Return an array in c with pointers

Question

I am trying to make a program that makes simple things. Actually, I know how to do it but in an easier way without pointers and stuff. However, I wondered how I could do it differently (like I did below). Obviously, there is something I miss about pointers, I did the math, but still I cannot get its philosophy. Thank you!

long *read_array(int n1, int n2)
{
    int i, j;
    long a[n1][n2];
    for(i=0;i<n1;i++)
        for(j=0;j<n2;j++)
            printf("Fill the table");
    a[i][j]=GetLong();
    return a;
}

long *Min_of_Rows(int m, int n, long *a)
{
    long B[];
    int i, j;
    for(i=0;i<m;i++)
        B[i]=a[i][0];
    for(j=0;j<n;j++)
        if (a[i][j]<B[i])
            B[i]=a[i][j];

    return B;
}

void *Print_B_array (int M, long *b)
{
    int i;
    for(i=0; i<M; i++)
        printf("%ld\n",b[i]);
}

main()
{
    long *a, *b;
    int n1, n2;

    printf("give rows");
    n1=GetInteger();

    printf("give columns");
    n2=GetInteger();

    a=read_array(n1, n2);
    b=Min_of_rows(n1, n2, a);
    Print_B_Array(n1, b);

}

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

long *read_array(int n1, int n2){
    int i, j;
    long a[n1][n2];

    printf("Fill the table\n");
    for(i=0;i<n1;i++){
        for(j=0;j<n2;j++){
            a[i][j]=GetLong();
        }
    }
    long *ret = malloc(sizeof(a));
    memcpy(ret, a, sizeof(a));
    return ret;
}

long *Min_of_Rows(int m, int n, long *a){
    long *B = malloc(m*sizeof(*B));
    int i, j;
    long (*A)[n] = (void*)a;

    for(i=0;i<m;i++){
        B[i]=A[i][0];
        for(j=0;j<n;j++){
            if(A[i][j]<B[i])
                B[i]=A[i][j];
        }
    }
    return B;
}

void *Print_B_Array (int M, long *b){
    int i;
    for(i=0; i<M; i++)
        printf("%ld\n", b[i]);
}

int main(void){
    long *a, *b;
    int n1, n2;

    printf("give rows:");
    n1=GetInteger();

    printf("give columns:");
    n2=GetInteger();

    a=read_array(n1, n2);
    b=Min_of_Rows(n1, n2, a);
    printf("min of rows\n");
    Print_B_Array(n1, b);
    free(a);
    free(b);
    return 0;
}