问题
In a serial communication, I need to send that ciphertext from python to UART which is able to read it by using C.
At first, In the python side use this technique:
ciphertext=(b'\x24\x70\xb4\xc5\x5a\xd8\xcd\xb7\x80\x6a\x7b\x00\x30\x69\xc4\xe0\xd8')
ser.write(ciphertext)
In the C side, I put the received ciphertext in a buffer. I test If the first byte 24, the Start of the packet:
if (buffer_RX[0]=='\x24')
{
for (i=1;i<=17;i++) //sizeof(buffer_RX)==17
{
printf("%x \n",buffer_RX[i]);
}
}
else {
printf("It is not a packet!");
}
The result that I have is:
70
b4
c5
5a
d8
cd
b7
80
6a
7b
0
30
69
c4
e0
d8
So I have two questions,
- The fist is why \x00
is displayed by that way, I mean just one zero 0?
- The second is how could I have this same representation:
const unsigned int ciphertext[4] = {0x70b4c55a, 0xd8cdb780, 0x6a7b0030,0x69c4e0d8};
If I suppose that is concatenation between bytes, I mean that is maybe:
unsigned int plaintext[1];
plaintext[1]= buffer_RX[3]|buffer_RX[4]|buffer_RX[5];
But I am not sure, I think that I have a syntax error.
This representation will help me to continue next steps. But I don't know how could I have it please.
回答1:
If you want to create a 4 bytes long data from your 1 byte chunks.
const unsigned int ciphertext[4] = {0x70b4c55a, 0xd8cdb780, 0x6a7b0030,0x69c4e0d8};
It is not enough to OR the bytes, you have to shift them into the right position.
What you are currently doing, (note that you are over indexing plaintext[1]
this should be plaintext[0]
):
unsigned int plaintext[1];
plaintext[1]= buffer_RX[3]|buffer_RX[4]|buffer_RX[5];
has this result in the memory, assuming int
is 32 bit, 4 byte long:
--------------------------------------------------------------------------------
| 3. byte | 2. byte | 1. byte | 0. byte |
--------------------------------------------------------------------------------
| 0x00 | 0x00 | 0x00 | buffer_RX[3] | buffer_RX[4] | buffer_RX[5] |
--------------------------------------------------------------------------------
so you have to shift the appropriate bytes to the right position with the left-shift operator first <<
, and OR the bytes after that.
Here is an example:
#include <stdio.h>
int main()
{
unsigned char buffer_RX[4];
buffer_RX[0] = 0x70;
buffer_RX[1] = 0xb4;
buffer_RX[2] = 0xc5;
buffer_RX[3] = 0x5a;
unsigned int plaintext[1];
plaintext[0] = (buffer_RX[0]<<24) | (buffer_RX[1]<<16) | (buffer_RX[2]<<8) | buffer_RX[3];
printf("plaintext: %08x\n", plaintext[0]);
return 0;
}
The output:
In memory:
----------------------------------------------------------------------
| 3. byte | 2. byte | 1. byte | 0. byte |
----------------------------------------------------------------------
| buffer_RX[0] | buffer_RX[1] | buffer_RX[2] | buffer_RX[3] |
----------------------------------------------------------------------
You can see that buffer_RX[0]
has been shifted by 24 which is 3 byte so it skips the first three cell, jumps to the last one.
buffer_RX[1]
by 16 which is 2 byte, so it skips to first two, buffer_RX[2]
by 8 which is 1 byte, so it skips the first. And buffer_RX[3]
was not shifted because it goes to the first place.
Regarding the 0x00
, it is zero represented on 8 bits. If your print it, that will be simply 0. If you print 0x0000
it will be 0 as well. It just how printf
prints 0 by default, it does not prints zeros unnecessarly. If you have 0x70
in a 32 bit variable then it is actually 0x00000070
but printf
will print 0x70
, because unnecessary zeros are chopped off, unless you tell otherwise. That is where %02x
comes in. In the %02x
the 02
tells printf
that you want to display 2 bytes no matter what so it will print two zeros in case of 0x00
printf("%02x \n",buffer_RX[i]);
If you want to print the whole 32 bit (4 byte) of 0x00000070
the following line will do that:
printf("%08x \n",buffer_RX[i]);
来源:https://stackoverflow.com/questions/40831263/how-to-have-the-correct-representation-of-words-in-cpythonuart