Ok I'm back!
The problem is
openssl dgst -sha256 -hex -sign chiave_priv_3.pem a.txt
In that way I do another SHA256! I did SHA256 3 times!
To resolve this issue, I can do something like
$ openssl pkeyutl -inkey chiave_priv_3.pem -sign -in a.txt -pkeyopt digest:sha256 | xxd -p -c 256
or I can do single SHA256 and apply another SHA256 with openssl
$ printf 0200000001e2a8148889a8ec60fd9d28564ed8996bf7ffd6b11388ed9c044d2c250088d83b000000001976a914d2bb7890f3f6356d89673367b44e9a7d0265009188acffffffff01c0e4022a010000001976a914824441111b374bec1952a5b3fa9dd4e3ed679b3888ac0000000001000000 | xxd -r -p | sha256sum -b | xxd -r -p > a.txt
$ openssl dgst -sha256 -hex -sign chiave_priv_3.pem a.txt
I prefer the first solution!
About
"mandatory-script-verify-flag-failed (Non-canonical signature: S value is unnecessarily high) (code 16)" it's more complicated than that.
I converted the S (DER signature) to base10. (it's another signature, not the same of thread, sorry but I have my notes)
For example:
$ s=`echo "ibase=16; $(printf 00f00e64e164ce4fee984165ba8205a8544ece37458006687cdaa53d4e6e1859bc | tr '[:lower:]' '[:upper:]')" | bc | tr -d '\n' | tr -d '\' | awk '{print $1}'`
$ echo $s
108580515770129610852831425129233053758690240817412348750872366071983533218236
Then convert N to base 10, and get N/2
$ N=`echo "ibase=16;FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141" | bc | tr -d '\n' | tr -d '\' | awk '{print $1}'`
$ echo $N
108580515770129610852831425129233053758690240817412348750872366071983533218236
$ N2=`echo "$N/2" | bc | tr -d '\n' | tr -d '\' | awk '{print $1}'`
$ echo $N2
57896044618658097711785492504343953926418782139537452191302581570759080747168
You can find very cool stuffs if u search
108580515770129610852831425129233053758690240817412348750872366071983533218236 or
57896044618658097711785492504343953926418782139537452191302581570759080747168 in google
Now I Check if s is greater than N/2, if it is I need to subtract it. (N-S)
$ s=`echo "$N - $s" | bc | tr -d '\n' | tr -d '\' | awk '{print $1}'`
$ echo $s
7211573467186584570739559879454854094147323461662555631732797069534628276101
Convert the result to base16
$ s=`echo "obase=16;$s" | bc`
$ echo $s
E2412F237BCDCA1AD1AD7DA1075D8C0AD258A07066D695F99DEA0AAEC7034A4
Sometimes you can get odd bytes, in that case I have 63 hex.
$ printf FF19B1E9B31B01167BE9A457DFA57AA6BE0A5A12F4237BEE52D213E621DE785 | wc -c
63
It's very similar when you get seed phrase, if you don't have a block of 11 bits, you need to add some "padding", then I add 0 at the beginning.
s=0FF19B1E9B31B01167BE9A457DFA57AA6BE0A5A12F4237BEE52D213E621DE785
Now I can make a "new" DER signature, replace the old s with the new one, calculate the length of it and the length of signature!
And it works!
I hope to help someone!
Thanks to Andrew Chow and BrewMaster for your time guys
(English is not my mother tongue; please excuse any errors on my part)
