4 ms·

msvc detects uninitialized stack variable usage as a run time check and throws an exception. I've always wondered why the compiler can't warn about this at com

msvc detects uninitialized stack variable usage as a run time check and throws an exception. I've always wondered why the compiler can't warn about this at compile time. It should know when uninitialized stack space is being used since it is, after all, the compiler artificially designating what the stack is!

gpderetta 10y ago
Compilers are certainly capable of warning about uses of uninitialized variables, but there are necessarily false positives and false negatives as the problem is in general undecidable.
- ci5er 10y ago
  Is it? Why isn't it as simple as a rule stating that a warning is generated whenever, in the AST, a variable is used on the RHS before it's ever used on the LHS of an expression? (Probably something to do with globals/scope, right?)
  - CJefferson 10y ago
    When you have ifs and for loops around, it maybe depend on arguments to the function, and also the return values of functions you call. You could say "warn if there are any inputs, and any return values for any function I call, that can lead to a problem" but I suspect that would produce too many false positives.
  - mnarayan01 10y ago
    That's easy enough to do (and plenty of compilers do it), the problem is determining it for things set conditionally (uncomputable in the the general case) e.g.: int y, z; if (x < 3) { y = 0; } if (x < 4) { z = y + 1; }
    - oldmanhorton 10y ago
      Well, in the most general case where x's value can only be determined at runtime, most any compiler can and will assume that the initial if could or could not be hit and will therefore assume that y could, conservatively, be uninitialized. Its not uncomputable, its just conservative.
      - mnarayan01 10y ago
        I mean it's undecidable even if x is computed from purely static values: `x` can be the result of an arbitrary C-function and C is Turing complete.
        nitrogen 10y ago
        If x is a 32-bit int, then there are only ~4bil possible values to consider. If any of them not ruled out already would lead to an uninitialized read, issue a warning.
    - CoolGuySteve 10y ago
      But this is still an error when x == 3, y will be uninitialized when z is set. There are plenty of warnings in -Wall that are conservative. For example in g++ C++, out of order constructor initialization even for primitive types is warned for the same reason. In this case the use of y on the RHS without explicit initialization should be warned. It sounds annoying, but having a bug's behaviour change between different build modes is much more irritating.
      - mnarayan01 10y ago
        I probably should have made the second conditional `(x < 2)` rather than `(x < 4)` so that the compiler should not generate a warning. Here a serious compiler would likely be able to statically determine that the variable was always initialized, but for a sufficiently complex function call (even an inlined one) it would not. The compiler could just be "conservative" like you said and warn whenever it couldn't statically determine validity, but note that initializing e.g. `y = 0` has a performance cost (cf. reordering constructor initializers).
gsg 10y ago
In general it's impossible. Consider this C program: void do_something(int *); int f() { int x; do_something(&x); return x; } Is the return of x a use of initialized memory? Maybe. Relink the program and the answer might change.
- smhenderson 10y ago
  Isn't x already initialized to zero as it's static? I googled it and this[0] seemed relevant: 1655 — if it has arithmetic type, it is initialized to (positive or unsigned) zero; [0] http://c0x.coding-guidelines.com/6.7.8.html http://c0x.coding-guidelines.com/6.7.8.html
  - nkurz 10y ago
    > Isn't x already initialized to zero as it's static? If x was static, this would be true. But in this case, x is just a default automatic variable. For x to be static, it would have to be declared outside the function, or have the keyword 'static'.
    - smhenderson 10y ago
      OK, thanks, appreciate the correction. I see what I had wrong now. For anyone else interested in a longer explanation these SO comments are pretty informative: http://stackoverflow.com/questions/14049777/why-are-global-variables-always-initialized-to-0-but-not-local-variables#14049791 http://stackoverflow.com/questions/14049777/why-are-global-v...
  - gsg 10y ago
    What you want is just a few lines up: 1652 If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. (x is an 'auto' variable in the given program.)
    - smhenderson 10y ago
      Yeah, that's what I get for replying without making sure I'm correct! :-)

gpderetta 10y ago

Compilers are certainly capable of warning about uses of uninitialized variables, but there are necessarily false positives and false negatives as the problem is in general undecidable.

ci5er 10y ago
Is it? Why isn't it as simple as a rule stating that a warning is generated whenever, in the AST, a variable is used on the RHS before it's ever used on the LHS of an expression? (Probably something to do with globals/scope, right?)
- CJefferson 10y ago
  When you have ifs and for loops around, it maybe depend on arguments to the function, and also the return values of functions you call. You could say "warn if there are any inputs, and any return values for any function I call, that can lead to a problem" but I suspect that would produce too many false positives.
- mnarayan01 10y ago
  That's easy enough to do (and plenty of compilers do it), the problem is determining it for things set conditionally (uncomputable in the the general case) e.g.: int y, z; if (x < 3) { y = 0; } if (x < 4) { z = y + 1; }
  - oldmanhorton 10y ago
    Well, in the most general case where x's value can only be determined at runtime, most any compiler can and will assume that the initial if could or could not be hit and will therefore assume that y could, conservatively, be uninitialized. Its not uncomputable, its just conservative.
    - mnarayan01 10y ago
      I mean it's undecidable even if x is computed from purely static values: `x` can be the result of an arbitrary C-function and C is Turing complete.
      - nitrogen 10y ago
        If x is a 32-bit int, then there are only ~4bil possible values to consider. If any of them not ruled out already would lead to an uninitialized read, issue a warning.
  - CoolGuySteve 10y ago
    But this is still an error when x == 3, y will be uninitialized when z is set. There are plenty of warnings in -Wall that are conservative. For example in g++ C++, out of order constructor initialization even for primitive types is warned for the same reason. In this case the use of y on the RHS without explicit initialization should be warned. It sounds annoying, but having a bug's behaviour change between different build modes is much more irritating.
    - mnarayan01 10y ago
      I probably should have made the second conditional `(x < 2)` rather than `(x < 4)` so that the compiler should not generate a warning. Here a serious compiler would likely be able to statically determine that the variable was always initialized, but for a sufficiently complex function call (even an inlined one) it would not. The compiler could just be "conservative" like you said and warn whenever it couldn't statically determine validity, but note that initializing e.g. `y = 0` has a performance cost (cf. reordering constructor initializers).

gsg 10y ago

In general it's impossible. Consider this C program: void do_something(int *); int f() { int x; do_something(&x); return x; } Is the return of x a use of initialized memory? Maybe. Relink the program and the answer might change.

smhenderson 10y ago
Isn't x already initialized to zero as it's static? I googled it and this[0] seemed relevant: 1655 — if it has arithmetic type, it is initialized to (positive or unsigned) zero; [0] http://c0x.coding-guidelines.com/6.7.8.html http://c0x.coding-guidelines.com/6.7.8.html
- nkurz 10y ago
  > Isn't x already initialized to zero as it's static? If x was static, this would be true. But in this case, x is just a default automatic variable. For x to be static, it would have to be declared outside the function, or have the keyword 'static'.
  - smhenderson 10y ago
    OK, thanks, appreciate the correction. I see what I had wrong now. For anyone else interested in a longer explanation these SO comments are pretty informative: http://stackoverflow.com/questions/14049777/why-are-global-variables-always-initialized-to-0-but-not-local-variables#14049791 http://stackoverflow.com/questions/14049777/why-are-global-v...
- gsg 10y ago
  What you want is just a few lines up: 1652 If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. (x is an 'auto' variable in the given program.)
  - smhenderson 10y ago
    Yeah, that's what I get for replying without making sure I'm correct! :-)