Another Question on implementing a matrix

I learnt here about a month ago that plain arrays are the best when making a matrix class(as opposed to linked lists ..etc...as I had a whim to do !!).

Which one of the following would be a better idea when I'd have to make the matrices do almost all sorts of operations and transformations..within my knowledge..?

Code:

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template<class X,int R,int C> //Type,No of Rows and Columns
class matrix
{
    public:
    typedef std::array<X,C> row_a;
    //Size of a row == no. of cols
    std::array<row_a,R> dat;
};

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OR

Code:

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template<class X,int R,int C> //Type,No of Rows and Coloumns
class matrix
{
    public:
    std::array<X,(R*C)> dat;
};

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If the former is better ....what is the way to access the individual X objects ..with iterators...?

Quote:

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If the former is better ....what is the way to access the individual X objects ..with iterators...?

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Why not just brackets?

Code:

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matrix[j][i]

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The reason for creating such types is not to inflict terrible syntax on everyone.

Would a 1d array respond same to matrix[i][j] compared to a array declared in a 2d fashion?

Quote:

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Originally Posted by manasij7479

Would a 1d array respond same to matrix[i][j] compared to a array declared in a 2d fashion?

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No, but the solution to the problem isn't difficult. The j,i'th element (row j, column i) is located at index "j*C+i" (C is your "number of columns" constant).

You could use an operator such as:

Code:

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X &operator()(int row, int col) { return dat[row*C + col]; }

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Then you address the matrix as mat(j, i)

..Thank You..
<finally> What is the difference between *your piece of code and

Code:

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X operator()(int row, int col) { return *( dat[row*C+col] ); }

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?

Quote:

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Originally Posted by manasij7479

..Thank You..
<finally> What is the difference between *your piece of code and

Code:

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X operator()(int row, int col) { return *( dat[row*C+col] ); }

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?

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The above code is incorrect, because you can't apply the dereferencing "*" operator to an array element (well, if it was a pointer you could, but the result would not be of type X). The reason I made the return type a reference is so that you can assign to the matrix by writing:

Code:

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mat(j, i) = val;

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