我正在使用嵌套的结构/结构,经过几个小时的伪代码和尝试后,我提出的最终结果不起作用或不能编译。
我想采用两个向量A和B,并将它们相互比较。我设置了嵌套结构来读取向量的起点和终点,以及向量结构本身。所以我想我可能在下面做了一些错误,但我被卡住了。
#include <iostream>
#include <cmath>
#include <string>
using namespace std;
struct Point // Reads in three coordinates for point to make a three dimensional vector
{
double x;
double y;
double z;
};
struct MathVector // Struct for the start and end point of each vector.
{
Point start;
Point end;
};
Point ReadPoint()
{
Point pt; // Letter distinguishes between vector A and vector B, or "letterA" and "letterB"
double x, y, z;
cout << "Please input the x-coordinate: " << endl;
cin >> pt.x;
cout << "Please input the y-coordinate: " << endl;
cin >> pt.y;
cout << "Please input the z-coordinate: " << endl;
cin >> pt.z;
return pt;
}
void DotProduct (MathVector letterA, MathVector letterB, double& a_times_b ) //formula to compute orthogonality
{
a_times_b = (letterA.end.x - letterA.start.x)*(letterB.end.x - letterB.start.x) + (letterA.end.y - letterA.start.y)*(letterB.end.y - letterB.start.y) + (letterA.end.z - letterA.start.z)*(letterB.end.z - letterB.start.z);
}
int main()
{
MathVector letterA;
MathVector letterB;
double a_times_b;
letterA = ReadPoint();
letterB = ReadPoint();
DotProduct (letterA, letterB, a_times_b);
cout << "The vector " << letterA << " compared with " << letterB << " ";
if ( a_times_b == 0)
cout << "is orthoganal." << endl;
else
cout << "is not orthoganal." << endl;
return 0;
}
答案 0 :(得分:2)
问题在于您的ReadPoint
返回类型为Point
,但您正在返回MathVector
的实例。此外,您还读取了最终忽略的变量的输入。
您应该将ReadPoint
写为:
Point ReadPoint()
{
Point p;
cout << "Please input the x-coordinate: " << endl;
cin >> p.x;
cout << "Please input the y-coordinate: " << endl;
cin >> p.y;
cout << "Please input the z-coordinate: " << endl;
cin >> p.z;
return p;
}
或者更好的版本:
Point ReadPoint()
{
Point p;
cout << "Please enter point-coordinate : " << endl;
cin >> p.x >> p.y >> p.z; //example input : 1 2 3
return p;
}
或者,更好的是,将>>
运算符重载为:
std::istream & operator>>(std::istream & in, Point & p)
{
cout << "Please enter point-coordinate : " << endl;
return cin >> p.x >> p.y >> p.z; //example input : 1 2 3
}
//Use this as
Point pointA, pointB;
cin >> pointA >> pointB;
现在阅读一本好的C ++书。如果你已经阅读一个,那么确保它真的好。以下列出了所有级别的真正好的C ++书籍:
答案 1 :(得分:0)
答案 2 :(得分:0)
Point ReadPoint()
{
MathVector letter; // Letter distinguishes between vector A and vector B, or "letterA" and "letterB"
double x, y, z;
cout << "Please input the x-coordinate: " << endl;
cin >> x;
cout << "Please input the y-coordinate: " << endl;
cin >> y;
cout << "Please input the z-coordinate: " << endl;
cin >> z;
return letter;
}
你没有解释你正在尝试做什么或者你得到了什么错误,但这段代码使没有感觉。您有三个变量x
,y
和z
。您可以使用从用户获得的值填充它们。然后,您不会对这些变量执行任何操作,并返回默认构造函数创建的MathVector
,即使您说您要返回Point
。这没什么意义。
答案 3 :(得分:0)
letterA
和letterB
的类型为MathVector
MathVector letterA;
MathVector letterB;
double a_times_b;
letterA = ReadPoint();
letterB = ReadPoint();
您应该创建另一种方法来阅读Mathvector
..正如您使用Point
一样。
和方法ReadPoint
返回类型必须是Point ..如果你读点,那么在这里进行计算以创建MathVector
go tet startpoint
和endpoint
格式的对象。
答案 4 :(得分:0)
'operator ='错误不匹配意味着没有将MathVector分配给Point的功能。您正在调用ReadPoint(),它返回一个Point并尝试将返回的值分配给MathVector类型的变量。编译器无法自动创建“转换”功能。你必须自己提供一个。也许你的意思是
letterA.start = ReadPoint();
letterA.end = ReadPoint();