My program was able to craft and send raw transactions to geth v1.3.3 before, but after I upgrade to geth v1.4.0, calling sendRawTransaction over RPC always returns invalid sender error.
Is transaction serialization (i.e. RLP) changed somehow from v1.3.3 to v1.4.0? Here is a dump of by raw transaction that triggers an invalid user error:
0x0000: F8 CA 80 85 0B A4 3B 74 00 83 01 5F 90 94 08 BE ......;t..._....
0x0010: 24 CD 8D CF 73 F8 FA 5D B4 2B 85 5B 43 70 BD 5C $...s..].+.[Cp.\
0x0020: 44 8B 80 B8 64 B0 70 B9 BA 00 00 00 00 00 00 00 D...d.p.........
0x0030: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0x0040: 00 00 00 00 00 00 00 00 01 87 44 2E B8 96 6A 07 ..........D...j.
0x0050: 0C 31 C1 E8 AE A3 60 F5 35 32 47 81 13 34 31 D4 .1....`.52G..41.
0x0060: 4B FA 0A 0B 1B 9F 13 C6 F5 00 00 00 00 00 00 00 K...............
0x0070: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0x0080: 00 00 00 00 00 00 00 00 00 1B A0 DE A4 6B 8C E8 .............k..
0x0090: 72 5A 31 49 92 EC 6B 6F C6 89 8C BB D7 A4 B9 8A rZ1I..ko........
0x00A0: 10 D2 F7 9E CE 6B D5 0F C5 19 E9 A0 8F 74 57 C2 .....k.......tW.
0x00B0: 1C DA CB 7D 7A 2B 46 58 98 53 31 C3 4B CF 50 1F ...}z+FX.S1.K.P.
0x00C0: 17 CE 16 80 95 30 38 9B 98 3C 5B B8 .....08..<[.
A more machine readable version of my transaction is:
F8CA80850BA43B740083015F909408BE24CD8DCF73F8FA5DB42B855B4370BD5C448B80B864B070B9BA000000000000000000000000000000000000000000000000000000000000000187442EB8966A070C31C1E8AEA360F535324781133431D44BFA0A0B1B9F13C6F500000000000000000000000000000000000000000000000000000000000000001BA0DEA46B8CE8725A314992EC6B6FC6898CBBD7A4B98A10D2F79ECE6BD50FC519E9A08F7457C21CDACB7D7A2B4658985331C34BCF501F17CE16809530389B983C5BB8
Log from geth gives
I0504 20:22:27.392581 9768 types.go:106] Generated response: *shared.ErrorResponse &{%!s(float64=1) 2.0 %!s(*shared.ErrorObject=&{-32603 Invalid sender})}
I0504 20:22:27.392886 9768 http.go:157] Sending payload: {
"id": 1,
"jsonrpc": "2.0",
"error": {
"code": -32603,
"message": "Invalid sender"
}
}
I believe the JSON RPC stuff changed in geth v1.4.0. I can't tell why this is happening without seeing the full sendRawTransaction you are calling but check out the docs: https://github.com/ethereum/wiki/wiki/JSON-RPC#eth_sendtransaction
Also, this should be migrated to https://ethereum.stackexchange.com/
Related
I tried to process payment with MasterCard / MIR.
I do read data from VISA successfully after send PDOL, but MS does not requires PDOL.
1st step in transaction:
Select 2PAY.SYS
[SEND] : 00 A4 04 00 0E 32 50 41 59 2E 53 59 53 2E 44 44 46 30 31 00
[READ] : 6F 23 84 0E 32 50 41 59 2E 53 59 53 2E 44 44 46 30 31 A5 11
BF 0C 0E 61 0C 4F 07 A0 00 00 00 04 10 10 87 01 01 90 00
2nd step:
[SEND] : 00 A4 04 00 07 A0 00 00 00 04 10 10 00
F
[READ] : 6F 44 84 07 A0 00 00 00 04 10 10 A5 39 50 0A 4D 41 53 54 45 52 43 41 52 44 5F 2D 04 72 75 65 6E 87 01 01 9F 11 01 01 9F 12 0A 4D 41 53 54 45 52 43 41 52 44 BF 0C 0F 9F 4D 02 0B 0A 9F 6E 07 06 43 00 00 30 30 00 90 00
Card does not requres PDOL
3rd step:
[SEND] : 80 A8 00 00 02 83 00 00
[READ] : 77 16 82 02 19 80 94 10 08 01 01 00 10 01 01 01 18 01 02 00 20 01 02 00 90 00
Recieved Application File Locator (AFL)
Step 4:
Read all available data.
[SEND] : 00 B2 01 0C 00
and other sectors
I red all sectors. But there are not tags requires for payment: 9F26 - Application Cryptogram, 9F37 - Unpredictable Number, 9F36 - Transatcion Counter.
To get this tags I could make command Generate AC with CDOL, but how to generate CDOL?
Card says about CDOL1 and CDOL2. And CDOL1 requires tags that card generate itself.
Card answers, contains CDOL1 and CDOL2:
70 81 A0 57 13 55 45 46 77 77 25 42 79 D2 01 12 01 58 11 10 00 00 79 0F 5A 08 55 45 46 77 77 25 42 79 5F 24 03 20 11 30 5F 25 03 17 11 01 5F 28 02 06 43 5F 34 01 01 8C 21 9F 02 06 9F 03 06 9F 1A 02 95 05 5F 2A 02 9A 03 9C 01 9F 37 04 9F 35 01 9F 45 02 9F 4C 08 9F 34 03 8D 0C 91 0A 8A 02 95 05 9F 37 04 9F 4C 08 8E 0E 00 00 00 00 00 00 00 00 42 03 1E 03 1F 03 9F 07 02 3D 00 9F 08 02 00 02 9F 0D 05 B4 50 84 00 00 9F 0E 05 00 00 00 00 00 9F 0F 05 B4 70 84 80 00 9F 42 02 06 43 9F 4A 01 82 90 00
How generate offline limit transaction using paypass? How to do Generate AC?
Thank you!
CDOL1 and CDOL2 splits like below. Tag and its lengths. You are supposed to provide the data alone in the same order and size as you would for PDOL. In the below case, apart from two towards the bottom, rest all are available in the terminal ready to use.
CDOL1
9F02 06 //transaction amount
9F03 06 //other amount, cashback
9F1A 02 //termial country
95 05 //TVR terminal has arrived after terminal risk management
5F2A 02 //currency code
9A 03 //Transaction date
9C 01 //transaction type
9F37 04 //unpredictable number
9F35 01 //terminal type
9F45 02 //data Authentication code from Transaction Related Data ODA
9F4C 08 //icc dynamic number from Transaction Related Data ODA
9F34 03 //cvm results
CDOL2
91 0A //issuer authentication data
8A 02 //ARC
95 05 //TVR
9F37 04 //unpredictable number
9F4C 08 //icc dynamic number
9F26 and 9F36 will be returned by the card in response to your GEN AC.
9F37 is generated by the terminal.
Question: Do I need to parse this? If so, how? If not, how can I iterate through this?
Issue: I am uploading two images from iOS to server. The files come in under request.files and looks this:
{ files:
[ {
name: 'file0',
data: <Buffer ff d8 ff e0 00 10 4a 46 49 46 00 01 01 00 00 48 00 48 00 00 ff e1 00 58 45 78 69 66 00 00 4d 4d 00 2a 00 00 00 08 00 02 01 12 00 03 00 00 00 01 00 06 ... >,
encoding: '7bit',
mimetype: 'image/jpg',
mv: [Function: mv] },
{
name: 'file1',
data: <Buffer ff d8 ff e0 00 10 4a 46 49 46 00 01 01 00 00 48 00 48 00 00 ff e1 00 58 45 78 69 66 00 00 4d 4d 00 2a 00 00 00 08 00 02 01 12 00 03 00 00 00 01 00 06 ... >,
encoding: '7bit',
mimetype: 'image/jpg',
mv: [Function: mv]
}
] }
I am trying to access each object and insert data in S3 and other information into Postgres. The issue is that I cannot parse this obj.
Object.keys(request.files).length // = 1, the length is clearly 2
request.files.length // = undefined
rqeuest.files.file1 // undefined
If I edit this json in a text editor to have ' ' around all words, I can iterate through the object fine. How can I work with this json Object?
It looks like request.files was already parsed from JSON and is a Javascript object.
If that is true, to get at the files property value, you need to use request.files.files.
Note: request.files.files.file1 would still not work, since request.files.files is an array.
I want to change many of parts of a unicode string in ollydbg.
Is there any plugin or trick that can be used to replace bulky of unicode string at once?
For example, if 100 unicode 'test' words exist in the string reference then I want to change all 'test' to 'test2'
00459FD0 5C 00 55 00 6C 00 74 00 72 00 61 00 4D 00 61 00 \test\.J.
00459FF0 5C 00 4D 00 69 00 63 00 72 00 6F 00 73 00 6F 00 \Microso
0045A000 66 00 74 00 20 00 53 00 68 00 61 00 72 00 65 00 ft Share
0045A010 64 00 5C 00 54 00 72 00 69 00 45 00 64 00 69 00 d\TriEdi
0045A020 74 00 5C 00 44 00 68 00 74 00 6D 00 6C 00 45 00 t\DhtmlE
0045A030 64 00 2E 00 6F 00 63 00 78 00 00 00 0C 00 00 00 d.ocx...
0045A040 3F 00 6C 00 61 00 6E 00 67 00 3D 00 00 00 00 00 ?lang=..
0045A050 22 00 00 00 5C 00 55 00 6C 00 74 00 72 00 61 00 ".\test.︁ൃᇏ
0045A080 89 40 00 A0 C9 05 42 28 10 9D 0B 2A 87 4B D3 11 䂉ꀀ⡂鴐⨋䮇ᇓ
Until now I was trying to find a trick or plugin but I have been unsuccessful.
How can I accomplish this task?
Thank you.
I struggled few hours to figure this, so here is the way I replaced all my strings with Olly (it was ascii, but no difference).
First we can't do hardcoding, except if the string are at the same size. So go find a free space in your executable, and paste it your string (you have to put a zero byte right after).
Keep your string offset in mind, and in your CPU view, search for all referenced text strings.
Then select the string you want to replace, press enter then space to open the assemble window, then replace your offset with the one of your new string.
Cheerz! :)
I have a binary file that I want to open as (or to convert to) jpeg image.
The file is only supposed by me to be a jpeg within a binary format, therefore I include a part of it so that You can tell me if I'm wrong ('cause unfortunately I never saw one of those before):
0000000000 0E 03 13 01 00 10 00 00 EC B0 00 1E 00 01 00 00 [................]
0000000016 00 CA 00 00 00 5C 00 0F 00 01 00 00 01 26 00 00 [.....\.......&..]
0000000032 00 00 01 2F 00 01 00 00 01 26 00 00 83 B2 00 6A [.../.....&.....j]
0000000048 00 01 00 00 84 D8 00 00 00 04 01 2C 00 01 00 00 [...........,....]
0000000064 84 DC 00 00 00 14 01 32 00 01 00 00 84 F0 00 00 [.......2........]
0000000080 00 08 07 AD 00 02 00 00 84 F8 00 00 00 19 00 0F [................]
0000000096 00 02 00 00 85 11 00 00 00 00 01 2F 00 02 00 00 [.........../....]
0000000112 85 11 00 00 0D D3 01 2C 00 02 00 00 92 E4 00 00 [.......,........]
0000000128 00 14 01 32 00 02 00 00 92 F8 00 00 00 08 07 AD [...2............]
0000000144 00 03 00 00 93 00 00 00 00 19 00 0F 00 03 00 00 [................]
0000000160 93 19 00 00 00 00 01 2F 00 03 00 00 93 19 00 00 [......./........]
0000000176 59 7B 01 2C 00 03 00 00 EC 94 00 00 00 14 01 32 [Y{.,...........2]
0000000192 00 03 00 00 EC A8 00 00 00 08 00 00 00 04 00 00 [................]
0000000208 00 02 00 00 00 00 4E 43 53 41 20 48 44 46 20 56 [......NCSA HDF V]
0000000224 65 72 73 69 6F 6E 20 34 2E 32 20 52 65 6C 65 61 [ersion 4.2 Relea]
0000000240 73 65 20 30 2C 20 44 65 63 65 6D 62 65 72 20 32 [se 0, December 2]
0000000256 2C 20 32 30 30 33 00 00 00 00 00 00 00 00 00 00 [, 2003..........]
0000000272 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 [................]
0000000288 00 00 00 00 00 00 FF D8 FF E0 00 10 4A 46 49 46 [............JFIF]
0000000304 00 01 01 00 00 01 00 01 00 00 FF DB 00 43 00 10 [.............C..]
0000000320 0B 0C 0E 0C 0A 10 0E 0D 0E 12 11 10 13 18 28 1A [..............(.]
0000000336 18 16 16 18 31 23 25 1D 28 3A 33 3D 3C 39 33 38 [....1#%.(:3=<938]
0000000352 37 40 48 5C 4E 40 44 57 45 37 38 50 6D 51 57 5F [7#H\N#DWE78PmQW_]
0000000368 62 67 68 67 3E 4D 71 79 70 64 78 5C 65 67 63 FF [bghg>Mqypdx\egc.]
0000000384 DB 00 43 01 11 12 12 18 15 18 2F 1A 1A 2F 63 42 [..C......./../cB]
0000000400 38 42 63 63 63 63 63 63 63 63 63 63 63 63 63 63 [8Bcccccccccccccc]
0000000416 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 [cccccccccccccccc]
0000000432 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 63 [cccccccccccccccc]
0000000448 63 63 63 63 FF C0 00 11 08 01 A0 01 C0 03 01 22 [cccc..........."]
0000000464 00 02 11 01 03 11 01 FF C4 00 1F 00 00 01 05 01 [................]
0000000480 01 01 01 01 01 00 00 00 00 00 00 00 00 01 02 03 [................]
0000000496 04 05 06 07 08 09 0A 0B FF C4 00 B5 10 00 02 01 [................]
How can I convert it to jpeg to view its content (that is supposed to be an image)?
Is this conversion necessary or would You reccomend a software that can open it as I like?
I hope someone could help me, thanks in advance.
ANSWER
(As I can't answer my own question right now, I'm doing it here...)
At the end I came up with a solution thanks to the input information of unwind.
I downloaded HEX editor (http://www.hhdsoftware.com/free-hex-editor) to edit my binary file. Then I searched for the string where the 0xff 0xd8 was (in may case in line 0000000288). This is supposed to be the beginning of a JPEG file. Then I deleted everything that came before of that (also the six pair of zeros within the same line). Then I saved m edits and tried again to open it with an image processing program (in my case, I'm usin ENVI), and....IT WORKS! Now the binary file is red as an image file!
The problem now is that I have plenty of those files (302), and I need to edit all of them. Moreover, they each contain more than one jpeg, so I need to modify each one ore times. Guess I need to improve my programming knowledge...
Well, according to this page JPEG files begin with the byte pair 0xff 0xd8, so you could search forwards for that sequence, and throw away the data before it.
In your file, it happens on the line starting 0000000288.
Suppose I wanted to write a program to display "hello world", and I wanted to write it in binary. How could I do this?
I have some idea that:
I'd need to determine what chip architecture I'm using
I'd need to find out what kind of binary it uses
I'd need some reference for that flavor of binary
I might need to change a setting in my editor (Vim)
Can anybody walk me through this?
It's bit more complicated, because actually printing "Hello, world!" to stdout is a system call, thus you need to know the correct kernel syscall number. Which of course varies by operating system. Also you need to know the binary format, which also tend to vary, although ELF (Executable and Linkable Format) is universal across few flavors of Unix and Linux.
See Hello, world! in assembler.
This is Linux assembler code:
section .text
global _start ;must be declared for linker (ld)
_start: ;tell linker entry point
mov edx,len ;message length
mov ecx,msg ;message to write
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg db 'Hello, world!',0xa ;our dear string
len equ $ - msg ;length of our dear string
... which on 32-bit Linux, compilation results in binary of 360 bytes, although it's mostly zeros:
00000000 7f 45 4c 46 01 01 01 00 00 00 00 00 00 00 00 00 |.ELF............|
00000010 02 00 03 00 01 00 00 00 80 80 04 08 34 00 00 00 |............4...|
00000020 c8 00 00 00 00 00 00 00 34 00 20 00 02 00 28 00 |........4. ...(.|
00000030 04 00 03 00 01 00 00 00 00 00 00 00 00 80 04 08 |................|
00000040 00 80 04 08 9d 00 00 00 9d 00 00 00 05 00 00 00 |................|
00000050 00 10 00 00 01 00 00 00 a0 00 00 00 a0 90 04 08 |................|
00000060 a0 90 04 08 0e 00 00 00 0e 00 00 00 06 00 00 00 |................|
00000070 00 10 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................|
00000080 ba 0e 00 00 00 b9 a0 90 04 08 bb 01 00 00 00 b8 |................|
00000090 04 00 00 00 cd 80 b8 01 00 00 00 cd 80 00 00 00 |................|
000000a0 48 65 6c 6c 6f 2c 20 77 6f 72 6c 64 21 0a 00 2e |Hello, world!...|
000000b0 73 68 73 74 72 74 61 62 00 2e 74 65 78 74 00 2e |shstrtab..text..|
000000c0 64 61 74 61 00 00 00 00 00 00 00 00 00 00 00 00 |data............|
000000d0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................|
*
000000f0 0b 00 00 00 01 00 00 00 06 00 00 00 80 80 04 08 |................|
00000100 80 00 00 00 1d 00 00 00 00 00 00 00 00 00 00 00 |................|
00000110 10 00 00 00 00 00 00 00 11 00 00 00 01 00 00 00 |................|
00000120 03 00 00 00 a0 90 04 08 a0 00 00 00 0e 00 00 00 |................|
00000130 00 00 00 00 00 00 00 00 04 00 00 00 00 00 00 00 |................|
00000140 01 00 00 00 03 00 00 00 00 00 00 00 00 00 00 00 |................|
00000150 ae 00 00 00 17 00 00 00 00 00 00 00 00 00 00 00 |................|
00000160 01 00 00 00 00 00 00 00 |........|
Since you want to "compile by hand", this basically means translating assembler mnemonics above to their opcodes, and then wrapping the result in correct binary format (ELF in the example above)
UPDATE: As this answer shows by #adam-rosenfield, the ELF binary for "Hello, world!" can be handcrafted down to 116 bytes. Original answer is now deleted, but still visible to moderators, so here's a copy:
Here's a 32-byte version using Linux system calls:
.globl _start
_start:
movb $4, %al
xor %ebx, %ebx
inc %ebx
movl $hello, %ecx
xor %edx, %edx
movb $11, %dl
int $0x80 ;;; sys_write(1, $hello, 11)
xor %eax, %eax
inc %eax
int $0x80 ;;; sys_exit(something) hello:
.ascii "Hello world"
When compiled into a minimal ELF file, the full executable is 116
bytes:
00000000 7f 45 4c 46 01 01 01 00 00 00 00 00 00 00 00 00 |.ELF............|
00000010 02 00 03 00 01 00 00 00 54 80 04 08 34 00 00 00 |........T...4...|
00000020 00 00 00 00 00 00 00 00 34 00 20 00 01 00 00 00 |........4. .....|
00000030 00 00 00 00 01 00 00 00 00 00 00 00 00 80 04 08 |................|
00000040 00 80 04 08 74 00 00 00 74 00 00 00 05 00 00 00 |....t...t.......|
00000050 00 10 00 00 b0 04 31 db 43 b9 69 80 04 08 31 d2 |......1.C.i...1.|
00000060 b2 0b cd 80 31 c0 40 cd 80 48 65 6c 6c 6f 20 77 |....1.#..Hello w|
00000070 6f 72 6c 64 |orld|
00000074
Normally, you'd use a hex editor for this. Figure out the assembly code, hand-assemble it, use the hex editor to enter the binary values, then save them to a file. Once you have your file, drop into your machine monitor and load the file at an available address, then jump to the first instruction. This was pretty common practice on single-board computers and is still done on microcontrollers today, but it's not something you're going to do on a contemporary OS. If you really want to do this, I'd recommend running a low-level emulator (SIMH will work) or working with a microcontroller (you can pick up a TI MSP430 development kit for less than five bucks).