How to initialize a struct using the c style while using the g++ compiler?

Question

Generally, in order to initialize a struct in c, we could only specify part of the fields. Like below:

static struct fuse_operations hello_oper = {
    .getattr    = hello_getattr,
    .readdir    = hello_readdir,
    .open       = hello_open,
    .read       = hello_read,
};

However, in C++, we should initialize the variables in the struct without naming the fields. Now, what if I would like to initialize a struct using the c style while using the g++ compiler, how to accomplish this? PS: the reason I need to do this is that the struct fuse_operations has too many fields in it.

Don Doerner · Answer 1 · 2013-01-14T17:49:49.543

You wrote:

   static struct fuse_operations hello_oper = {
       .getattr    = hello_getattr,
       .readdir    = hello_readdir,
       .open       = hello_open,
       .read       = hello_read,
   };

Generally, in order to initialize a struct in c, we could only specify part of the fields [...] However, in C++, we should initialize the variables in the struct without naming the fields. Now, what if I would like to initialize a struct using the c style while using the g++ compiler, how to accomplish this? PS: the reason I need to do this is that the struct fuse_operations has too many fields in it.

My solution was to specialize the struct with a constructor:

struct hello_fuse_operations:fuse_operations
{
    hello_fuse_operations ()
    {
        getattr    = hello_getattr;
        readdir    = hello_readdir;
        open       = hello_open;
        read       = hello_read;
    }
}

Then declare a static instance of the new struct:

static struct hello_fuse_operations hello_oper;

Testing worked OK for me (but this depends on the memory layout of the C-struct and C++-struct to be the same -- not sure that's guaranteed)

* UPDATE *

Though this approach worked fine in practice, I have subsequently converted my code to use a utility class, i.e., a class with a single static 'initialize' method that takes a reference to a fuse_operation struct and initializes it. This avoids any possible uncertainty regarding memory layout, and would be my recommended approach in general.

score 3 · Accepted Answer · answered Aug 25 '12 at 13:11

3

Unfortunately, even the C++11 version of the C++ standard lacks the designated initializers feature of C99.

answered Aug 25 '12 at 13:11

Sergey Kalinichenko

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@MimiEAM You would need to put code in a C file, and compile it with a essentially a different compiler, even if it's from the same manufacturer. I would definitely avoid that. – Sergey Kalinichenko Aug 25 '12 at 13:16
so it's a bad practice in general ? because it is so widely used – MimiEAM Aug 25 '12 at 13:19
1

@MimiEAM Including C headers and linking with libraries that are built entirely in C is indeed used widely. Mixing C and C++ inside the same module, on the other hand, would be unusual. – Sergey Kalinichenko Aug 25 '12 at 13:33
1

@MimiEAM: `extern "C"` does not magically switch the compiler into C mode and does not somehow allow you to insert C code into C++ translation unit. `extern "C"` can only affect how C++ symbols are linked. – AnT stands with Russia Jan 14 '13 at 17:53

score 0 · Answer 3 · answered Sep 25 '12 at 04:48

maybe you can write a variable param function which takes as input the function pointers and assigns the rest of the attribs as NULL. Since you only have one struct - fuse_operations, you can implement the function for only one struct. something like init_struct(int no_op, ...) wherein you pass function pointers to implementations. Its too complex and laborious, but I suppose you can write it once and use it always...

How to initialize a struct using the c style while using the g++ compiler?

3 Answers3

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