Yeah my own first reaction was “Oh, gcc doesn’t warn about overflow assignment? That’s a definite bug.”

I code in C/C++ too, although not exclusively. I facepalmed a few seconds later, but it goes to show the value of having this be a compiler warning.

these are signed, so you‘re getting -2^16... so many warnings to emit!

/edit: had that mixed up in my head. Interestingly, clang does warn about the overflow:

    main.c:6:17: warning: implicit conversion from 'int'
      to 'int16_t' (aka 'short') changes value from
      65536 to 0 [-Wconstant-conversion]
  int16_t s16 = 65536;

Oops, you meant -2^15 I guess.

No, they meant -2^16, ie, -(0)

-2^16 is not -0. The unary minus binds more tightly than the binary xor, so this comes out as

  -2^16 = (-2) ^ 16 = 0b111...1110 ^ 0b10000 = 0b111...11011110 = -18

if I'm not mistaken.

No, we mean the power operator, 2 to the power of 16 would be '0' in a short int, since the value overflows...

But since philosophically a signed int has '1's in every bit from the MSB to infinity towards the left, it should thus be -0 not 0.

In two's complement (the usual binary representation of signed ints), -0 is the same as 0.