Cannot declare a void function before the main() in C - c

Question

I'm a beginner in C language and in my previous question I have asked about proper function declaration order in C. I was told that in standard C, it is necessary to declare functions before calling them.
But for the following example code I cannot declare the the function times2p before the main(). I tried to declare it as: void times2(int in_data); right before the main(), but I get errors. Here below is the example code:
#include <stdio.h>
#include <stdlib.h>
void times2p(int *in_data);
void times2(int in_data);
int main()
{
int y = 5;
int s = times2(y);
printf("%d\n", y);
printf("%d\n", s);
printf("-------------\n");
int yp = 5;
times2p(&yp);
printf("%d\n", yp);
return 0;
}
//Multiplies the input argument by two
void times2(int in_data){
in_data = in_data*2;
}
void times2p(int *in_data){
*in_data = *in_data*2;
}
How and where should the times2p function be declared in this case? If I don't declare the code still compiles without error but I was told that I have to declare the functions in C in my previous question.
Here is the error:
||=== Build: Debug in test1 (compiler: GNU GCC Compiler) ===|
C:\Users\nnn\Documents\CodeBlocks\test1\main.c||In function 'main':|
C:\Users\nnn\Documents\CodeBlocks\test1\main.c|16|error: void value not ignored as it ought to be|
C:\Users\nnn\Documents\CodeBlocks\test1\main.c|13|warning: unused variable 'read_x' [-Wunused-variable]|
||=== Build failed: 1 error(s), 1 warning(s) (0 minute(s), 0 second(s)) ===|

Answer 1

Your code in the current version of the question is incorrect, but not for the reason you're asking about.
There's nothing wrong with the void function declarations before main, and the compiler doesn't complain about them. The problem is that a void function doesn't return a value, and you're trying to call them as if they did:
int s = times2(y);
This would be correct only if times2(y) returned an int value (or a value of some type implicitly convertible to int).
Also, looking at the definition of the function:
void times2(int in_data){
in_data = in_data*2;
}
The parameter in_data is a local variable within the function. Changing its value effectively does nothing.
Your times2p function looks OK, and it should work. If you want times2() to be useful, it needs to return a value:
int times2((int in_data) {
return in_data * 2;
}

Answer 2

(The question has changed, so this answer may not be perfect anymore, sorry).
You need to declare at least the prototype of the functions before you use it. Like this:
#include <stdio.h>
#include <stdlib.h>
//How should the function times2p be declared here?
void times2p(int *in_data);
int main()
{
int y = 5;
times2p(&y);
printf("%d\n", y);
return 0;
}
//Multiplies the input argument by two
void times2p(int *in_data){
*in_data = *in_data*2;
}
Or declare the function itself, like this:
#include <stdio.h>
#include <stdlib.h>
//Multiplies the input argument by two
void times2p(int *in_data){
*in_data = *in_data*2;
}
int main()
{
int y = 5;
times2p(&y);
printf("%d\n", y);
return 0;
}