The function is used once, and because it occurs before the actual definition (and there was no forward declaration) an implicit declaration occurs (the erroneous int bar()). When the function definition is later reached by the compiler, its signature doesn't match the implicit declaration's signature, causing the problem. If the function signature matches the implicit declaration, then there is no problem:

  int main(int argc, char* argv[]) {
    printf("%d\n", foo(10));
    return 0;
  }
  int foo(int n) {
    return n - 1;
  }

Will work just fine, giving you only a warning but will print out "9" as expected.

> an implicit declaration occurs (the erroneous int bar()

I still don't understand; why does the implicit declaration assume int bar() is the signature when bar is being passed an int and returning a value to a char?

Because C doesn't do type inference.

You don't need type inference to handle this correctly, you just need to wait until you've parsed the rest of the file and seen the signatures of all functions.

Yikes! The implicit declaration, if used before (as in a lower line number than the function definition?) sounds like a trap, and at the root of this.

Yes.

But that's C. You can't fix it without fundamentally altering the language.