这是我的代码的一部分,当我编译它时,它说 1:不匹配operator = 2:参数1从'Matrix'到'Matrix&'没有已知的转换 但如果我删除运算符+部分它的工作原理 问题出在哪儿?! :|
gcc错误: “在'z = Matrix :: operator +(Matrix&)('*& y))'''''''''''''''''''''''''''''' 候选人是: ATRIX和放大器;矩阵算子:: =(矩阵&安培) 参数1从“Matrix”到“Matrix&”没有已知的转换“
class Matrix {
//friend list:
friend istream& operator>>(istream& in, Matrix& m);
friend ostream& operator<<(ostream& in, Matrix& m);
int** a; //2D array pointer
int R, C; //num of rows and columns
static int s1, s2, s3, s4, s5;
public:
Matrix();
Matrix(const Matrix&);
~Matrix();
static void log();
Matrix operator+ (Matrix &M){
if( R == M.R && C == M.C ){
s4++;
Matrix temp;
temp.R = R;
temp.C = C; temp.a = new int*[R];
for(int i=0; i<R; i++)
temp.a[i] = new int[C];
for(int i=0; i<R; i++)
for(int j=0; j<C; j++)
temp.a[i][j] = a[i][j] + M.a[i][j];
return temp;
}
}
Matrix& operator = (Matrix& M){
s5++;
if(a != NULL)
{
for(int i=0; i<R; i++)
delete [] a[i];
delete a;
a = NULL;
R = 0;
C = 0;
}
R = M.R;
C = M.C;
a = new int*[R];
for(int i=0; i<R; i++)
a[i] = new int[C];
for(int i=0; i<R; i++)
for(int j=0; j<C; j++)
a[i][j] = M.a[i][j];
return *this;
}
};
答案 0 :(得分:2)
Matrix operator+ (Matrix &M){
Matrix& operator= (Matrix &M){
他们都有同样的问题 - 参数类型应该是const Matrix&
(就像在复制构造函数中一样)。否则,您无法将临时对象传递给运算符。