You can't return arrays from functions in C. You also can't (shouldn't) do this:
char *returnArray(char array []){
char returned [10];
//methods to pull values from array, interpret them, and then create new array
return &(returned[0]); //is this correct?
}
returned is created with automatic storage duration and references to it will become invalid once it leaves its declaring scope, i.e., when the function returns.
You will need to dynamically allocate the memory inside of the function or fill a preallocated buffer provided by the caller.
Option 1:
dynamically allocate the memory inside of the function (caller responsible for deallocating ret)
char *foo(int count) {
char *ret = malloc(count);
if(!ret)
return NULL;
for(int i = 0; i < count; ++i)
ret[i] = i;
return ret;
}
Call it like so:
int main() {
char *p = foo(10);
if(p) {
// do stuff with p
free(p);
}
return 0;
}
Option 2:
fill a preallocated buffer provided by the caller (caller allocates buf and passes to the function)
void foo(char *buf, int count) {
for(int i = 0; i < count; ++i)
buf[i] = i;
}
And call it like so:
int main() {
char arr[10] = {0};
foo(arr, 10);
// No need to deallocate because we allocated
// arr with automatic storage duration.
// If we had dynamically allocated it
// (i.e. malloc or some variant) then we
// would need to call free(arr)
}