C语言编程问题：因式分解 参数返回问题

1.因式分解

#include <iostream>#include <math.h>using namespace std;bool IsPrime(int n);void Resolve(int n);int main(){int n;cout<<"Input a integer num n(n>0):";cin>>n;if(n==1)cout<<"1=1"<<endl;else if(IsPrime(n))cout<<n<<"="<<"1*"<<n<<endl;else //  >1  and not primeResolve(n);cout<<endl;return 0;}bool IsPrime(int n){int k,i;k=static_cast<int>(sqrt((double)n));if(n==1)return false;for(i=2;i<=k;i++){if(n%i==0)break;}if(i>k)return true;else return false;}void Resolve(int n){int i;cout<<n<<"=";i=2;while(1){if(IsPrime(i)&&n%i==0){cout<<i<<"*";n=n/i;}elsei++;if(n==1)break;}}

2.函数返回类型为指针需要注意的问题

#include <stdio.h>#include <string.h>typedef struct student{char sex;char name[10];}stu;char *get();int main(){printf("1. %s\n",get());printf("2. %s\n",get());return 0;}char *get(){stu doctor;doctor.sex='M';strcpy(doctor.name,"wang");//方法1：分配堆（由程序员自己释放），返回局部变量的地址char *s;int len=strlen((char *)&doctor);s=new char[len+1];strcpy(s,(char *)&doctor);return s;//return (char *)&doctor;}

char *get(){static stu doctor;//方法2：申明为static类型，返回局部变量的地址doctor.sex='M';strcpy(doctor.name,"wang");return (char *)&doctor;}