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0 pointsCyberDildonics3y ago0 comments

while C only once

You listed C twice

0 comments

That's a typo. You can still see they're correct about the ratio (two "C"s for every three "A"s and "B"s).

There's no ratio. It's even across all of them, as long as the integer keeps incrementing. One more number (9) instead of stopping at 8 and there would be an even spread.

ahmedtd3y ago

Not when the counter overflows back to 0. If it's a 3 bit counter, 0 is A again, not C.

manigandham3y ago

The comment says a 32-bit signed int. Where is the 3-bit assumption coming from?

1 more reply

andreareina3y ago

Edited, thanks

j / k navigate · click thread line to collapse

0 comments

woodruffw3y ago

That's a typo. You can still see they're correct about the ratio (two "C"s for every three "A"s and "B"s).

manigandham3y ago

There's no ratio. It's even across all of them, as long as the integer keeps incrementing. One more number (9) instead of stopping at 8 and there would be an even spread.

ahmedtd3y ago

Not when the counter overflows back to 0. If it's a 3 bit counter, 0 is A again, not C.

manigandham3y ago

The comment says a 32-bit signed int. Where is the 3-bit assumption coming from?

1 more reply

andreareina3y ago

Edited, thanks

j / k navigate · click thread line to collapse