what is the return type of typeof?

Question

Hi,

I want to provide the type of an element as parameter to an initialization of an array of pointers to element of an unknown types

something like

void* init(type t)
  void* array = malloc(sizeof_type(t)*10));
  return array;
}

and later call for example

   init(typeof(int))

But I was not able to figure what is the return type of typeof.

I guess the sizeof_type think can be achieved using

   malloc((type) 0);

Thanks in advance

PS: this if for a vector implementatin if someone can point me to some flexiblecode i would be very thankful as well

Answer 1

I know your question wasn't tagged as such, but if you happen to be able to use C++, you may want to look into using template classes. Libraries like STL use templates to provide (type) universal containers and algorithms.

http://www.cplusplus.com/doc/tutorial/templates/

http://en.wikipedia.org/wiki/Template_%28programming%29

Answer 2

I don't usually use gnu (I think it is only in gnu C), but I don't think it's an expression in the normal sense that you can assign its value to a variable and later use it. I think it can only be used in very specific contexts as a type.

From the docs:

The syntax of using of this keyword looks like sizeof, but the construct acts semantically like a type name defined with typedef.

A typeof-construct can be used anywhere a typedef name could be used. For example, you can use it in a declaration, in a cast, or inside of sizeof or typeof.

Answer 3

What you want is not possible.

typeof() does not really return a type, since it is a keyword which is evaluated at compile-time. It is replaced by the name for the type of its parameter.

Maybe you can You can not use the stringification operator of the preprocessor to make it a string, like this ##typeof(a), but then it would still be impossible to malloc/cast something when you have the type in a string.

Answer 4

This:

#define INIT(t) (malloc(sizeof(t) * 10))

may be used like this:

int* a = INIT(int);

or possibly like this:

void* b = INIT(typeof(c));

or even this:

void* b = INIT(c);

since sizeof(c) == sizeof(typeof(c)).

Answer 5

Why can't you just use sizeof? something like:

int c;
void *p = calloc( 10,sizeof(c) );

Answer 6

You can use a macro for this purpose:

#define INIT(t) ((t*) init_1(sizeof(t)))

void *init_1(size_t sz) {
  void *array = malloc(sz * 10);
  return array;
}

Then you use like:

int *array = INIT(int);