Saving matrices with data structures-"get" and "set" confusion

Hi everyone,
I´m taking a programming class and currently we´re doing data structures. Today, we discussed how to save square matrices column-wise with data structures. The teacher said that after the first few steps where you declare the structure, allocate the matrix, free it and get the dimension (which mostly make sense to me I think), you have to "get" and "set" the matrix by putting in something like

Code:

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void setMatrixEntry(Matrix* A, int i, int j, double Aij)
{

A->entries[i+j*A->m] = Aij;

}


double getMatrixEntry(Matrix* A, int i, int j)
{

return A->entries[i+j*A->m]
}

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I´m really confused here because I can´t figure out what "A->entry[i+j*A->m] = Aij;" and "return A->entry[i+j*A->m]" do exactly or how you get to them.

I´d be really grateful for any help!

You should show the definition of Matrix to make it easier to explain.

The data structure is

Code:

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typedef struct _Matrix_ {

int m;

int n;

double* entry;

} Matrix;



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and the allocation is:

Code:

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Matrix* newMatrix(int m, int n) {

int i = 0;

Matrix* A = malloc(sizeof(Matrix));

A->m = m;

A->n = n;

A->entry = malloc(m*n*sizeof(double));

for (i=0; i<m*n; ++i) {

A->entry[i] = 0;

}

return A;

}



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Is that what you meant?

Yes. Let's consider getMatrixEntry:

Code:

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return A->entries[i + j * A->m];

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The A-> part is simple: A is a pointer, so we use -> to dereference A while accessing the entries member of the Matrix object.

Now, consider the entries member: this is a pointer to double. Obviously, we cannot use normal 2D array notation. Therefore, we need to compute the index of the element. This is what the i + j * A->m expression is about. In C, we normally use row-major ordering, so matrix[i][j] would correspond to A->entries[i * A->m + j]. The idea here is that we skip i * A->m number of elements in the dynamic array (i.e., skip i number of rows), then skip j number of elements (i.e., j number of columns within a row) to get to the element corresponding to matrix[i][j].

In this case though, you're using column-major ordering ("save square matrices column-wise"), hence we use the opposite: skip j * A->m number of elements (i.e., j number of columns) then skip i number of elements (i.e., i number of rows within a column), thus A->entries[i + j * A->m].

Thank you, I think I understand how to get to it now!

I´ve still got one question, though...
What exactly is
return A->entries[i + j * A->m]; supposed to do? The idea is that a user puts in entries and they get saved in the matrix (I think). Why do we need A->entries[i + j * A->m]; ?

Quote:

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Originally Posted by Suuper Susi

What exactly is
return A->entries[i + j * A->m]; supposed to do? The idea is that a user puts in entries and they get saved in the matrix (I think). Why do we need A->entries[i + j * A->m]; ?

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Return the value of the entry at position i, j in the matrix.

Ok, thank you!

And just to make sure I´ve understood this...does setMatrixEntry give every entry the value/name Aij?

Quote:

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Originally Posted by Suuper Susi

Ok, thank you!

And just to make sure I´ve understood this...does setMatrixEntry give every entry the value/name Aij?

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Every entry? Can you explain why you think that? It sounds like you are just guessing and you don't understanding laserlight's explanations at all. I'm being genuine, you have a misunderstanding that needs to be sorted out.