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.

score 4 · Accepted Answer · answered Sep 19 '23 at 15:58

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).

Why is char being "extended" to an int?

1 Answers1