How to Get Fired Using Switch Statements and Statement Expressions

Re:How to Get Fired Using Switch Statements and Statement Ex... by jjnoakes on Friday October 28, 2016 @09:02AM (#12815289)

> there is a (slightly) saner interpretation: branch there to execute the case, then "return" 10 from the statement expression and compare that to i, continuing onward from there as usual [...] and that does not cause any uninitialised reads or writes

The code you wrote did one additional thing: it moved the "switch" starting point from before the "i=0" initialization to after the "i=0" initialization, which is more than just a different interpretation, it's completely different semantics.

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Re:How to Get Fired Using Switch Statements and Statement Ex... by yorwba on Friday October 28, 2016 @09:25AM (#12815504)

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.

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Re:How to Get Fired Using Switch Statements and Statement Ex... by jjnoakes on Friday October 28, 2016 @12:00PM (#12817405)

Yep, you are right, I missed that "i" was initialized twice; when I see "i=0" in a for loop I assume it isn't initialized right above (otherwise, why put the "i=0" in the for loop at all?).

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