1001. NumCalClass

Time Limit: 1sec    Memory Limit:256MB

Description

補充完成類numCal的定義

(注意：所有運算符的重載都必須符合運算符原本的語義，比如++A是“先加後使用”)

class numCal {
public:
numCal(int a[], int size); //構造函數，size是數組a的長度
numCal(const numCal& other) //拷貝構造函數
~numCal(); //析構函數，要完成釋放內存的操作
int sum(); //求動態數組arr中size個元素值和
numCal& operator=(const numCal& other) //=運算符重載
numCal& operator++(); //前序++運算符重載，使得++A後，A.arr的每個元素值加1(A爲numCal類的對象)
numCal operator++(int); //後序++運算符重載，使得A++後，A.arr的每個元素值加1(A爲numCal類的對象)
//add any other function or property needed here
private:
int size; //數組arr元素個數
int *arr; //動態數組
};
調用示例：
int main()
{
int a[3] = {1, 3, 5};
numCal n1(a, 3);
cout << n1.sum() << endl;

numCal n2(n1);
cout << n2.sum() << endl;

int b[4]={2, 4, 6, 8};
numCal n3(b,4);
n2=n3;

cout << (n2++).sum() << endl;
cout << n2.sum() << endl;

cout << (++n2).sum() << endl;
cout << n2.sum() << endl;

n2 = n2;
return 0;
}

輸出結果：
9
9
20
24
28
28

注意：你只需要提交類定義。

Problem Source: 12級期末機考題

一發WA：(*@ο@*)  作大死，一小時了還是WA；

#include<iostream>
using namespace std;
class numCal {
public:
	numCal(int a[], int size){ //構造函數，size是數組a的長度
		arr = a;
		/*for (int i = 0; i < size; i++){
			arr[i] = a[i];
		}*/
		this->size = size;
	}
	numCal(const numCal& other){ //拷貝構造函數
		this->arr = other.arr;
		this->size = other.size;
	}
	~numCal(){//析構函數，要完成釋放內存的操作
		
	
	}
	int sum(){ //求動態數組arr中size個元素值和
		int sum = 0;
		for (int i = 0; i<size; i++){
			sum += arr[i];
		}
		return sum;
	}
	numCal& operator=(const numCal& other){ //=運算符重載
		size = other.size;
		arr = other.arr;
		return *this;
	}
	numCal& operator++(){ //前序++運算符重載，使得++A後，A.arr的每個元素值加1(A爲numCal類的對象)
		for (int i = 0; i<size; i++){
			arr[i]+=1;
		}
		return *this;
	}
	numCal operator++(int ize){//後序++運算符重載，使得A++後，A.arr的每個元素值加1(A爲numCal類的對象)
		numCal temp(*this);
		for (int i = 0; i<size; i++){
		    ++arr[i];
		}
		cout << (*this).sum() << endl;
		return temp;
	}
	//add any other function or property needed here
private:
	int size;	//數組arr元素個數
	int *arr;	//動態數組
};
int main(){
	int a[3] = { 1, 3, 5 };
	numCal n1(a, 3);
	cout << n1.sum() << endl;

	numCal n2(n1);
	cout << n2.sum() << endl;

	int b[4] = { 2, 4, 6, 8 };
	numCal n3(b, 4);
	n2 = n3;
	 
	cout << (n2++).sum() << endl;
	cout << (n2).sum() << endl;

	cout << (++n2).sum() << endl;
	cout << n2.sum() << endl;

	n2 = n2;
	system("pause");
	return 0;
}

最終AC：構造函數出現問題

#include<iostream>
using namespace std;
class numCal {
public:
	numCal(int a[], int size1)
	{
		int*list = new int;
		this->size = size1;
		for (int i = 0; i < size1; i++)
		{
			list[i] = a[i];
		}
		arr = list;
	}
	numCal(const numCal& other)
	{
		this->size = other.size;
		int* list = new int;
		for (int i = 0; i < size; i++)
		{
			list[i] = other.arr[i];
		}
		this->arr = list;
	}
	~numCal(){//析構函數，要完成釋放內存的操作


	}
	int sum(){ //求動態數組arr中size個元素值和
		int sum = 0;
		for (int i = 0; i<size; i++){
			sum += arr[i];
		}
		return sum;
	}
	numCal& operator=(const numCal& other){ //=運算符重載
		size = other.size;
		arr = other.arr;
		return *this;
	}
	numCal& operator++(){ //前序++運算符重載，使得++A後，A.arr的每個元素值加1(A爲numCal類的對象)
		for (int i = 0; i<size; i++){
			arr[i] += 1;
		}
		return *this;
	}
	numCal operator++(int ize){//後序++運算符重載，使得A++後，A.arr的每個元素值加1(A爲numCal類的對象)
		numCal temp(*this);
		for (int i = 0; i<size; i++){
			++arr[i];
		}
	
		return temp;
	}
	//add any other function or property needed here
private:
	int size;	//數組arr元素個數
	int *arr;	//動態數組
};

conclusion：=_=醉了，做了4個多小時才ac所有的題目。。還百度了很多問題，期末要掛的感覺。開始擼期末理論試題