Call #define multiple times based on a specific value

user_16290700 0
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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])?

macros c
2 Answers

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]);
}
user_6433830

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});
}
Allan_Wind_97060
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