Convert new to malloc

[artmansoft]

Hello all,

I have a struct called GFSPPair. I made a reference to a GFSPPair instance called "pair" by:

GFSPPair *p_pair = new GFSPPair( pair );

It works fine for c++, but I want to convert it to ansi c by replacing new with malloc.

What should I write instead?

Thanks,
artmansoft

[imi]

malloc takes the number of bytes and returns a void* to the new memory. You can cast the void* to your struct using plain C-casts. You can get the size necessary using "sizeof( any struct )"

Ciao, Imi.

[moorecm]

Is it possible to leave it in C++ and give it external linkage with extern "C" { /*...*/ }?

I don't like the idea of stepping backwards. For one, I don't think malloc calls constructors.

[artmansoft]

For some reasons, I cannot use extern "C". Also, I know that I need sizeof( any struct ) and a cast to GFSPPair. But the problems is the argument of GFSPPair, pair: new GFSPPair( pair )!

Does anybody have any idea how to exactly convert it to malloc calls?

[chrisname]

You call malloc like this: struct GFSPPair* pair = malloc(sizeof(struct GFSPPair));
You shouldn't have to cast it -- malloc returns a void pointer for a reason.
Edit: the prefix "struct" is important -- you can't just do this:

struct foo {
        int a;
        int b;
};

int main()
{
        foo bar;

struct foo {
        int a;
        int b;
};

int main()
{
        struct foo bar;

[imi]

But the problems is the argument of GFSPPair, pair: new GFSPPair( pair )!

together with:

I have a struct called GFSPPair. I made a reference to a GFSPPair instance called "pair" by:

... hm.. you don't mean by any chance that you just want a pointer pointing to the same object? Then it's in C++ and C the same: GFSPPair * p_pair = &pair;

If you want a real and full copy (not just a reference to the existing object), you have to use memcpy after you got your memory to copy the content of pair over to the memory where p_pair points to. Of course, your GFSPPair struct must noch contain any functions or constructors (C does not support functions in structs).

You shouldn't have to cast it -- malloc returns a void pointer for a reason.

ANSI C does implicit casting from void* to any pointer? That's odd, are you sure? I could swear it's only the other way around.

Edit: the prefix "struct" is important -- you can't just do this:

Or use the C-typical typedef idiom typedef struct { ... } foo;.

[artmansoft]

I found a solution by:

GFSPPair *p_pair = (GFSPPair*)malloc(sizeof(GFSPPair))
*p_pair = pair

In this way, we can make a new reference (as opposed to GFSPPair * p_pair = &pair) which refers to pair.

Thanks all,
artmansoft

[chrisname]

@imi,
Excerpt from linux/Documentation/CodingStyle:

Chapter 14: Allocating memory

The kernel provides the following general purpose memory allocators:
kmalloc(), kzalloc(), kcalloc(), and vmalloc(). Please refer to the API
documentation for further information about them.

The preferred form for passing a size of a struct is the following:

p = kmalloc(sizeof(*p), ...);

The alternative form where struct name is spelled out hurts readability and
introduces an opportunity for a bug when the pointer variable type is changed
but the corresponding sizeof that is passed to a memory allocator is not.

Casting the return value which is a void pointer is redundant. The conversion
from void pointer to any other pointer type is guaranteed by the C programming
language.

I know that's kmalloc() and not malloc() but still, I'm fairly sure the Linux programmers know what they're talking about. You don't have to cast a void pointer to another type of pointer because the compiler does that for you. I mean, I guess if you want to it's ok -- but it's redundant code. I don't see that a compiler would optimize that out for you, either.

@artmansoft,
Huh?

[Disch]

You don't have to cast a void pointer to another type of pointer because the compiler does that for you.

C++ requires you explicitly cast.

C does not.

[chrisname]

He's trying to make his code ANSI C compliant.

[Disch]

Ah you're right he did say that didn't he.

So yeah. You don't need to cast as long as you're compiling it as C and not as C++.