CODEWITHSUNDEEP
cvoid-pointers
Sagi2313
Sagi2313

Reputation: 105

void pointer : what is wrong?

I have a 3-byte variable, but since there are no 3byte variables in C , I use a long 32bit instead. Then I want to send ONLY 3 bytes on the bus. I access the bus as ext Ram 8-bit width. To send only 3 bytes, i need to break the long value into 2-byte(a short) and 1-byte ( a char).

what I do is: 

typedef union
{
  unsigned char b[3];
  unsigned long lng;
} lng_array;

void SendLong(unsigned long d)
{   
  volatile void* c =(void*)Dat; // Dat is a #defined long number equal to the address on the bus that data must be sent
  lng_array tmp;
  tmp.lng=d;
  *c=*(unsigned short*)(&tmp.b[0]);  // Error
  *c=*(unsigned char*)(&tmp.b[2]);   // Error
}

for the 2 "*c=" lines I get an error :"Error 36 invalid use of void expression "

What am I doing wrong here?

Upvotes: 0

Views: 142

Answers (2)

hlovdal
hlovdal

Reputation: 28180

Since the variable c is of type pointer to void, you cannot dereference it (e.g. reading/writing *c). If you want to access the memory it points to you must add a type cast.

The C standard states that

6.3.2.2 void

The (nonexistent) value of a void expression (an expression that has type void) shall not be used in any way, ...

and since c is of type pointer to void dereferencing it produces a void expression which then cannot be used for anything. Whatever there is on the other side of the equal sign is irrelevant with regards to this.

Upvotes: 2

Ferenc Deak
Ferenc Deak

Reputation: 35408

*c makes no sense (cannot be derefenced) if the c is a void* since the compiler would need to know the size of it. Cast it again to a unsigned char*, like: *((unsigned char*)(c)) = (unsigned char*)(&tmp.b[2]);

Upvotes: 3

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