Why is char being "extended" to an int?

Question

I know I'm missing something so simple here, but I can't figure it out.

If I do char letter = 'A'; then letter = letter << 8, letter = 0, as one should expect, since we are shifting the bits "off the shelf" to the left.

But, if we do this:

char letter = 'A';
void setup() 
{
 Serial.begin(9600);
 int twoBytes = letter << 8;
 Serial.println(twoBytes, BIN);
}
void loop() {}

The result is 100000100000000, but why? << has precedence over =, so it should be bitshifting letter left by 8 first, then assigning that value (which should be 0) to twoBytes.

• Nick Gammon: Isn't there an implication that it will work (for some definition of "work")? If you were expecting zero why not just write: int twoBytes = 0; - the fact that you didn't implies that you expect letter to be treated as 16 bits.
• Nick Gammon: I just want to point out that although you posted it here, this is entirely a C++ question. The Arduino compiler is the GNU C++ compiler. Nothing "Arduino" about the way it treats code you write, with the exception of some pre-processing that the IDE does. See here for more details about that.
• AJ_Smoothie: @NickGammon .....because I need letter to be in high byte of twoBytes, and I'm putting another char into the low byte of twoBytes. Even though the code I wrote worked, being the inquisitive person I am I realized it didn't make sense, hence the question.

Answer 1

This is a funny, non-intuitive rule of the C++ language called "integral promotion". Simply stated, it says that no computation is ever performed on a type smaller than int: anything smaller gets implicitly promoted to int as soon as it appears in an arithmetic or logic expression.

See Implicit conversions in cppreference.com (warnings: it's hairy).

• AJ_Smoothie: Thank you, I'll take it on faith! Would it be good practice to say int twoBytes = letter; twoBytes << 8; instead for clarity, or does it even matter?
• Edgar Bonet: I would say that, for an experienced C or C++ programmer, it may not really matter. For someone unfamiliar with these rules, anything that can make the code clearer is a good thing. You may also write int twoBytes = int(letter) << 8;, or you may stick with letter << 8 and add a comment.
• Nick Gammon: In this sort of case, where you only want the right-most 8 bits I would personally "and" it, like this: int twoBytes = (letter << 8) & 0xFF; Since you want the low-order byte, make it explicit.
• timemage: I'm not sure how what you're saying is connected to what they're saying, but int twoBytes = (letter << 8) & 0xFF; is either complicated way of putting zero in twoBytes or potentially undefined behavior, depending on the value and system.
• Nick Gammon: The OP was surprised that the result wasn't zero, so I was pointing out that if you are doing a shift (left or right) and you aren't sure how many bits are being retained in the result, then anding it would make it clear. I don't see how the results could possibly be undefined.