i is initialized twice in the example code

  int i = 0;
    switch(i){
     for(i = 0; i < ({case 0:; 10;}); i++){

The compiler probably tried to be intelligent and optimized it away.

EDIT: Okay I tried reading the assembly and I think I understand whats going on:

  movl    $0, -8(%rbp)                     # translation of i = 0
  ...                                      # the following is jumped over
  movl    -8(%rbp), %eax
  movl    %eax, -16(%rbp) # 4-byte Spill   # i is moved to somewhere else in memory
  ...                                      # the switch jumps between these
  movl    -16(%rbp), %eax # 4-byte Reload  # tries to reload i from its "new" position

Whoever wrote the code to place variables into memory obviously didn't consider this kind of insane control flow.