Why is the result of sha256 the same for inputs 0 to 15?

[curiosity81]

Just for curiosity:

Why does

Code:

echo "obase=16;ibase=10;$var" | bc | xxd -p -r | sha256sum

result in

Code:

e3b0c44298fc1c149afbf4c8996fb92427ae41e4649b934ca495991b7852b855

where $var is in {"0", ..., "15"}?

Thx.

[Foxpup]

Because converting hex digits into bytes requires an even number of the former, since each byte requires two hex digits. In particular, xxd drops the last hex digit if there are an odd number of them, and if this results in a zero-length input, it will produce a zero-length output (compare your hash to the output of "echo -n | sha256sum").

[curiosity81]

Because converting hex digits into bytes requires an even number of the former, since each byte requires two hex digits. In particular, xxd drops the last hex digit if there are an odd number of them, and if this results in a zero-length input, it will produce a zero-length output (compare your hash to the output of "echo -n | sha256sum").

I see. This makes sense!

Thank you very much.

[curiosity81]

So, to do it correctly, if bc returns an odd number of digits, would it be correct to add a 0 at the beginning of the hex-string?

[Foxpup]

Yes.

[curiosity81]

Thx again.