[SOLVED] Throw an error/warning when supplying wrong argument type to C function

Issue

I have this code:

#include <stdint.h>

void something(float a);

int main()
{
   uint8_t a = 28;
   
   something(a);

   return 0;
}

void something(float a)
{
   
   printf("%f\n", a);
}

I am using a similar function to log variables of different types to a file and I would like to get an error/warning message since I am calling function something with a wrong argument type (uint8_t instead of float).

How can I achieve this?

Solution

The old school trick is to change the function to use pointers, since pointers in C have much stricter typing rules than integers and floating point.

#include <stdio.h>
#include <stdint.h>

void something(const float* a);

int main()
{
   uint8_t a = 28;
   
   /* gcc -std=c11 -pedantic-errors */
   something(&a); // error: passing argument 1 of 'something' from incompatible pointer type
   something(a);  // error: passing argument 1 of 'something' makes pointer from integer without a cast

   return 0;
}

void something(const float* a)
{
   printf("%f\n", *a);
}

The modern C version:

#include <stdio.h>
#include <stdint.h>

void something_float (float a);

#define something(x) _Generic((x), float: something_float)(x)

int main()
{
   uint8_t a = 28;
   
   something(a); // error: '_Generic' selector of type 'unsigned char' is not compatible with any association

   return 0;
}

void something_float (float a)
{
   printf("%f\n", a);
}

Answered By - Lundin

Answer Checked By - Dawn Plyler (WPSolving Volunteer)