C->C++ Automatically cast void pointer into Type pointer in C++ in #define in case of type is not given (C-style) [MSVS]

9
votes

Hi!

I've used the following C macro, But in C++ it can't automatically cast void* to type*.

#define MALLOC_SAFE(var, size) { \
    var = malloc(size); \
    if (!var) goto error; \
}

I know, I can do something like this:

#define MALLOC_SAFE_CPP(var, type, size) { \
    var = (type)malloc(size); \
    if (!var) goto error; \
}

But I don't want to rewrite a big portion of code, where MALLOC_SAFE was used.

Is there any way to do this without giving the type to the macro? Maybe some MSVC 2005 #pragma/__declspec/other ?

p.s.: I can't use C compiler, because my code is part (one of hundreds modules) of the large project. And now it's on C++. I know, I can build my code separately. But it's old code and I just want to port it fast.

The question is about void* casting ;) If it's not possible, I'll just replace MACRO_SAFE with MACRO_SAFE_CPP

Thank You!

4
c++cvisual-c++castingvoid-pointers
How is your memory freed? using free directly, or a macro? – Steve Townsend
Why can't you compile the code as a C program? Visual C++ will compile C. – James McNellis
@zxcat: Which version of MSVC? – kennytm
The C++ compilers I'm familiar with compile C as well, if the file is specified as C. In the meantime, if you're replacing the macro, you might consider changing to new/delete instead of malloc/free. – David Thornley
I know you said you don't want to do this, that's why I'm not providing it as an answer, just a consideration. Would it really be that bad to rewrite the code? I mean, it's terrible even as C code. It certainly doesn't belong in a C++ program. If it were me, even if it took a few full working days to rewrite it, I would do it. In my time off if I had to. – Benjamin Lindley

4 Answers

36
votes

To makes James' answer even dirtier, if you don't have decltype support you can also do this:

template <typename T>
class auto_cast_wrapper
{
public:
    template <typename R>
    friend auto_cast_wrapper<R> auto_cast(const R& x);

    template <typename U>
    operator U()
    {
        return static_cast<U>(mX);
    }

private:
    auto_cast_wrapper(const T& x) :
    mX(x)
    {}

    auto_cast_wrapper(const auto_cast_wrapper& other) :
    mX(other.mX)
    {}

    // non-assignable
    auto_cast_wrapper& operator=(const auto_cast_wrapper&);

    const T& mX;
};

template <typename R>
auto_cast_wrapper<R> auto_cast(const R& x)
{
    return auto_cast_wrapper<R>(x);
}

Then:

#define MALLOC_SAFE(var, size)                      \
{                                                   \
    var = auto_cast(malloc(size));                  \
    if (!var) goto error;                           \
}

I expanded on this utility (in C++11) on my blog. Don't use it for anything but evil.

15
votes

I do not recommend doing this; this is terrible code and if you are using C you should compile it with a C compiler (or, in Visual C++, as a C file)

If you are using Visual C++, you can use decltype:

#define MALLOC_SAFE(var, size)                      \
{                                                   \
    var = static_cast<decltype(var)>(malloc(size)); \
    if (!var) goto error;                           \
}
5
votes

For example, like this:

template <class T>
void malloc_safe_impl(T** p, size_t size)
{
    *p = static_cast<T*>(malloc(size));
}

#define MALLOC_SAFE(var, size) { \
    malloc_safe_impl(&var, size); \
    if (!var) goto error; \
}
3
votes

Is there a reason nobody just casts var, your argument to SAFE_MALOC()? I mean, malloc() returns a pointer. You're storing it somewhere that accepts a pointer... There are all sorts of neat type-safe things that other folks have already pointed out... I'm just wondering why this didn't work:

#define MALLOC_SAFE(var,size)  {  \
    (* (void **) & (var)) = malloc(size); \
    if ( ! (var) ) goto error;    \
    }

Yeah... I know. It's sick, and throws type-safety right out the window. But a straight ((void *)(var))= cast wouldn't always work.