overload function call operator

[jr.freester]

I've written a Matrix container class and overloaded the function call
operator to return values at a specified index.
Below is the member function

double operator()(int a , int b)
{

col))

{
cerr << "Invalid index for Matrix" <<endl;
}

return this->data[(a-1)*this->col + (b-1)];
}

It is invoked
int row, col;
double d;
Matrix M;
d = M(row,col);

My question is thus, can I accomplish the reverse task of assigning a
double d to a Matrix M at power row,col ie.
M(row,col) = d;

Justin

 

[Kai-Uwe Bux]

I've written a Matrix container class and overloaded the function call
operator to return values at a specified index.
Below is the member function

double operator()(int a , int b)

Probably, it is not the best idea to allow signed arguments.

{

Probably, it would be better to go with the C convention to start indexing
at 0. This will get rid of many "-1" in the code, which are prone to error.

{
cerr << "Invalid index for Matrix" <<endl;

This should be an assert().

In an at()-method, it you would throw something.

}

return this->data[(a-1)*this->col + (b-1)];
}

It is invoked
int row, col;
double d;
Matrix M;
d = M(row,col);

My question is thus, can I accomplish the reverse task of assigning a
double d to a Matrix M at power row,col ie.
M(row,col) = d;

Yes, return a reference

double & operator()( size_type a, size_type b );


Best

Kai-Uwe Bux

 

[jr.freester]

I've written a Matrix container class and overloaded the function call
operator to return values at a specified index.
Below is the member function

double operator()(int a , int b)

Probably, it is not the best idea to allow signed arguments.

{

Probably, it would be better to go with the C convention to start indexing
at 0. This will get rid of many "-1" in the code, which are prone to error.

    {
        cerr << "Invalid index for Matrix" <<endl;

This should be an assert().

In an at()-method, it you would throw something.

    }

    return this->data[(a-1)*this->col + (b-1)];
}

It is invoked
int row, col;
double d;
Matrix M;
d = M(row,col);

My question is thus,  can I accomplish the reverse task of assigning a
double d to a Matrix M at power row,col ie.
M(row,col) = d;

Yes, return a reference

  double & operator()( size_type a, size_type b );

Best

Kai-Uwe Bux

Kai-Uwe, thank you for your quick response, but I am having trouble
understanding the last statement.

Yes, return a reference

double & operator()( size_type a, size_type b );

what would the body of this function look like? I don't understand
how adding the reference operator changes the order of assignment of
M(a,b) -> d TO d -> M(a,b). Any additional help would be
appreciated.

Justin

 

[Kai-Uwe Bux]

I've written a Matrix container class and overloaded the function call
operator to return values at a specified index.
Below is the member function

double operator()(int a , int b)

Probably, it is not the best idea to allow signed arguments.

{
if((a < 1) || (b < 1) || ((a + 1) > this->row) || ((b + 1) > this-
col))

Probably, it would be better to go with the C convention to start
indexing at 0. This will get rid of many "-1" in the code, which are
prone to error.

{
cerr << "Invalid index for Matrix" <<endl;

This should be an assert().

In an at()-method, it you would throw something.

}

return this->data[(a-1)*this->col + (b-1)];
}

It is invoked
int row, col;
double d;
Matrix M;
d = M(row,col);

My question is thus,  can I accomplish the reverse task of assigning a
double d to a Matrix M at power row,col ie.
M(row,col) = d;

Yes, return a reference

double & operator()( size_type a, size_type b );

Best

Kai-Uwe Bux

Kai-Uwe, thank you for your quick response, but I am having trouble
understanding the last statement.

Yes, return a reference

double & operator()( size_type a, size_type b );

what would the body of this function look like?

e.g.:

double & operator()( size_type a, size_type b ) {
assert( a < this->row );
assert( b < this->col );
return( this->data[ a * this->col + b ] );
}

I don't understand
how adding the reference operator changes the order of assignment of
M(a,b) -> d TO d -> M(a,b).

It doesn't. It affects how expressions in the return statement are
interpreted. In this case,

this->data[ a * this->col + b ]

is interpreted as a reference. Note that by and in itself,

this->data[ a * this->col + b ]

is an lvalue. In particular,

this->data[ a * this->col + b ] = some_thing;

would be well-formed. That magic can be wrapped up in a reference and be
returned.


Best

Kai-Uwe Bux