我在C++中创建了一个带有重载操作符的矩阵类,重载了操作符,<< (输出,带有ostream库),操作符+,它添加了矩阵,operator =,用于将一个矩阵分配给另一个矩阵。问题是当我用
cout<<m1+m2<<endl;我犯了个错误
E0349: No operator << matches these operands type are: std::ostream << Matrix但如果我这样做的话:
Matrix m = m1 + m2;
cout<<m<<endl;效果很好。下面是<<操作符:
ostream& operator<< (ostream &os,const Matrix& m)
{
if (m.isValid())
{
os << '|';
for (int i = 0; i < m.getRows(); i++)
{
for (int j = 0; j < m.getCols(); j++)
{
os << m.getMatrix()[i][j];
if (j < m.getCols() - 1)
{
os << ',';
}
}
os << '|';
}
}
else
{
os << "invalid matrix!";
}
return os;
}+操作符:
Matrix Matrix::operator+ (Matrix &m)
{
Matrix* answer = new Matrix(m); //allocating space
if (valid && m.valid)//if they are both valid
{
if (colNum == m.colNum&&rowNum == m.rowNum) //if are from same size
{
answer->valid = true; //valid set to be true
for (int i = 0; i < rowNum; i++) //going over the matrix
{
for (int j = 0; j < colNum; j++)
{
answer->matrix[i][j] += matrix[i][j]; //adding data
}
}
}
else
{
//clearing data
delete answer;
answer = new Matrix();
}
}
else
{
//clearing data
delete answer;
answer = new Matrix();
}
return *answer;
}和=操作符:
Matrix Matrix::operator= (Matrix &m)
{
int rows = m.rowNum; //putting cols and rows from the data
int cols = m.colNum;
if (m.valid)
{
matrix = new int*[rows]; //defining the matrix - first allocatin space for rows
for (int i = 0; i < rows; i++) //now allocating space for cols
{
matrix[i] = new int[cols];
}
for (int i = 0; i < rows; i++) //now going over the matrix and putting the data in
{
for (int j = 0; j < cols; j++)
{
matrix[i][j] = m.matrix[i][j];
}
}
}
//putting the rows and cols data
rowNum = m.rowNum;
colNum = m.colNum;
valid = m.valid; //setting to the right valid type
return *this;
}以及类变量:
class Matrix
{
private:
bool valid;
int** matrix;
int rowNum;
int colNum;有一个复制构造函数、一个字符串构造函数和一个构造函数,根据算法获取字符串输入并将其转换为矩阵。
发布于 2018-05-24 11:55:00
实际上,您面临的是Can't pass temporary object as reference (链接指向StackOverflow上的另一个问题)。
就你的情况而言,解决办法应该是取代:
ostream& operator<< (ostream &os,Matrix& m)通过以下方式:
ostream& operator<< (ostream &os,const Matrix& m)作为重载的运算符签名。当前需要一个lvalue引用,如果是
Matrix m = m1 + m2;
cout << m << endl;这就是使用m (一个命名变量)调用函数的方法,它将超过命令的执行。在失败的情况下,您使用m1+m2调用函数--这是一个加号操作的结果,它是一个临时对象。只是和你的过载不匹配。
顺便说一句--这个改变也是有意义的,因为你没有在打印矩阵时修改它--你把它当作const。
https://stackoverflow.com/questions/50508675
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