c++language-lawyerone-definition-ruleinline-functionstranslation-unit

inline function in different translation units with different compiler flags undefined behaviour?


in visual studio you can set different compiler options for individual cpp files. for example: under "code generation" we can enable basic runtime checks in debug mode. or we can change the floating point model (precise/strict/fast). these are just examples. there are plenty of different flags.

an inline function can be defined multiple times in the program, as long as the definitions are identical. we put this function into a header and include it in several translation units. now, what happens if different compiler options in different cpp files lead to slightly different compiled code for the function? then they do differ and we have undefined behaviour? you could make the function static (or put it into an unnamed namespace) but going further, every memberfunction defined directly in a class is implicit inline. this would mean that we may only include classes in different cpp files if these cpp files share the identical compiler flags. i can not imagine this to be true, because this would basically be to easy to get wrong.

are we really that fast in the land of undefined behaviour? or will compilers handle this cases?


Solution

  • As far as the Standard is concerned, each combination of command-line flags turns a compiler into a different implementation. While it is useful for implementations to be able to use object files produced by other implementations, the Standard imposes no requirement that they do so.

    Even in the absence of in-lining, consider having the following function in one compilation unit:

    char foo(void) { return 255; }
    

    and the following in another:

    char foo(void);
    int arr[128];
    void bar(void)
    {
      int x=foo();
      if (x >= 0 && x < 128)
         arr[x]=1;
    }
    

    If char was a signed type in both compilation units, the value of x in the second unit would be less than zero (thus skipping the array assignment). If it were an unsigned type in both units, it would be greater than 127 (likewise skipping the assignment). If one compilation unit used a signed char and the other used unsigned, however, and if the implementation expected return values to sign-extended or zero-extended in the result register, the result could be that a compiler might determine that x can't be greater than 127 even though it holds 255, or that it couldn't be less than 0 even though it holds -1. Consequently, the generated code might access arr[255] or arr[-1], with potentially-disastrous results.

    While there are many cases where it should be safe to combine code using different compiler flags, the Standard makes no effort to distinguish those where such mixing is safe from those where it is unsafe.