¿Cómo codifico en base58 el ID de la cadena usando Python?

Question

¿Cómo codifico en base58 el ID de la cadena usando Python?

- 7
- 199 2019-02-16
Cuando Tezosnecesitafirmar una carga útil de operación de horneado oendoso,elbyte 0 será un 0x01para una operación de horneado o 0x02para una operación deendoso. Losbytes 1-5en la carga útil de la operación contienenel ID de cadena (también conocido como Red),que sepuede codificaren base58 usando estaespecificación :

let chain_id="\ 087 \ 082 \ 000" (* Net (15) *)

El ID de cadena resultante será una cadena como NetXdQprcVkpaWU (el ID de cadena demainnet actualmente activo al 16 defebrero de 2019).

¿Cómo conviertoel campo de 4bytesen losbytes 1-5 alformato Net (15) codificadoen base58 usandoelmódulopybitcointools de Vitalik?

Intenté usarel scripten scripts/b58_prefix.py para determinarelmagicbyte apropiadoparapasar a lafunción bitcoin.bin_to_b58check ,perono obtengoel resultado apropiado:

$ ./b58_prefix.py Net 4 15 5722583 [87L,81L,215L]

Aquíestámi código:

defget_chain_id (self): chainid=bytes.fromhex (self.payload [2:10]) devolverbitcoin.bin_to_b58check (identificador de cadena,magicbyte=5722583)

En zeronet,una carga útil de respaldo se ve así: 023bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae ,donde 3bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae ,donde 3bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae ,donde 3bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae ,donde 3bb717ee debe ser NetXSzLHKwSumh7 en zeronet.
When Tezos needs to sign a baking or endorsement operation payload, byte 0 will be a 0x01 for a baking operation or 0x02 for an endorsement operation. Bytes 1-5 in the operation payload contain the chain ID (otherwise known as Network), which can be encoded in base58 using this specification: let chain_id = "\087\082\000" (* Net(15) *) The resulting chain ID will be a string like NetXdQprcVkpaWU (the currently active mainnet chain ID as of Feb. 16th, 2019). How do I convert the 4-byte field in bytes 1-5 into the base58 encoded Net(15) format using Vitalik's pybitcointools module? I've tried using the script in scripts/b58_prefix.py to determine the appropriate magicbyte to pass to the bitcoin.bin_to_b58check function, but I'm not getting the appropriate result: $ ./b58_prefix.py Net 4 15 5722583 [87L, 81L, 215L] Here is my code: def get_chain_id(self): chainid = bytes.fromhex(self.payload[2:10]) return bitcoin.bin_to_b58check(chainid, magicbyte=5722583) On zeronet, an endorsement payload looks like this: 023bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae, where 3bb717ee is the chain id, but that converts to Net1BPz7FKbUqsY and it should be NetXSzLHKwSumh7 on zeronet. protocol encoding base58 python fuente Luke Youngblood


                                    
                                        
                                            
                                                
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                                                                            2019-02-17
                                                                        
                                                                    
                                                                    
                                                                         Tumagicbytepareceestarequivocado.Sitoma los valores debytes decimales del original,conviértalos a hexadecimal y luego rellene con un cero a laizquierda,obtendrá

 >>> struct.unpack('>L', b'\x00\x57\x52\x00')[0]
5722624
 

 Este valor deberíaproducirel resultadoesperado: 

 >>> payload = '023bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae'
>>> mb = struct.unpack('>L', b'\x00\x57\x52\x00')[0]
>>> chainid = bytes.fromhex(payload[2:10])
>>> bitcoin.bin_to_b58check(chainid, magicbyte=mb)
'NetXSzLHKwSumh7'
 
                                                                        Your magicbyte seems to be wrong. If you take the decimal byte values from the original, convert them to hex, then pad it with a leading zero, you get

>>> struct.unpack('>L', b'\x00\x57\x52\x00')[0]
5722624


This value should produce the expected result:

>>> payload = '023bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae'
>>> mb = struct.unpack('>L', b'\x00\x57\x52\x00')[0]
>>> chainid = bytes.fromhex(payload[2:10])
>>> bitcoin.bin_to_b58check(chainid, magicbyte=mb)
'NetXSzLHKwSumh7'


                                                                        
                                                                        
                                                                        
                                                                        
                                                                            
                                                                                
                                                                                    
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                                                                                    cayblood
                                                                                
                                                                            
                                                                            
                                                                                
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                
                                                                            
                                                                            
                                                                        
                                                                    
                                                                    
                                                                
                                                            
                                                            
                                                                
                                                                    
                                                                        
                                                                            
                                                                                
                                                                                    Sí,5722624eselmagicbyte correcto:
                                                                                    Yeah 5722624 is the right magicbyte:
                                                                                    
                                                                                    
                                                                                    
                                                                                
                                                                                
                                                                                    
                                                                                        
                                                                                    
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                                                                                    2019-02-17
                                                                                    
                                                                                        
                                                                                            
                                                                                            Stephen Andrews
                                                                                        
                                                                                    
                                                                                    
                                                                                
                                                                                
                                                                            
                                                                        
                                                                    
                                                                
                                                            
                                                        
                                                    
                                                        
                                                            

                                                                
                                                                    
                                                                        
                                                                            
                                                                            
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                                                                            2019-02-16
                                                                        
                                                                    
                                                                    
                                                                         Parece queen realidad soloestá obteniendo 2bytes de datos (4 caracteres hexadecimales).Verifiquéesto decodificandoel resultado que obtuviste,y solo devolvió dosbytes de datosparaelbytemágico dado. 

 Intente realizarel siguiente cambio: 

 def get_chain_id(self):
    chainid = bytes.fromhex(self.payload[2:10])
    return bitcoin.bin_to_b58check(chainid, magicbyte=5722583)
 
                                                                        It looks like you are only actually grabbing 2 bytes of data (4 hex chars). I verified this by decoding the result you got, and it only returning two bytes of data for the given magic byte.

Try making the following change:

def get_chain_id(self):
    chainid = bytes.fromhex(self.payload[2:10])
    return bitcoin.bin_to_b58check(chainid, magicbyte=5722583)


                                                                        
                                                                        
                                                                        
                                                                        
                                                                            
                                                                                
                                                                                    
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                                                                                    Stephen Andrews
                                                                                
                                                                            
                                                                            
                                                                                
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                    
                                                                                        
                                                                                    
                                                                                
                                                                            
                                                                            
                                                                        
                                                                    
                                                                    
                                                                
                                                            
                                                            
                                                                
                                                                    
                                                                        
                                                                            
                                                                                
                                                                                    Gracias,estoesparte delproblema -necesitotomar 8 caracteres (4bytes hexadecimales),o "self.payload [2:10]".En zeronet,una carga útil deendoso se ve así: `023bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae`,donde` 3bb717ee`eselidentificador de la cadena,pero que se convierteen `NetFK` y debe ser.

Edité unpoco lapregunta anterior.Gracias,denuevo,portoda su ayuda.
                                                                                    Thanks, this is part of the problem - I need to grab 8 characters (4 hex bytes), or `self.payload[2:10]`. On zeronet, an endorsement payload looks like this: `023bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae`, where `3bb717ee` is the chain id, but that converts to `Net1BPz7FKbUqsY` and it should be `NetXSzLHKwSumh7` on zeronet.

I edited the question above a bit. Thanks again for all your help.
                                                                                    
                                                                                    
                                                                                    
                                                                                
                                                                                
                                                                                    
                                                                                        
                                                                                    
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                                                                                    2019-02-17
                                                                                    
                                                                                        
                                                                                            
                                                                                            Luke Youngblood
                                                                                        
                                                                                    
                                                                                    
                                                                                
                                                                                
                                                                            
                                                                        
                                                                    
                                                                
                                                            
                                                        
                                                    
                                                        
                                                            

                                                                
                                                                    
                                                                        
                                                                            
                                                                            
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                                                                            2019-02-26
                                                                        
                                                                    
                                                                    
                                                                         Enprimer lugar,¡muchasgracias!Me has ayudado a resolverelmisterio de lafirma delbloque :)
Puede utilizarelpaquete pytezos.encoding : 

 from pytezos.encoding import base58_encode

def get_chain_id(self):
    chainid = bytes.fromhex(self.payload[2:10])
    return base58_encode(chain_id, b'Net')
 
                                                                        First of all, many thanks! You've helped me with solving the block signature mystery :)
You can use pytezos.encoding package:

from pytezos.encoding import base58_encode

def get_chain_id(self):
    chainid = bytes.fromhex(self.payload[2:10])
    return base58_encode(chain_id, b'Net')


                                                                        
                                                                        
                                                                        
                                                                        
                                                                            
                                                                                
                                                                                    
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                                                                                    Michael Zaikin

cayblood · Accepted Answer · 2019-02-17

5

2019-02-17

Tumagicbytepareceestarequivocado.Sitoma los valores debytes decimales del original,conviértalos a hexadecimal y luego rellene con un cero a laizquierda,obtendrá

 >>> struct.unpack('>L', b'\x00\x57\x52\x00')[0]
5722624

Este valor deberíaproducirel resultadoesperado:

 >>> payload = '023bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae'
>>> mb = struct.unpack('>L', b'\x00\x57\x52\x00')[0]
>>> chainid = bytes.fromhex(payload[2:10])
>>> bitcoin.bin_to_b58check(chainid, magicbyte=mb)
'NetXSzLHKwSumh7'

Your magicbyte seems to be wrong. If you take the decimal byte values from the original, convert them to hex, then pad it with a leading zero, you get

>>> struct.unpack('>L', b'\x00\x57\x52\x00')[0]
5722624

This value should produce the expected result:

>>> payload = '023bb717ee882891d7be5b881cefa98946800e3d67b5d01b4237b3618709defb51ec37c3e100000518ae'
>>> mb = struct.unpack('>L', b'\x00\x57\x52\x00')[0]
>>> chainid = bytes.fromhex(payload[2:10])
>>> bitcoin.bin_to_b58check(chainid, magicbyte=mb)
'NetXSzLHKwSumh7'

Sí,5722624eselmagicbyte correcto:

Yeah 5722624 is the right magicbyte:
- 1
- 2019-02-17
- Stephen Andrews

¿Cómo codifico en base58 el ID de la cadena usando Python?

3 respuestas