CODEWITHSUNDEEP
cfunctionpointers
Alex Stiles
Alex Stiles

Reputation: 161

How to pass pointer of unknown type to function?

I want to pass a variable by value (I used pointers only to avoid giving the parameter a type) into a function so that it can be used by another function contained within. The inner function can be given a variable of any type.

I have already tried using a generic pointer (void*) but it tells me that "'void*' is not a pointer-to-object type". I don't want to cast the pointer because the whole point of using a pointer is that it is generic.

void print_string(char *string, void *variable) {
  Serial.print(string);
  Serial.print(": ");
  Serial.print(*variable); # This takes a variable of any type (int, float, double, char *);
  Serial.println();
}

I want to be able to provide an int, double, float or string (char*).

Upvotes: 0

Views: 3625

Answers (2)

yellowtape
yellowtape

Reputation: 3

You have to know the type, you could do something like:

  enum DataType{
    TYPE_FLOAT,
    TYPE_U32,
    TYPE_I32,
  }
  
  void print_string(char *string, void *variable, DataType type) {
  Serial.print(string);
  Serial.print(": ");
  
  switch(type) {
     case TYPE_I32:
     Serial.print(*(int32_t *)variable);
     break;

     case TYPE_U32:
     Serial.print(*(uint32_t *)variable);
     break;
  }
  Serial.println();
}

uint32_t data = 1234;
print_string("hello", &data, TYPE_U32);

Upvotes: 0

bigwillydos
bigwillydos

Reputation: 1371

The answer is that you have to cast it despite not wanting to. You are trying to dereference a pointer which points to some memory location, but you aren't giving the compiler any semblance of what the size of that information is when it is void * so it has no way of translating that to the proper assembly instruction for accessing that memory.

Upvotes: 3

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