Initialize struct with single field

Question

I have a struct:

 struct MY_TYPE {
      boolean flag;
      short int value;
      double stuff;
    };

I know I can intialize it by:

MY_TYPE a = { .flag = true, .value = 123, .stuff = 0.456 };

But, now I need to create a pointer variable My_TYPE* and I only want to initialize one field there? I tried e.g.:

MY_TYPE *a = {.value = 123};

But I get compiler error "Designator in intializer for scalar type 'struct MY_TYPE *'".

Is it possible to initialize the struct with one field?

Answer 1

First of all, you are mixing up struct MY_TYPE and typedef. The code posted won't work for that reason. You'll have to do like this:

typedef struct 
{
  bool flag;
  short int value;
  double stuff;
} MY_TYPE;

You can then use a pointer to a compound literal, to achieve what you are looking for:

MY_TYPE* ptr = &(MY_TYPE){ .flag = true, .value = 123, .stuff = 0.456 };

But please note that the compound literal will have local scope. If you wish to use these data past the end of the local scope, then you have to use a pointer to a statically or dynamically allocated variable.