这是我们的C++考试题,英语的,能不能麻烦各位高手帮我做一下啊,据说很简单的

用中文回答就行,越规范越简洁正确越好,我要打印下来的。网页上没法显示黑体,我就用引号了一.1.Describethedifferencesbetweenthe"point... 用中文回答就行,越规范越简洁正确越好,我要打印下来的。网页上没法显示黑体,我就用引号了

一.1.Describe the differences between the "pointer" and "reference".
2.A class mainly contains two kinds of members.What are they?
3.Compare the two ways to define a constant .Decide which way is better and tell the reason:
#define BUFSIZE 256
Const int bufsize=256
4.Please tell the difference among "public","protected" and "private" in a class.
5.Describe the "constructor" and "destructor" of class.
6.How to define the "copy constructor" ? Give an exemple to explain.
7.Give an example to implement the "polymorphism" in object-oriented programming.
8.How to define a "pure virtual function" ? Give an example.
9.How to define a "template function" ? Give an exemple to explain.
10.What is the "inline function" ?

三.Find out the error in the following code and give the reason.
1.class X{
private:
int a;
};
Class Y: public X{
public:
void set(int c){
This->a=c;
}
};
int main(){
Y y1;
y1.set(10);
return 0;
}

2.class X{
public :
int func(void){
return a++;
}
int a;
};
class Y:public X{
public :
int describe () const{
return func();
}
};
四.1.Please put the content in the following array ia into a vector and sort the data in ascend ,then output the result into standard stream.
int ia[]={46,4,89,3,5,78,12}

2.Please design a complex(复数) class which can be presented as a+bi .It implements the complex operations :+ ,-,x,÷.The rule for complex addition as follow:
suppose:C1:a1+b1i
C2:a2+b2i
C3=C1+C2
then
C3=(a1+a2)+(b1+b2)i
You reasonably use the "cpevator overloading".
就是,太鄙视一楼的翻译了。顶二楼,不过还是希望看看有没有其他更好的答案
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一.1.Describe the differences between the "pointer" and "reference".
指针和引用的区别:1.指针是一个实体,而引用仅是个别名;
2.引用只能在定义时被初始化一次,之后不可变;指针可变;
3.引用不能为空,指针可以为空;
2.A class mainly contains two kinds of members.What are they?
一个累包含的2种成员:1:数据成员 2:成员函数.
3.Compare the two ways to define a constant .Decide which way is better and tell the reason:
#define BUFSIZE 256
Const int bufsize=256
答: Const int bufsize=256更好,因为使用#define没有只是代码替换没有数据
类型检查,可能会出错
4.Please tell the difference among "public","protected" and "private" in a class.
答:1公有成员public
在程序的任何地方都可以被访问
2私有成员private
只能被成员函数和类的友元访问
3被保护成员protected
可以被子类访问
对其他程序则表现得像private
5.Describe the "constructor" and "destructor" of class.
构造函数是实例化一个对象。
析构函数销毁一个对象。
6.How to define the "copy constructor" ? Give an exemple to explain.
例子:
Example::Example(const Example& t) //拷贝构造函数的定义  
 {  
 nSize=t.nSize; //复制常规成员
  pBuffer=new char[nSize]; //复制指针指向的内容
  memcpy(t.pBuffer,pBuffer,nSize*sizeof(char));
  }
7.Give an example to implement the "polymorphism" in object-oriented programming.
#include <iostream>
using namespace std;
class B{
public:
void virtual vf() { cout << "This is class B" << endl; }
};
class D: public B{
public:
void vf() { cout << "This is class D" << endl; }
};
main(){
B b, *pb;
D d, *pd;
pb = &b; pb->vf();
pb = &d; pb->vf();
pd = (D*)&b; pd->vf();
pd = &d; pd->vf();
}
8.How to define a "pure virtual function" ? Give an example.

class B{
public:
void virtual vf() { } = 0;
};

9.How to define a "template function" ? Give an exemple to explain.

template<class T>
void fu(T,T){ cout < < "template version fun(T,T) called " < <endl;}

10.What is the "inline function" ?
内联函数:编译时,类似宏替换,使用函数体替换调用处的函数名。一般在代码
中用inline修饰,但是否能形成内联函数,需要看编译器对该函数定义的具体处理。
三.Find out the error in the following code and give the reason.
1.class X{
private: //这里错,,要用protected;
int a;
};
Class Y: public X{
public:
void set(int c){
This->a=c;
}
};
int main(){
Y y1;
y1.set(10);
return 0;
}

2.class X{
public :
int func(void){
return a++;
}
int a;
};
class Y:public X{
public :
int describe () const{ //这里错,不能有const
return func();
}
};
四.1.Please put the content in the following array ia into a vector and sort the data in ascend ,then output the result into standard stream.
int ia[]={46,4,89,3,5,78,12}

2.Please design a complex(复数) class which can be presented as a+bi .It implements the complex operations :+ ,-,x,÷.The rule for complex addition as follow:
suppose:C1:a1+b1i
C2:a2+b2i
C3=C1+C2
then
C3=(a1+a2)+(b1+b2)i
You reasonably use the "cpevator overloading".

答:
class RealNum{
public:

RealNum(double temp_x,double temp_y){
x = temp_x ;
y = temp_y ;
}

virtual double oper() = 0;

protected:
double x;
double y;
};

class Plus:public RealNum{
public:
Plus(double temp_x,double temp_y):RealNum(temp_x,temp_y){
}
double oper(){
return this ->x + this ->y;
}
};

class subtract:public RealNum{
public:
subtract(double temp_x,double temp_y):RealNum(temp_x,temp_y){
}
double oper(){
return this ->x - this ->y;
}
};

class multiply:public RealNum{
public:
multiply(double temp_x,double temp_y):RealNum(temp_x,temp_y){
}
double oper(){
return this ->x * this ->y;
}
};

class division:public RealNum{
public:
division(double temp_x,double temp_y):RealNum(temp_x,temp_y){
}
double oper(){
return this ->x / this ->y;
}
};

打了很久!希望你采纳!可能不是全对,但还是可信的
匿名用户
2011-07-20
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是清华的ACM试题吧,我刚做完,你试试看。

#include <iostream>
using std::cout;
using std::cin;
using std::endl;

struct item{
char key;
int priority;
struct item** smaller;
struct item** greater;
};

typedef struct item Item;

int placeToAdd(Item** a,char b,int n){
int i;
for(i=0;i<n;i++){
if((a[i]->key)==b) return i;
else if((a[i]->key)==' ') {
a[i]->key=b;
return i;
}
}
if(i>=n) return -1;
}

int hasCompared(Item** a,int i,int j,int n){
int m=0;
if(i==j)return -1;
while((((a[i]->greater[m])!=0||a[i]->smaller[m])!=0)&&m<n){
if((a[i]->greater[m])==a[j]) return 1;
else if((a[i]->smaller[m])==a[j]) return -1;
m++;
}
if(m<n){
a[i]->greater[m]=a[j];
m=0;
while((a[j]->smaller[m])!=0&&m<n) m++;
if(m<n) a[j]->smaller[m]=a[i];
return 0;
}
}

int keySetting(int i,int j,Item** a,int n){
int m=hasCompared(a,i,j,n);
if(m==-1) return -1;
else if(m==1) return 1;
else{
if(a[i]->priority>=a[j]->priority) a[j]->priority=a[i]->priority+1;
int k=0;
while((a[j]->greater[k])!=0&&k<n){
a[j]->greater[k]->priority+=1;
k++;
}
return 0;
}
}

int main(){
int l,m,n,i,j;
int unsorted=0,inconsistent=0;
char sign,first,second;
cin>>m>>n;
if(m==0||n==0){
cout<<"Sorted sequence cannot be determined."<<endl;
return 1;
}
Item** a=new Item *[m];
for(i=0;i<m;i++){
a[i]=new Item;
a[i]->key=' ';
a[i]->priority=0;
a[i]->smaller= new Item *[m];
a[i]->greater=new Item *[m];
for(j=0;j<m;j++){
a[i]->smaller[j]=0;
a[i]->greater[j]=0;
}
}
l=n;
while(n>0){
cin>>first>>sign>>second;
i=placeToAdd(a,first,m);
j=placeToAdd(a,second,m);
if(i!=-1&&j!=-1){
if(keySetting(i,j,a,m)==-1) inconsistent=1;
}
else unsorted=1;
n--;
}
if(unsorted) cout<<"Sorted sequence cannot be determined."<<endl;
else if(inconsistent) cout<<"Inconsistency found after "<<l<<" relations."<<endl;
else{
int min;
for(i=0;i<m-1;i++){
min=i;
for(j=i+1;j<m;j++){
if((a[j]->priority)<(a[min]->priority)) min=j;
else if((a[j]->priority)==(a[min]->priority)) unsorted=1;
}
Item* temp=a[i];
a[i]=a[min];
a[min]=temp;
}
if(unsorted) cout<<"Sorted sequence cannot be determined."<<endl;
else{
cout<<"Sorted sequence determined after "<<l<<" relations: ";
for(i=0;i<m;i++)
cout<<(a[i]->key);
cout<<endl;
}
}
for(i=0;i<m;i++){
delete[] (a[i]->smaller);
delete[] (a[i]->greater);
delete a[i];
}
delete[] a;
return 0;
}
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自动检测中英文中译英英译中 百度翻译
翻译结果(英 > 中)复制结果

1.describe之间的差异的“指针”和“参考”。

2类。主要包含2种members.what他们?

3.compare的方法来定义一个常数。决定哪种方式更好,告诉原因:

#定义bufsize256

整型常量bufsize=256

4.please告诉之间的差异,“公共”,“保护”和“私人”在一个班级。

5.describe“建设者”和“破坏者”级。

6.how界定“复制构造函数”?为例说明。

7.give例子执行“多态性”在面向对象编程。

8.how定义一个“纯虚拟函数”?举一个例子。

9.how定义一个“模板”功能?为例说明。

10.what是“内联函数”?

三。找出错误在下面的代码和给予的原因。

1.class×{

私人的:

返回一个;

};

类:公共×{

公开:

设置(第三){

这- >=丙;

}

};

诠释主体(){

你日圆;

第一集(10);

返回0;

}

2.class×{

公开:

国际功能(无效){

返回一个+ +;

}

返回一个;

};

类:公共×{

公开:

在描述()常量{

返回函数();

}

};

四。1.please把内容在以下数组是成为一个向量和排序数据上,然后将结果输出到标准流。

国际自动化[]={46,4,89,3,5,78,12}

2.please设计一个复杂(复数)类,可以作为一个联系。它实现了复杂的操作:+,-,×,÷。规则复杂除了如下:

假设:C 1:A 1+b1i

芹菜:2+b2i

c=C 1+C 2

然后

c=(1+2)+(A+B)i

你合理地使用“cpevator超载”。
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顶楼上...另外,,,鄙视一楼的...严重鄙视
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