MPI in C and pointer transmissions

Question

I am having a new little problem;

I have a little pointer called:

int *a;

Now ..somewhere inside my main method I allocate some space for it using the following lines and assign a value:

a = (int *) malloc(sizeof(int));
*a=5;

..and then I attempt to transmit it (say to process 1):

MPI_Bsend(a, 1, MPI_INT, 1, 0, MPI_COMM_WORLD);

On the other end, if I try to receive that pointer

int *b;
MPI_Recv(b, 1, MPI_INT, 0, MPI_ANY_TAG, MPI_COMM_WORLD, &status);

printf("This is what I received: %d \n", *b);

I get an error about the buffer!

However if instead of declaring 'b' a pointer I do the following:

int b;
MPI_Recv(&b, 1, MPI_INT, 0, MPI_ANY_TAG, MPI_COMM_WORLD, &status);

printf("This is what I received: %d \n", b);

...all seems to be good! Could someone help me figure out what's happening and how to only use pointer?

Thanks in advance!

Answer 1

The meaning of the line

MPI_Bsend(a, 1, MPI_INT, 1, 0, MPI_COMM_WORLD);

is the following: "a is a point in memory where I have 1 integer. Send it.`

In the code you posted above, this is absolutely true: a does point to an integer, and so it is sent. This is why you can receive it using your second method, since the meaning of the line

MPI_Recv(&b, 1, MPI_INT, 0, MPI_ANY_TAG, MPI_COMM_WORLD, &status);

is "receive 1 integer, and store it at &b. b is a regular int, so it's all good. In the first receive, you're trying to receive an integer into an int* variable there is no allocated memory that b is pointing to, so Recv has nowhere to write to. However, I should point out:

NEVER pass a pointer's contents to another process in MPI

MPI processes cannot read each others' memory, and virtual addressing makes one process' pointer completely meaningless to another.