Topic: Why is the result of sha256 the same for inputs 0 to 15? (Read 86 times)
Thx again.
Yes.
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?
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.
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").
Just for curiosity:
Why does
result in
where $var is in {"0", ..., "15"}?
Thx.
Why does
Code:
echo "obase=16;ibase=10;$var" | bc | xxd -p -r | sha256sum
result in
Code:
e3b0c44298fc1c149afbf4c8996fb92427ae41e4649b934ca495991b7852b855
where $var is in {"0", ..., "15"}?
Thx.
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