If you call something like this in C:
#include <stdio.h>
#define REPS ,a
...
int a = 1;
printf("%d" REPS);
It will work, but is it possible to call the REPS macro multiple times based on an unknown value, like for example, that I want to have five inputs in a scanf, yet I want my code to automate it (for example, if #define REPS ,a[i] then: ... ,a[1] ,a[2])?
It will work, but is it possible to call the REPS multiple times based in an unknown value
No. #define creates a preprocessor macro that you can use in your code, but when the compiler compiles your code, the actual value is substituted for the macro. If you have:
#define FOO 7
for example, then every occurrence of FOO in your code is replaced by 7 before the code is compiled; by the time the compiler sees your code, there's no #define and no FOO, only 7 wherever FOO was. Although there are some other preprocessor commands (e.g. #if) that can control whether a given #define is evaluated at all, there are no other preprocessor control structures (loops etc.).
I want to have five inputs in a scanf, yet I want my code to automate it (for example, if #define REPS ,a[i] then: ... ,a[1] ,a[2])?
You can certainly automate something like that; it's just that the preprocessor isn't the right tool for the job. Consider:
int reps = 5
//...
for (int i = 0; i < reps; i++) {
scanf(" %d", &a[i]);
}
REPS is evaluated at compile time, so it cannot depend on run-time values in this case a. There are hacks but in general you cannot do compile loops with macros.
I suggest you write a function along these lines instead:
#include <stdio.h>
void print_int_array(size_t n, int a[n]) {
for(int i = 0; i < n; i++)
printf("%d%s", a[i], i + 1 < n ? ", " : "\n");
}
int main() {
print_int_array(0, (int []) {});
print_int_array(1, (int []) {1});
print_int_array(2, (int []) {1, 2});
}