Printing values of an array using a function in C language - arrays

Question

I am pretty new to programming and I was working on printing arrays using functions and I ran to the following error.
Test.c: In function ‘main’:
Test.c:21:54: error: incompatible type for argument 1 of ‘theta’
21 | printf("The theta values are = %lf\n", x[i], theta(x[i]));
| ~^~~
| |
| double
Test.c:5:21: note: expected ‘double *’ but argument is of type ‘double’
5 | double theta(double x[N])
| ~~~~~~~^~~~
And here is the code.
#include <stdio.h>
#include <stdlib.h>
#define N 50
double theta(double x[N]);
int main(){
int i;
double x[i];
printf("The theta values are = %lf\n", x[i], theta(x[i]));
return 0;
}
double theta(double x[N]){
int i;
for(i = 0; i < 50; i++){
x[i] = (double)(i)/ ((double)(N) - 1.0);
}
return x[i];
}
I just want to print 50 values between 0 and 1. Just like linspace(0:1:50) in MATLAB.
Thanks for the help.

Answer 1

First, in declare function, you cannot set size of array.
change to
double theta(double* X)
Second, do double theta(double x[N]); in main function.
and this is compiler didn`t advise you but I found bug in your code.
First, i in theta is different with i in main
Second, use for to print
I think you want this code
#include<stdio.h>
#include<stdlib.h>
#define N 50
double* theta()
{
int i;
double x[N];
for(i = 0; i < N; i++){
x[i] = (double)(i)/ ((double)(N) - 1.0);
}
return x;
}
int main(){
int i;
double* x=theta();
for(i=0;i<N;i++){
printf("The theta values are = %lf\n", x[i]);
}
return 0;
}
And I like this code more
#include<stdio.h>
#include<stdlib.h>
#define N 50
int main(){
int i;
for(i=0;i<N;i++){
printf("%lf\n",i/(N-1.0));
}
return 0;
}

Answer 2

I found the solution. Here it is.
#include <stdio.h>
#include <stdlib.h>
#define N 50
double* theta(double x[], int nb_elements){
int i;
for(i = 0; i < nb_elements; i++){
x[i] = (double)(i)/ ((double)(nb_elements) - 1.0);
}
return &x[0];
}
int main(){
int i;
double x[N];
theta(x, N);
for(i = 0; i < N; i++){
printf("%lf\n", x[i]);
}
return 0;
}