Multiplication of c++ matrices [closed]
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Improve the questionI noticed that my function skips the last row if the matrix is not square when multiplying. I don't see the error yet, so please help me.
void multiplication(int (&a1)[4][4], int (&b)[4][4], int (&c)[4][4], int n, int l){
for (int i=0; i<n; i++){
for (int j=0; j<l; j++){
for(int k=0; k<n; k++){
int save = a1[i][k]*b[k][j];
c[i][j] += save;
save = 0;
}
}
}
}
Condition for matrix multiplication (the number of rows of one=the number of columns of the second) is checked in another function.
1 answers
You don't need a new loop, you need the function to know about all the sizes of the matrices, now you pass n and l this is only enough to multiply the square matrices by the type 4 на 4.
When multiplying the matrix 3 на 4 with 4 на 2, note that you have 3 numbers that are 3, 4, 2, and your function can only know about two of them at the moment.
Here is a proper example of your function:
//row1 - кол-во строк в матрице a1, col1 - кол-во столбцов в 1
матрице, col2 - кол-во столбвоц во 2 матрице
void multiplication(int(&a1)[4][4], int(&b)[4][4], int(&c)[4][4], int row1, int col1, int col2) {
for (int i = 0; i < row1; i++) {
for (int j = 0; j < col2; j++) {
c[i][j] = 0;
for (int k = 0; k < col1; k++)
c[i][j] += a1[i][k] * b[k][j];
}
}
}
For more understanding, I renamed the variables and signed in the comments that they are denote.
Now to multiply the matrix 3 на 4 and 4 на 2, you need to call the function like this:
multiplication(a, b, c, 3, 4, 2);