Why i cant make this series by C code? - c

Question

can you tell me why this doesn't work.
int function(int);
int main()
{
int g[20],N;
printf("Type N");
scanf("%d",&N);
g[20]=function(N);
printf("s[0] is %d\n",g[0]);
printf("s[1] is %d\n",g[1]);
printf("s[2] is %d\n",g[2]);
}
int function(int N){
int s[20];
s[0]=1;
s[1]=3;
s[2]=5;
return s[20];
}
I just want that my function return this numbers 1,3,5 but it returns some weird numbers, i thinks it's adresses or something.
PS. I just began to learn C.

Answer 1

You need to something like this.
void function(int *);
int main()
{
int g[20],N; // array g local to main funciton
printf("Type N");
scanf("%d",&N);
function(g); // invoke function by passing array base address as an argument
printf("s[0] is %d\n",g[0]); // the first three positions of the g array
printf("s[1] is %d\n",g[1]); // have not been set by function
printf("s[2] is %d\n",g[2]); // they are also all unknown values
}
void function(int *s){
s[0]=1; //*(s+0)
s[1]=3;
s[2]=5;
}

Answer 2

From your code, you seem to have some misunderstanding about arrays and scope. The two arrays you declare start with unknown values until you set them. see my comments to your original code:
int function(int);
int main()
{
int g[20],N; // array g local to main funciton
printf("Type N");
scanf("%d",&N);
g[20]=function(N); // invoke function with int typed in and return int
printf("s[0] is %d\n",g[0]) // the first three positions of the g array
printf("s[1] is %d\n",g[1]); // have not been set by function
printf("s[2] is %d\n",g[2]); // they are also all unknown values
}
int function(int N){ // you never use N in this function, why is it a parameter
int s[20]; // declare a int array local to function
s[0]=1; // you only set the first three items in array, rest are unknown values.
s[1]=3;
s[2]=5;
return s[20]; // return the 20th int item of the array, (unknown memory contents)
}