Why does this assignment break my program?
c
Solution
It breaks because the array `states` is empty. Make it the size of `argc` (that's allowed in C99) to fix the problem:
char *states[argc];
The reason for this is as follows: `char *states[] = {};` makes an array of zero elements, so any dereference `states[i]` is undefined behavior.
Problem
I am learning C and I'm not sure how to phrase this, but why does uncommenting line 11 in the following code break this program? ``` #include <stdio.h> int main(int argc, char *argv[]) { printf("argc: %d\n", argc); char *states[] = {}; int i = 0; while(i < argc) { printf("arg %d: %s\n", i, argv[i]); //states[i] = "test"; i++; } return 0; } ``` When I uncomment this line and run the program I get this: ``` greggery@Lubu:~/code$ ./myprog aaa bbb ccc argc: 4 arg 0: ./lc arg 1: aaa ``` Why is `states[i] = "test";` breaking the `while` loop? When I comment it out I see all the arguments printed.