Asked  7 Months ago    Answers:  5   Viewed   38 times

If I have a struct in C++, is there no way to safely read/write it to a file that is cross-platform/compiler compatible?

Because if I understand correctly, every compiler 'pads' differently based on the target platform.

 Answers

94

No. That is not possible. It's because of lack of standardization of C++ at the binary level.

Don Box writes (quoting from his book Essential COM, chapter COM As A Better C++)

C++ and Portability


Once the decision is made to distribute a C++ class as a DLL, one is faced with one of the fundamental weaknesses of C++, that is, lack of standardization at the binary level. Although the ISO/ANSI C++ Draft Working Paper attempts to codify which programs will compile and what the semantic effects of running them will be, it makes no attempt to standardize the binary runtime model of C++. The first time this problem will become evident is when a client tries to link against the FastString DLL's import library from a C++ developement environment other than the one used to build the FastString DLL.

Struct padding is done differently by different compilers. Even if you use the same compiler, the packing alignment for structs can be different based on what pragma pack you're using.

Not only that if you write two structs whose members are exactly same, the only difference is that the order in which they're declared is different, then the size of each struct can be (and often is) different.

For example, see this,

struct A
{
   char c;
   char d;
   int i;
};

struct B
{
   char c;
   int i;
   char d;
};

int main() {
        cout << sizeof(A) << endl;
        cout << sizeof(B) << endl;
}

Compile it with gcc-4.3.4, and you get this output:

8
12

That is, sizes are different even though both structs have the same members!

The bottom line is that the standard doesn't talk about how padding should be done, and so the compilers are free to make any decision and you cannot assume all compilers make the same decision.

Tuesday, June 1, 2021
 
Manju
answered 7 Months ago
90

Jeremiah is right - MPI_Type_create_struct is the way to go here.

It's important to remember that MPI is a library, not built into the language; so it can't "see" what a structure looks like to serialize it by itself. So to send complex data types, you have to explicitly define its layout. In a language that does have native support for serialization, a set of MPI wrappers can concievably make use of that; mpi4py for instance makes use of python's pickle to transparently send complex data types; but in C, you have to roll up your sleeves and do it yourself.

For your structure, it looks like this:

#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>
#include <stddef.h>

typedef struct car_s {
        int shifts;
        int topSpeed;
} car;

int main(int argc, char **argv) {

    const int tag = 13;
    int size, rank;

    MPI_Init(&argc, &argv);
    MPI_Comm_size(MPI_COMM_WORLD, &size);

    if (size < 2) {
        fprintf(stderr,"Requires at least two processes.n");
        exit(-1);
    }

    /* create a type for struct car */
    const int nitems=2;
    int          blocklengths[2] = {1,1};
    MPI_Datatype types[2] = {MPI_INT, MPI_INT};
    MPI_Datatype mpi_car_type;
    MPI_Aint     offsets[2];

    offsets[0] = offsetof(car, shifts);
    offsets[1] = offsetof(car, topSpeed);

    MPI_Type_create_struct(nitems, blocklengths, offsets, types, &mpi_car_type);
    MPI_Type_commit(&mpi_car_type);

    MPI_Comm_rank(MPI_COMM_WORLD, &rank);
    if (rank == 0) {
        car send;
        send.shifts = 4;
        send.topSpeed = 100;

        const int dest = 1;
        MPI_Send(&send,   1, mpi_car_type, dest, tag, MPI_COMM_WORLD);

        printf("Rank %d: sent structure carn", rank);
    }
    if (rank == 1) {
        MPI_Status status;
        const int src=0;

        car recv;

        MPI_Recv(&recv,   1, mpi_car_type, src, tag, MPI_COMM_WORLD, &status);
        printf("Rank %d: Received: shifts = %d topSpeed = %dn", rank,
                 recv.shifts, recv.topSpeed);
    }

    MPI_Type_free(&mpi_car_type);
    MPI_Finalize();

    return 0;
}
Wednesday, June 2, 2021
 
samayo
answered 7 Months ago
42

Is it faster to create these objects as class or as struct?

You are the only person who can determine the answer to that question. Try it both ways, measure a meaningful, user-focused, relevant performance metric, and then you'll know whether the change has a meaningful effect on real users in relevant scenarios.

Structs consume less heap memory (because they are smaller and more easily compacted, not because they are "on the stack"). But they take longer to copy than a reference copy. I don't know what your performance metrics are for memory usage or speed; there's a tradeoff here and you're the person who knows what it is.

Is it better to create these objects as class or as struct?

Maybe class, maybe struct. As a rule of thumb: If the object is :
1. Small
2. Logically an immutable value
3. There's a lot of them
Then I'd consider making it a struct. Otherwise I'd stick with a reference type.

If you need to mutate some field of a struct it is usually better to build a constructor that returns an entire new struct with the field set correctly. That's perhaps slightly slower (measure it!) but logically much easier to reason about.

Are objects on the heap and the stack processed equally by the garbage collector?

No, they are not the same because objects on the stack are the roots of the collection. The garbage collector does not need to ever ask "is this thing on the stack alive?" because the answer to that question is always "Yes, it's on the stack". (Now, you can't rely on that to keep an object alive because the stack is an implementation detail. The jitter is allowed to introduce optimizations that, say, enregister what would normally be a stack value, and then it's never on the stack so the GC doesn't know that it is still alive. An enregistered object can have its descendents collected aggressively, as soon as the register holding onto it is not going to be read again.)

But the garbage collector does have to treat objects on the stack as alive, the same way that it treats any object known to be alive as alive. The object on the stack can refer to heap-allocated objects that need to be kept alive, so the GC has to treat stack objects like living heap-allocated objects for the purposes of determining the live set. But obviously they are not treated as "live objects" for the purposes of compacting the heap, because they're not on the heap in the first place.

Is that clear?

Saturday, June 5, 2021
 
exxed
answered 7 Months ago
66

There are times when the second form won't work. Say you want to create a linked list of Monsters. With the first form, you can add a pointer to the next Monster in the struct.

struct Monster {
    Object proto;
    int hit_points;
    struct Monster* next;
};

You can't do that in the second form since the struct doesn't have a name.

Friday, August 27, 2021
 
beny23
answered 4 Months ago
92

No. tagMyStruct is the name of the struct. In C, unlike C++, you must explicitly use the struct keyword every time you use the struct type. For example

tagMyStruct x; //error
struct tagMyStruct x; //OK

To avoid writing struct all the time, struct tagMyStruct is typedef'd to MYSTRUCT. Now you can write

MYSTRUCT x; //ok, same as struct tagMyStruct x;

What you thought this was (a variable definition) would be without the typedef keyword, like this

struct tagMyStruct {
    int numberOne;
    int numberTwo;
} MYSTRUCT;

BTW

MYSTRUCT pStruct = new MYSTRUCT; //error cannot convert MYSTRUCT* to MYSTRUCT

is not valid C or C++ anyway. Maybe you mean

MYSTRUCT* pStruct = new MYSTRUCT; //valid C++ (invalid C - use malloc instead of new in C)

hth

Thursday, November 4, 2021
 
Exoon
answered 1 Month ago
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