如何将struct *作为函数的参数

Question

I want to pass alloc as an argument but don't know how, can somebody help me?

void parameter(unsigned int argnum, struct resistor* alloc)
{
/*...*/
}

struct resistort
{
  const double E6[6];
  double E12[12];
  const double E24[24];
  char e[3];
  double value;
  double log10val;
  double val;
  double serielval[2];
  double reset;
}rv;

int main(int argc, char *argv[])
{
  struct resistor *alloc = NULL;
  alloc = (struct resistor *)malloc(sizeof(struct resistor));

  parameter(argc, alloc);
}

in parameter I want to free(alloc)

I have hoped that it would function this way:

void parameter(unsigned int argnum, struct resistor* alloc);

but then I get this

warning: passing argument 2 of 'parameter' from incompatible pointer type [-Wincompatible-pointer-types]|
note: expected 'struct resistor *' but argument is of type 'struct resistor *'
error: conflicting types for 'parameter'

Answer 1

You are getting warning incompatible pointer type because you are using struct resistor before declaration:

void parameter(unsigned int argnum, struct resistor* alloc)
                                    ^^^^^^^^^^^^^^^

In your program, the declaration of struct resistor is after parameter() function.
You can solve this problem by either moving the struct resistor declaration before function parameter() or just do the forward declaration of struct resistor before parameter() function, like this:

struct resistor; //forward declaration 

void parameter(unsigned int argnum, struct resistor* alloc)
{
/*...*/
}

struct resistor
{
    const double E6[6];
    double E12[12];
    const double E24[24];
    char e[3];
    double value;
    double log10val;
    double val;
    double serielval[2];
    double reset;
}rv;

int main(int argc, char *argv[])
{
    struct resistor *alloc = NULL;
    alloc = (struct resistor *)malloc(sizeof(struct resistor));

    parameter(argc, alloc);
}