C default arguments

Another trick using macros:

#include <stdio.h>

#define func(...) FUNC(__VA_ARGS__, 15, 0)
#define FUNC(a, b, ...) func(a, b)

int (func)(int a, int b)
{
    return a + b;
}

int main(void)
{
    printf("%d\n", func(1));
    printf("%d\n", func(1, 2));
    return 0;
}

If only one argument is passed, b receives the default value (in this case 15)

YES

Through macros

3 Parameters:

#define my_func2(...) my_func3(__VA_ARGS__, 0.5)
#define my_func1(...) my_func2(__VA_ARGS__, 10)
#define VAR_FUNC(_1, _2, _3, NAME, ...) NAME
#define my_func(...) VAR_FUNC(__VA_ARGS__, my_func3, my_func2, my_func1)(__VA_ARGS__)

void my_func3(char a, int b, float c) // b=10, c=0.5
{
    printf("a=%c; b=%d; c=%f\n", a, b, c);
}

If you want 4th argument, then an extra my_func3 needs to be added. Notice the changes in VAR_FUNC, my_func2 and my_func

4 Parameters:

#define my_func3(...) my_func4(__VA_ARGS__, "default") // <== New function added
#define my_func2(...) my_func3(__VA_ARGS__, (float)1/2)
#define my_func1(...) my_func2(__VA_ARGS__, 10)
#define VAR_FUNC(_1, _2, _3, _4, NAME, ...) NAME
#define my_func(...) VAR_FUNC(__VA_ARGS__, my_func4, my_func3, my_func2, my_func1)(__VA_ARGS__)

void my_func4(char a, int b, float c, const char* d) // b=10, c=0.5, d="default"
{
    printf("a=%c; b=%d; c=%f; d=%s\n", a, b, c, d);
}

Only exception that float variables cannot be given default values (unless if it is the last argument as in the 3 parameters case), because they need period ('.'), which is not accepted within macro arguments. But can figure out a work around as seen in my_func2 macro (of 4 parameters case)

Program

int main(void)
{
    my_func('a');
    my_func('b', 20);
    my_func('c', 200, 10.5);
    my_func('d', 2000, 100.5, "hello");

    return 0;
}

Output:

a=a; b=10; c=0.500000; d=default                                                                                                                                                  
a=b; b=20; c=0.500000; d=default                                                                                                                                                  
a=c; b=200; c=10.500000; d=default                                                                                                                                                
a=d; b=2000; c=100.500000; d=hello  

you don't need to use VARARGS with just C. Here is an example.

int funcA_12(int a1, int a2) { ... }

#define funcA(a1) funcA_12(a1, 0)

This answer is very similar to the two functions method above but in this case, you're using a macro for the function name that defines the arguments.

https://github.com/cindRoberta/C/blob/master/structure/function/default_parameter.c

#include<stdio.h>

void f_impl(int a, float b) {
  printf("%d %g\n", a, b);
}

#define f_impl(...) f_macro(__VA_ARGS__, 3.7)
#define f_macro(a, b, ...) f_impl(a, b)

int main(void) {
  f_impl(1);
  f_impl(1, 2, 3, 4);

  return 0;
}

I know how to do this in a better manner. You simply assign NULL to a parameter, so, you will have no value. Then you check if the parameter value is NULL, you change it to the default value.

void func(int x){
if(x == NULL)
  x = 2;
....
}

Though, it will cause warnings. a better choice would be to assign a value that will do nothing if the parameter value is that:

void func(int x){
if(x == 1)
  x = 2;
....
}

In the example above, if x is 1 the function changes it to 2;

Thanks to @user904963, EDIT: if you have to cover all ranges of numbers, it's not hard to add another argument only to say to the function whether it would set the parameter to default or not

void func(int x, bool useDefault){
if(useDefault) //useDefault == true
  x = 2;
....
}

However, remember to include stdbool.h