implicit declaration of function

早过忘川 提交于 2019-12-11 12:25:28

问题


GCC typically yields this warning when the proper header file is not included. This link --> www.network-theory.co.uk/docs/gccintro/gccintro_19.html says that because the function declaration is implicit (rather than explicitly declared via a header) the wrong argument types could actually be passed to the function, yielding incorrect results. I don't understand this. Does this mean the compiler generates code that pushes something, of the machine's word size, onto the stack for the callee to consume, and hopes for the best?

Detail is appreciated.


回答1:


In classic K&R C, that's pretty much what happened; there were default coercions (anything smaller than (int) was promoted to (int), for example), and for backwards compatibility any function without a prototype is still called that way, but by and large the only indication you got for passing the wrong type was a weird result or maybe a core dump. Which is where you get in trouble, as when the function has a prototype the exact (not coerced/promoted) value is pushed. So if you're passing a (char), if there's a prototype in scope then a single byte is pushed by the caller, otherwise 4 bytes (on most current platforms). If the caller and callee disagree about this, Bad Things happen.




回答2:


If the compiler doesn't have specific information about how the argument should be passed, such as when there's no prototype or for arguments that are passed where the prototype have an ellipsis ('...'), the compiler follows certain rules for passing the arguments. These rule basically follow what occurred in pre-standard (or K&R) C - before prototypes were used. Paraphrased from C99 6.5.2.2/6 "Function calls":

* the integer promotions are applied
* if the argument has float type it's promoted to double

After these default argument promotions are applied, the argument is simply copied to wherever the compiler normally copies arguments (generally, the stack). So a struct argument would be copied to the stack.

If the actual function implementation doesn't match how the compiler creates the parameters, then you get undefined behavior (with exceptions for signed/unsigned mismatch if the value can be represented or pointers to char and pointers to void can be mixed/matched).

Also in C90, if the function is implicitly declared (which C99 doesn't permit, though it does permit functions without prototypes), the return value is defaulted as int. Once again, the the actual function returns something else, undefined behavior results.



来源:https://stackoverflow.com/questions/6488429/implicit-declaration-of-function

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