#include<iostream.h>
void main()
{
int a=1,sum=1;
int i,j,array[100]; // 宣个 i , j都为int 型态变数 , array配置 100个int型态大小的元素
int c,d,you[10][10]; //c,d 皆为int型态 ,you[10][10]二维阵列 元素都是int相当一维you[100]
char e;
long* pointer1 = NULL; //将pointer1初始化先指向NULL就是没有指向任何有效记忆体位址
long data1 = 10,data2 = 20;
char A[4][30] = { "天","下","为","公"}; //二维阵列char[4][30] 4 代表有字串个数 30代表长度
char* B[] = { "天","下","为","公"}; //指标阵列 元素是char *型态 可以指向静态记忆体位址字串之类的
pointer1 = &data1; //pointer1指向data1
*pointer1 += 30; //*pointer1相当于指向data1记忆体中取值从data1 + 30
cout << "A=" << data1;
pointer1 = &data2; //pointer1改指向data2
data1 = *pointer1*2; //现在*pointer1指向 data2 所以是取址 *2 = 20*2 ; 最后data1 = 40;
cout << endl <<"B="<< dec <<data1;
cout << endl <<"C="<< dec <<*pointer1;
cout << endl <<"E="<< B[2]<< endl;
while(a<5)
{
sum =sum * a;
a++;
}
cout<<"The sum is : "<<sum<<"\n";
for(i=10;i<12;i++){
j=i-1;
array[j]=i;
cout<<"array["<<j<<"]="<<array[j]<<"\n";
}
for(c=0;c<2;c++){
for(d=0;d<1;d++){
you[c][d]=c*10+d+1;
cout<<"you["<<c<<"]["<<d<<"]="<<you[c][d]<<"\n";
}
}
e='d'
if(e == 'a')
{cout <<"立志贵坚，坚而有恒，其学必成。"<< endl;}
else
if(e == 'b')
{cout <<"等待是偷懒的藉口，怠惰是推托。"<< endl;}
else
if(e == 'c')
{cout <<"君子以同道为友，小人以同利为友。"<< endl; }
else
if(e == 'd')
{cout <<"成功的唯一秘诀－坚持最后一分钟。"<< endl; }
}
我只将 * and & 做一些注解其他都是基本的东西所以我没在注解
这题目算是很基本题目 , 如果想学好程式应该自己练习看看
