重载运算符时二进制表达式的无效操作数
invalid operands to binary expression when overloading a operator
我正在尝试编写一个 C++ 程序,它使用 mul 和 add 运算符重载实现一个简单的矩阵。我的矩阵存储在一个简单的 Vector 中。我还重载了 () 运算符,以便轻松地将值存储在我的矩阵中。
template < typename T1, int rows, int cols>
class Matrix{
public:
int row = rows;
int line = cols;
vector<T1> local_matrix;
Matrix(){
local_matrix = vector<T1>(rows*cols, 0);
}
T1 &operator () (int i, int j){
return local_matrix.at(i * cols +j);
}
const Matrix<T1, rows, cols> operator* (const Matrix<T1, rows, cols> &other) const{
Matrix<T1, other.row, other.line> result; // store the result here
int sum_elements_i = 0;
int left_counter = 0; // counter to iterate over the left Matrix
int right_counter = 0; // and for the right one
for(int i = 0; i < rows -1;++i){
for(int j = 0; j < other.spalte -1; ++j){
sum_elements_i = 0;
for(int k = 0; k < other.reihe -1; ++k){
sum_elements_i += local_matrix.at(left_counter) * other.local_matrix.at(right_counter);
left_counter++; // get the next value of the left matrix
right_counter += other.line; // and the next one of the right here we need to skip n-elements
// so its the 1st/2nd/.. value of the next line
}
right_counter = j+1;
result(i, j) = sum_elements_i;
}
left_counter = i+1;
}
return result;
}
};
我对 * 运算符的问题是,每次我想将 4x3 矩阵与 3x3 矩阵相乘时,程序都会崩溃并出现以下错误消息:
invalid operands to binary expression ('Matrix<int, 4, 3>' and 'Matrix<int, 3, 3>')
E = D*A;
~^~
这是我调用函数的代码:
Matrix<int, 3, 3> A; // Datatype, Rows, Columns
A(0,0) = 1; A(0,1) = 2; A(0,2) = 3;
A(1,0) = 1; A(1,1) = 2; A(1,2) = 3;
A(2,0) = 1; A(2,1) = 2; A(2,2) = 3;
Matrix<int, 4, 3> D, E;
D(0,0) = 1; D(0,1) = 2; D(0,2) = 3;
D(1,0) = 4; D(1,1) = 5; D(1,2) = 6;
D(2,0) = 7; D(2,1) = 8; D(2,2) = 9;
D(3,0) = 7; D(3,1) = 8; D(3,2) = 9;
std::cout << std::endl << "D:\n" << D << std::endl;
E = D*A;
非常感谢任何帮助,谢谢。
您必须为 operator *
使用正确的矩阵大小:
template <typename T, std::size_t N, std::size_t K, std::size_t M>
Matrix<T, N, M> operator* (const Matrix<T, N, K>& lhs, const Matrix<T, K, M>& rhs);
您定义函数的方式,您的函数具有签名 Matrix<int, 4, 3> Matrix<int, 4, 3>::operator(const Matrix<int, 4, 3> &) const
,这意味着您必须尝试将 4 x 3 矩阵与 4 x 3 矩阵相乘,并且它会以某种方式给出你另一个 4 x 3 矩阵。相反,您至少想要的是将 operator*
声明为
template <int cols2>
Matrix<int, rows, cols2> operator*(const Matrix<int, cols, cols2> & other) const {...}
这将允许您将一个 4 x 3 矩阵与一个 3 x 3 矩阵相乘,这将正确地输出一个 4 x 3 矩阵。
为了让更多类型的矩阵相乘,您可以进行的另一个更改是使签名
template <class T2, int cols2>
Matrix<typename std::common_type<T, T2>::type, rows, cols2>
operator*(const Matrix<T2, cols, cols2> & other) const {...}
我正在尝试编写一个 C++ 程序,它使用 mul 和 add 运算符重载实现一个简单的矩阵。我的矩阵存储在一个简单的 Vector 中。我还重载了 () 运算符,以便轻松地将值存储在我的矩阵中。
template < typename T1, int rows, int cols>
class Matrix{
public:
int row = rows;
int line = cols;
vector<T1> local_matrix;
Matrix(){
local_matrix = vector<T1>(rows*cols, 0);
}
T1 &operator () (int i, int j){
return local_matrix.at(i * cols +j);
}
const Matrix<T1, rows, cols> operator* (const Matrix<T1, rows, cols> &other) const{
Matrix<T1, other.row, other.line> result; // store the result here
int sum_elements_i = 0;
int left_counter = 0; // counter to iterate over the left Matrix
int right_counter = 0; // and for the right one
for(int i = 0; i < rows -1;++i){
for(int j = 0; j < other.spalte -1; ++j){
sum_elements_i = 0;
for(int k = 0; k < other.reihe -1; ++k){
sum_elements_i += local_matrix.at(left_counter) * other.local_matrix.at(right_counter);
left_counter++; // get the next value of the left matrix
right_counter += other.line; // and the next one of the right here we need to skip n-elements
// so its the 1st/2nd/.. value of the next line
}
right_counter = j+1;
result(i, j) = sum_elements_i;
}
left_counter = i+1;
}
return result;
}
};
我对 * 运算符的问题是,每次我想将 4x3 矩阵与 3x3 矩阵相乘时,程序都会崩溃并出现以下错误消息:
invalid operands to binary expression ('Matrix<int, 4, 3>' and 'Matrix<int, 3, 3>')
E = D*A;
~^~
这是我调用函数的代码:
Matrix<int, 3, 3> A; // Datatype, Rows, Columns
A(0,0) = 1; A(0,1) = 2; A(0,2) = 3;
A(1,0) = 1; A(1,1) = 2; A(1,2) = 3;
A(2,0) = 1; A(2,1) = 2; A(2,2) = 3;
Matrix<int, 4, 3> D, E;
D(0,0) = 1; D(0,1) = 2; D(0,2) = 3;
D(1,0) = 4; D(1,1) = 5; D(1,2) = 6;
D(2,0) = 7; D(2,1) = 8; D(2,2) = 9;
D(3,0) = 7; D(3,1) = 8; D(3,2) = 9;
std::cout << std::endl << "D:\n" << D << std::endl;
E = D*A;
非常感谢任何帮助,谢谢。
您必须为 operator *
使用正确的矩阵大小:
template <typename T, std::size_t N, std::size_t K, std::size_t M>
Matrix<T, N, M> operator* (const Matrix<T, N, K>& lhs, const Matrix<T, K, M>& rhs);
您定义函数的方式,您的函数具有签名 Matrix<int, 4, 3> Matrix<int, 4, 3>::operator(const Matrix<int, 4, 3> &) const
,这意味着您必须尝试将 4 x 3 矩阵与 4 x 3 矩阵相乘,并且它会以某种方式给出你另一个 4 x 3 矩阵。相反,您至少想要的是将 operator*
声明为
template <int cols2>
Matrix<int, rows, cols2> operator*(const Matrix<int, cols, cols2> & other) const {...}
这将允许您将一个 4 x 3 矩阵与一个 3 x 3 矩阵相乘,这将正确地输出一个 4 x 3 矩阵。
为了让更多类型的矩阵相乘,您可以进行的另一个更改是使签名
template <class T2, int cols2>
Matrix<typename std::common_type<T, T2>::type, rows, cols2>
operator*(const Matrix<T2, cols, cols2> & other) const {...}