This isn't quite right. In invalid UTF8, a continuation byte can also emit a replacement char if it's the start of the byte sequence. Eg, `0b01100001 0b10000000 0b01100001` outputs 3 chars: a�a. Whether you're at the beginning of an output char depends on the last 1-3 bytes.

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rockwotj 4 days ago | parent [-]

> outputs 3 chars

You mean codepoints or maybe grapheme clusters?

Anyways yeah it’s a little more complicated but the principle of being able to truncate a string without splitting a codepoint in O(1) is still useful

	▲	jridgewell 4 days ago \| parent [-]
		Yah, I was using char interchangeably with code point. I also used byte instead of code unit. > truncate a string without splitting a codepoint in O(1) is still useful Agreed!