Question

I use gsl random generator to generate 2 big matrices, and use gsl cblas to multiply them, but I always got Segmentation fault when the cblas operation begins. When I can’t solve this, I then write the code below, using the very basic idea to do matrix multiplication, and I still got Segmentation Fault, but all the two can work all right when matrix is really a small one, I’m really puzzled about this.

#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <time.h>
#define PI 3.1415926

void GenerateKey(int m, int n, int l, int q, float alpha)
{
    // initialization
    int i;
    int j;
    int k;

    float *A;
    float *S;
    float *E;
    float *B;

    float sigma = (alpha * q ) / sqrt(2 * PI);
    A=(float*)malloc(sizeof(float)*(m*n));
    S=(float*)malloc(sizeof(float)*(n*l));
    B=(float*)malloc(sizeof(float)*(m*l));
    E=(float*)malloc(sizeof(float)*(m*l));

    // init A
    for(i = 0; i < m*n; i++)
    {
        A[i]=0;
    }
    printf("\n");

    // init S
    for(i = 0; i < n*l; i++)
    {
        S[i]=0;
    }

    printf("\n");

    // init E
    for(i = 0; i < m*l; i++)
    {
        E[i]=0;
    }

    printf("\n");
    float po;
    for(i = 0; i < m; i++)
    {
        for(j=0; j<l; j++)
        {      
            po=0;
            for(k=0; k<n; k++)
            {
                po +=A[i*m+k]*S[k*n+j];
            }
            po += E[i*m +j];
            B[i*m+j]=((int)po) % q;
        }
    }

    printf("Game over");

    printf("\n");
    free(A);
    free(B);
    free(S);
    free(E);
}

int main()
{
    GenerateKey(2680,191,64,72973,0.000551);

    return 0;
}

Answer 1

When you’re doing i*m+j, shouldn’t that be i*l+j ? Similarly with i*m+k should be i*l+k and k*n+j should be k*l+j

The reason being, take for example E = (float*)malloc(sizeof(float)*(m*l)), so you have m rows and l columns (or vise versa), so if you are iterating over the m dimension you need to be multiplying your m iterator (i in this case) by the stride of your matrix in that dimension, which is l.