Because I'm still figuring out 'Arduino C' I leave all warnings on.
Recently I got this one:
narrowing conversion of '(int)(((unsigned char)bx_0) | ((unsigned char)bx_1))' from 'int' to 'const byte... etc.
...referring to this line:
const byte pinMask[3] = {bx_0, bx_1, bx_0 | bx_1};
...pointing to the or symbol.
The code in question (much simplified to recreate this):
const int latch[2] = {2, 4};
const byte bx_0 = 1 << latch[0];
const byte bx_1 = 1 << latch[1];
const byte pinMask[3] = {bx_0, bx_1, bx_0 | bx_1};
(This happens whether I use byte, uint8_t, or unsigned char everywhere.)
Without the or operator, no warning. If bx_0 and bx_1 are simple bytes, not shifted into existence, no warning. If I do this:
const byte pinMask[3] = {bx_0, bx_1, (byte)(bx_0 | bx_1)};
...again, no warning.
However, I get no complaints about this:
const byte notMask = !(bx_1 | bx_0);
...so a further condition is that the problem or has to be in an initializer list. Yikes.
I don't get it. There's no way an or of two bytes can force widening, so where did the (int) cast come from? It apparently has something to do with the 1 << latch[0] code, but the result of that is still a byte, by definition.
I have a solution -- the (byte) cast -- but in the real code it's ugly and feels unnecessary. Everything is already just bytes.
