C macro adding to a typename

Question

I am trying to create a C macro, that given a typename will append _info to it, take the address of that and a call a function with it. Example code (doesn't work):

#define new(X) __new(&(X_info))

struct classInfo
{
    size_t size;
    void* (*constructor)(void* p);
};

void* __new(struct classInfo* info, ...)
{
    return info->constructor(malloc(info->size));
}

void* queue_constructor(void* p)
{
    return p;
}

typedef struct
{
    uint64_t data;
} queue_t;

const struct classInfo queue_t_info = { .size        = sizeof(queue_t),
                                        .constructor = &queue_constructor};

int main(int argc, char** argv)
{
    queue_t* p = new(queue_t);
    return 0;
}

The preprocessor doesn't seem to want to expand X, as it's error'ing about an undefined symbol X_info. Not sure what I should change on the macro to fix this.

You shouldn't use leading underscore for your global identifiers, those are reserved for extensions by the compiler and standard library. I also would refrain from using C++ keywords (like new), as people coming from C++ background might find it confusing, and also some compilers really don't have a proper pure C mode (like MS Visual C++). — Some programmer dude

Aasmund Eldhuset Aasmund Eldhuset · Accepted Answer · 2013-02-26T09:43:21

You need to use the token concatenation feature of the preprocessor; otherwise it thinks that X_info is a token by itself:

#define new(X) __new(&(X ## _info))

The reason the preprocessor won't expand the X in X_info is that this would have created big problems if there actually was an identifier called X_info that you needed to refer to in the macro.

As a side note, there is also the stringification feature: if X is e.g. Person, "X" will remain "X", while #X will expand to the string constant "Person".

C macro adding to a typename

1 Answers