line of numeric data(separated by space), press

4、Enter to end: n" do cin >> data;num+;p_data = new doublenum;if (p_data = NULL) cout << "Not enough memory from heap.n" return 0; if(p_temp = NULL) *p_data =data;else for (i=0; i < num-1; i+) *(p_data+i) = *(p_temp+i); p_datanum-1 = data; delete p_temp;p_temp = p_data; w

5、hile (cin.get()!='n');cout << "there are " << num << " numbers" << endl;cout<<"the max number is "<<max(p_data,num)<<endl;cout<<"the min number is "<<min(p_data,num)<<endl;cout<<"th

6、e average is "<<average(p_data,num)<<endl;cout<<"the postive number is "<<postive(p_data,num)<<endl; delete p_data; return 0;运行结果2. 编写一个程序，用同一个函数名求任意2个或3个整数中的最大公约数。要求用重载函数实现。#include<iostream>using namespace std;int commer(int a,int b) /求两个数的最大公约数in

7、t com=0;if(a<b)com=a;elsecom=b;while(com)if(a%com)|(b%com)com-;elsebreak;return com;int commer(int a,int b,int c) /求三个数的最大公约数int com=0;if(a<b&&a<c)com=a;elseif(b<a&&b<c)com=b;elsecom=c;while(com)if(a%com)|(b%com)|(c%com)com-;elsebreak;return com;int main() /主函数int num=

8、0,i,data=0;int *p_data,*p_temp=NULL;cout<<" 输入两个或者三个正数n" do cin >> data;num+;p_data = new intnum;if (p_data = NULL) cout << "Not enough memory from heap.n" return 0; if(p_temp = NULL) *p_data =data;else for (i=0; i < num-1; i+) *(p_data+i) = *(p_temp+i); p_datanum-1 = data; delete p_temp;p_temp = p_data; while (cin.get()!='n'); if(num=2) cout<<"the max commer divisor is "<<commer(*p_data,*(p_data+1)<<endl; else co