Re: Don't understand this compiler warning

On Jun 28, 2005, at 06:39 , Markus Ruggiero wrote:

Folks,

This is some very old private code (from the mid-80ies, pre ANSI-C) that I try to port to the Mac. The code isn't mine but is supposed to compile under Windows 3.x (!) with MSVisualStudio 4.

I have no clue what the compiler tries to tell me:

    auto unsigned long    lcl_l_flags = 0;

    lcl_l_flags = pgf_setdef_flag_(HYPHEN_PGF,

                                   DDIF_M_T_HYPHENATION_ALLOWED,

                                   DDIF_M_T_HYPHENATION_ALLOWED);

pgf_setdef_flag is a macro defined as follows:

    #define pgf_setdef_flag_(flag,val,def)                    \

        ( (!spgfp) ? lcl_l_flags |= def                    \

               : (spgfp->flags & flag) ? lcl_l_flags |= val        \

                           : 0                \

        )

Thus the above code after preprocessing looks like:

    lcl_l_flags = ( (!spgfp) ? lcl_l_flags |= 1 : (spgfp->flags & (0x02)) ? lcl_l_flags |= 1 : 0 );

And then the compiler says:

    warning: operation on 'lcl_l_flags' may be undefined

This is a warning, not an error.  It indicates something that the compiler (writer) thinks may be problematic, but then again, may be all right.

In your case, the warning is correct: the code as written can lead to unexpected behavior.

Since 'lcl_l_flags' is an automatic variable, it is allocated on the stack.  Because of that, the variable will retain whatever value resides at that location in memory (you don't initialize the variable at the beginning of your routine).

But I *do* init the variable! So still the question is open: what could be wrong with this code fragment? I know, warnings are not errors - and sometimes the compiler is just plain wrong and the code is OK. I am just curious as to why the compiler thinks there might be problems.

Thanks for any more insights

---markus---

 

Your macro "updates" the value of the variable by doing the equivalent of

      lcl_l_flags = lcl_l_flags | 1;

If you are only interested in the low-order bit of the variable, then this isn't a problem (the macro will set that to either 1 or 0).  Otherwise, your code may be wrong.

Regards,

Justin

--

Justin C. Walker, Curmudgeon at Large

Institute for General Semantics

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My wife 'n kids 'n dog are gone,

I can't get Jesus on the phone,

But Ol' Milwaukee's Best is my best friend.

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Markus Ruggiero

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