如何输入向量? [英] How to make an input of vector?

问题描述

你好，请帮助我.

我写了一个简单的程序来创建和编辑矢量.有必要对这两个向量进行估算.如您所见，在我的代码中，默认情况下设置了向量:
vect x(5,6,7,8,9)，y(4,3,2,1,7);

Hello, please, help me someone.

I wrote a simple program for creating and editing vector. It is necessary to make an imput of these two vectors. As you see, in my code, vectors are set by default:
vect x(5,6,7,8,9),y(4,3,2,1,7);

#include <conio.h>
#include <iostream.h>

class vect {
public:
   vect(int=0,int=0,int=0,int=0,int=0);   
   vect operator+(const vect &) const;   
   vect operator-(const vect &) const;
   vect operator*(const vect &); 
   vect operator*(const int &); 
   
   void print() const;
   private:
   int a,b,c,d,e;
};
vect::vect(int a1,int b1,int c1,int d1,int e1)
{
   a = a1;
   b = b1;
   c = c1;
   d = d1;
   e = e1;
}
vect vect::operator+(const vect &operand2) const
{
   vect sum;
   sum.a = a + operand2.a;
   sum.b = b + operand2.b;
   sum.c = c + operand2.c;
   sum.d = d + operand2.d;
   sum.e = e + operand2.e;
   return sum;
}
vect vect::operator-(const vect &operand2) const
{
   vect diff;
   diff.a = a - operand2.a;
   diff.b = b - operand2.b;
   diff.c = c - operand2.c;
   diff.d = d - operand2.d;
   diff.e = e - operand2.e;
   return diff;
}
vect vect::operator*(const vect &operand2)
{
   vect proizv;
   proizv.a = a * operand2.a;
   proizv.b = b * operand2.b;
   proizv.c = c * operand2.c;
   proizv.d = d * operand2.d;
   proizv.e = e * operand2.e;
   return proizv;
}
vect vect::operator*(const int &operand2)
{
   vect proizv;
   proizv.a = a * operand2;
   proizv.b = b * operand2;
   proizv.c = c * operand2;
   proizv.d = d * operand2;
   proizv.e = e * operand2;
   return proizv;
}
void vect::print() const
{
   cout<<''(''<<a<<", "<<b<<", "<<c<<", "<<d<<", "<<e<<'')'';
}
int main(int argc, char* argv[])
{
   vect x(5,6,7,8,9),y(4,3,2,1,7);
   vect z;
   int scl = 8;
   cout<<"\n Vector 1:   ";
   x.print();
   cout<<"\n Vector 2:   ";
   y.print();
   z = x + y;
   cout<<"\n Slozhenie:   ";
   z.print();
   z = x - y;
   cout<<"\n Vichitanie:  ";
   z.print();
   z = x * y;
   cout<<"\n Vector*Vector:   ";
   z.print();
   z = x * scl;
   cout<<"\n Scalar1:   ";
   z.print();
   z = y * scl;
   cout<<"\n Scalar2:   ";
   z.print();
   getch();
   return 0;
}

推荐答案

我猜你的意思是:

I guess you mean something like:

    integer a,b,c,d,e;

    cout << "Enter vector 1 values: " << endl;
    cin >> a;
    cin >> b;
    cin >> c;
    cin >> d;
    cin >> e;
   vect x(a, b, c, d, e);
// repeat for vector y

显然不是很复杂，但是我相信您会明白.

Obviously not very sophisticated but I trust you get the idea.

看看如何实现插入和提取运算符，它将使您的代码更容易阅读.例如，如果您实现operator<<和运算符>>您可能最终会写出类似以下内容的内容:

Have a look at how you implement the insertion and extraction operators, it''ll make your code a lot easier to read. For example if you implement operator<< and operator>> you could end up writing something like:

int main()
{
    std::cout << "Please enter 10 integers, white space delimited to initialise my vectors: ";

    vect a, b;
    std::cin >> a >> b;

    std::cout << "Sum of your vectors is: " << a + b << std::endl;
    std::cout << "Difference of your vectors is: " << a - b << std::endl;
}

现在那里没有很多错误处理，但是它表明您的代码可以简单得多.

无论您用来学习C ++的文本是什么，都应包含有关如何重载运算符的更多详细信息.呆呆的去吧，如果您有问题，请提出更具体的问题.

干杯，

灰

Now there''s not a lot of error handling there but it shows how much simpler your code could be.

Whatever text you''re using to learn C++ from should have some more details on how to overload operators. Have a gawp there give it a go and if you''re having problems ask a more specific questions.

Cheers,

Ash

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