Passing pointer argument to function error?

[dissectcode]

Hello - Isn't this right? :

void Function(int *d, const int c, int n);
 
void Function(int *d, const int c, int n)
{
   while(n) { *(d++) = c; }
}
 
Function(&var1, 0, sizeof(thing));
 
 

Why do I get a warning : "passing arg1 of "Function" from incompatible pointer type" ??

thanks!

[JosAH]

What is the type of 'var1'?

kind regards,

Jos

[dissectcode]

What is the type of 'var1'?

kind regards,

Jos

oh - var1 is a structure....

typedef struct st_var
{
   Athing arr[max];
   int limit1;
   int limit 2;
} Var;
 
Var var1;
 

please advise..
thank you.

[boxfish]

Hi,
I don't get it. How can you pass a value of type st_var* into an argument of type int*? Is that really what you want to do? I would complain too, if I were the compiler. If this function has to work with any type, you could use a void*, but I don't think you can increment those. Why is the parameter of type int*?

[Banfa]

boxfish is right you can't increment points of type void *. To be able to increment a pointer to a type you have to know the types size and by definition the size of void is 0.

This looks to me like a self implementation of the library function memset. Assuming you can't use memset then you need to use void * as memset does and boxfish suggests. You will then be able to pass a pointer to anything to your function.

Then internally to the function copy the void * to a char * variable and use that in you loop.

[arnaudk]

OP: Even when you get your types right, that Function will either run forever if n is non-zero or do strictly nothing otherwise, the only exception being if c=0 and the address of n is eventually reached by incrementing d. I'm not sure this is what you intended.