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declare variable of struct with flexible array member at compile time

Alexander Polleti

Hello everybody,

is it possible to use a struct with flexible array members and declare a variable of it at compile time?

e.g.:

struct sFoo
{
  uint32_t const num_Elements;
  uint8_t  baz[];
}

#define DECLAREFOO (name, nElements) \ 
  struct sFoo name = { (nElements), ..?}

what I am using is the following:

struct sFoo
{
  uint32_t const num_Elements;
  uint8_t * const baz;
}

#define DECLAREFOO (name, nElements)                            \ 
  uint8_t name##_container[ (sizePerElement) * (numElements)];  \ 
  struct sFoo name = { (nElements), & name##_container[0]}

Can anyone think of a smarter way to do this without the container or malloc?

Thanks for your replies
Alexander

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