To be more specific as to the reason, this is the C idiom - it you pass it as a pointer to a pointer, you have to cast it as a void pointer to a pointer. If it is just a void *, C views this as an anonymous pointer, and you can use the address - as in int *intp; MPI_Alloc_mem(1000, MPI_INFO_NULL, &intp); If baseptr is a void**, you need to do this: MPI_Alloc_mem(1000, MPI_INFO_NULL, (void **)&intp); I don’t like it either, but this is the choice in C. Bill William Gropp Director and Chief Scientist, NCSA Thomas M. Siebel Chair in Computer Science University of Illinois Urbana-Champaign

On Oct 14, 2020, at 12:01 PM, Anthony Skjellum via mpi-forum <[email protected]> wrote:

Folks, I know we have had this function for a long time, and I've implemented ports of MPI that actually use it (e.g., with pre-pinned memory). But, I am trying to understand the logic for why baseptr is passed by value, instead of by reference. In C, everything is by value, so the last argument in normal C programs would be void **baseptr.

The standard has: int MPI_Alloc_mem(MPI_Aint size, MPI_Info info, void *baseptr); Now, MPICC/GCC takes this with void *memory = (void *)0; int error = MPI_Alloc_mem(1024, MPI_INFO_NULL, &memory); and you get the memory allocated properly. What is more, this is incorrect: int error = MPI_Alloc_mem(1024, MPI_INFO_NULL, memory); although it compiles fine because that is, indeed, the API. Why would we have the pointer going in by value, when it is coming out as an OUT argument? Isn't this plain wrong? void **baseptr means ---> I am passing you the address of a void pointer. Every viable implementation must do *(void **)baseptr = ... when providing the "malloc'd/special" memory. So... why did we fudge the C API. Is there some tie-in with the Fortran API? Thanks in advance, Tony Skjellum

-- Anthony Skjellum, PhD [email protected] <mailto:[email protected]> Cell: +1-205-807-4968

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