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Is signature char an interface problem?

Suppose I have a function

void foo(char *)

which internally must treat its input as a block of bytes with a trailing NUL (say, this is a hash function for strings). I can specify an unsigned char* argument in a function. I could also change the announcement to

 void foo(unsigned char *)

Now, given that char , signed char and unsigned char are three different types , will this change the interface with any reasonable definition of the term "interface" in C?

(This question is intended to solve the discussion raised by another question. I have my own opinions, but I will not accept the answer until the “winner” appears with the votes of others.)

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5 answers

In accordance with ISO / IEC 9899: TC3,

  • calling a function through an expression of an incompatible type is undefined behavior (6.5.2.2 §9)
  • compatible function types must have compatible parameter types (6.7.5.3 §15)
  • compatible pointer types must point to compatible types (6.7.5.1 §2)
  • char , signed char and unsigned char are different base types (6.2.5 §14) and therefore incompatible (6.2.7 § 1), which is also explicitly mentioned in footnote 35 on page 35

So this is clearly a change in the programming interface.

However, since char * , signed char * and unsigned char * will have identical representations and alignment requirements in any valid C language implementation, the binary interface will remain unchanged.

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Yes it is. Client code that was previously compiled will no longer compile (or, in any case, may generate new warnings), so this is a violation.

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I choose "C - none of the above."

Although this is not a direct answer to the question you asked, the correct solution to the situation seems quite simple and obvious to me: you should not use any of the above.

At least IMO, you really have a reason to do otherwise, your function should accept void * or (preferably) void const * . What you are looking for is basically an opaque pointer, and exactly what void * provides. The user does not need to know anything about the internal components of your implementation, and since any other type of pointer will implicitly convert to void * , this is one of the few features that also does not violate existing code.

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No, it is not. Any change in the client code is trivial (especially if it avoids the warning), and in practice in almost any C implementation, you will find that you do not even need to recompile, since pointers to char* and unsigned char* will be transmitted in the same way in the calling convention .

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Is char * an implicit conversion to unsigned char *?

  • Yes - you have not violated your interface
  • No - you missed the implementation
    the details.
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