I need help pertaining to binary files

Question

When I use writeByte(346); to a binary file and then do readByte(); it returns 90, not 346. I'm perplexed and would greatly appreciate some help.

Answer 1

346 doesn't fit in a single byte.

So 346 overflows and you get 346 - 256 (2^8) = 90.

You need at least 2 bytes to store 346.

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writeByte(int) takes a signed int as parameter, so the limits for one byte are -128 to 127