Segmentation fault on 2D arrays of certain size

Question

I'm creating a program with multiple threads where one of threads creates 2 matrices based on input. These inputs are n and seed. Input specifies the square size of the matrix (n x n size) while seed fills the matrices with the values starting from seed incremementally. The problem I'm having is that this function runs fine for inputs n = 1,2,3,4 but 5 and above causes a segmentation fault.

void *creatematrices(void *arg)
{

   pthread_mutex_lock(&mutex1); 
   pthread_mutex_lock(&mutex2); 
   pthread_mutex_lock(&mutex3);
   pthread_mutex_lock(&mutex4);  

   int seed;

   printf("To create two nxn matrices, A and B,first please enter the value for n\n");
   scanf("%d", &n);
   printf("Now please enter the seed integer for the first element of matrix A\n");
   scanf("%d", &seed);

   A = (int**)malloc(n * sizeof(int));
   for(int i = 0; i < n; i++)
   {
       A[i] = (int*)malloc(n * sizeof(int*));
   }

   B = (int**)malloc(n * sizeof(int));
   for(int i = 0; i < n; i++)
   {
       B[i] = (int*)malloc(n * sizeof(int*));
   }

   C = (int**)malloc(n * sizeof(int));
   for(int i = 0; i < n; i++)
   {
       C[i] = (int*)malloc(n * sizeof(int*));
   }

   for(int i = 0; i < n; i++)
   {
       for(int j = 0; j < n; j++)
       {
           A[i][j] = seed++;
       }
   }
   for(int i = 0; i < n; i++)
   {
       for(int j = 0; j < n; j++)
       {   
           B[i][j];
       }
   }

   pthread_mutex_unlock(&mutex1);
   pthread_mutex_unlock(&mutex2);

   pthread_exit(0);
}

I'm using dynamic allocation for the 2d Arrays. I've realized this is probably a stack overflow error so I've already entered ulimit -s unlimited to no avail.

Can you show us how you call this function? It's not clear why the declaration takes void* arg, among other things. — nanofarad
– nanofarad, Commented Oct 24, 2019 at 1:00
it's a separate thread so pthread_create(&th1, &attr, creatematrices, NULL); is used to call it. If this were implemented not as a thread it would probably take the pointers A, B, and C as arguments. — Brandon Phan
– Brandon Phan, Commented Oct 24, 2019 at 1:02
You've got the malloc backwards. The first malloc should be allocating pointers using sizeof(int*), and the second malloc should be allocating integers using sizeof(int). — user3386109
– user3386109, Commented Oct 24, 2019 at 1:05
Thank you you are 100% correct. Very silly of me. Do you have any idea why this would result in correct answers for smaller values though? Like I said this was working for n = 4. — Brandon Phan
– Brandon Phan, Commented Oct 24, 2019 at 1:13

Serge · Accepted Answer · 2019-10-24 02:20:07Z

2

Presumably A is of int **A

then you made a mistake in the mallocs:

A = (int**)malloc(n * sizeof(int)); // << wrong. It has to be sizeof (int*)
for(int i = 0; i < n; i++)
{
   A[i] = (int*)malloc(n * sizeof(int*)); // << wrong. It has to be sizeof (int)
}

In the above A[i] points to a memory pool which contains n pointers to int, so it must be dealt accodingly.

Same story with all other allocations.

answered Oct 24, 2019 at 2:20

Serge

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