Segmentation fault when accessing a c string in a another function - c

Question

I am writing a small program to get familiar with C again.
I have now a part of the program where I am stuck and can not go forward. I have a function where I create a c string and a pointer to the 2nd element of that string. (When it is finished it should receive a pointer to another string)
When I pass this strings to another function which should remove some elements on the first one I always get a segmentation fault. Which should mean, that I can't access that memory. But, if I change the order I pass the strings to the function I get the same error... .
Here is the code:
int analyze_sudoku(const void *self, Sudoku *sudoku) {
for(int i = 1; i < 82; i++)
{
int success = 0;
// All possible values of a column
// This can be rewritten of course but this way
// it makes the intention what this var is for
// very clear in my opinion
char options[10] = {'1','2','3','4','5','6','7','8','9'};
char *chars_in_row;
// The same problem...
// chars_in_row = sudoku->get_row_for_column(sudoku, i);
chars_in_row = &options[2];
printf("In field %d we have this chars in the row: %s\n", i,chars_in_row);
printf("length der chars: %d\n", strlen(chars_in_row));
printf("addresse of the char pointers: %p\n", (void *)chars_in_row);
// After receiving all chars from one row, we remove
// those from the one we have in our options
// HERE IS THE FUNCTION CALL
remove_from_options(options, chars_in_row);
// ... more code follows
}
}
And here is the function where I get my seg fault:
char *remove_from_options(char *options, char *already_in_use) {
puts("Welcome");
printf("Your options: %s\n", options);
// HERE THE SEG FAULT HAPPENS
// as already mentioned the error happens no matter what I give this function
printf("pointer address: %p", (void *)already_in_use);
printf("already in use: %s", already_in_use);
printf("in use länge: %d", strlen(already_in_use));
for(int i = 0; i < strlen(already_in_use); i++)
{
// some code...
}
}

Answer 1

This
char options[10] = {'1','2','3','4','5','6','7','8','9'};
is not a string but array of characters. In C language, strings are actually one-dimensional array of characters terminated by a null character '\0'.
The %s format specifier, in printf(), writes every byte up to and not including the first null terminator. Since the char array option does not have the null character, the printf() must be accessing beyond the size of option array which is leading to segmentation fault.
Try adding the null character at the last of option array, like this:
char options[10] = {'1','2','3','4','5','6','7','8','9','\0'};
EDIT
I totally missed the point that the last index of array option has been initialized with 0. There is no need to explicitly give the null character at the end of option array. OP has accepted this as the answer because this, somehow, resolve the OP's problem. Hence, I can not delete this post. Minimal complete and verifiable example is required to identify the root cause of the problem mentioned by OP.