HEX Multiplication

Using Arduino Programming Questions
#1

Hi

I want to know why when I Serial print below code I am not getting the whole result:

EX: 0x9e3779b9*32 = 0x13C6EF3720 but I am only getting 0xC6EF3720 which is missing the 9bit

uint32_t sum = 0x9e3779b9*32;
Serial.println(sum ,HEX);

Thank you

#2

The multiplication results exceeds the the highest number that can be represented by a 32-bit integer.

Also,

uint32_t sum = 0x9e3779b9*32;

The multiplication is only being done with 32-bit arithmetic.

#3 
void setup() 
{
  Serial.begin(9600);
  union
  {
    uint64_t x;
    uint16_t myData[4];
  } data;

  data.x = (uint64_t) 0x9e3779b9 * 32; //0x13C6EF3720
  Serial.print(data.myData[2], HEX); 
  Serial.print(data.myData[1], HEX); 
  Serial.println(data.myData[0], HEX);
}

void loop() 
{
  
}
#4

This:

 Serial.print(data.myData[2], HEX);
  Serial.print(data.myData[1], HEX);
  Serial.println(data.myData[0], HEX);

won't print leading 0s for data.myData[1] and data.myData[0] so it probably won't look right.

#5

The Serial Monitor shows:
smar.png

smar.png

#6

Try different numbers that have leading zeros in data.myData[1] and data.myData[0] meaning the 16 bit value is less than 0x1000. It's not a very good solution if it only works for some numbers.

#7

How about this one?
smat.png

void setup() 
{
  Serial.begin(9600);
  union
  {
    uint64_t x;
    uint8_t myData[8];
  } data;

  data.x = (uint64_t) 0x9e3779b9 * 32; //0x13C6EF3720
  for(int i=7; i>=0; i--)
  {
    byte x = data.myData[i];
    if(x <0x10)
    {
      Serial.print('0');
    }
    Serial.print(x, HEX);
  }
  Serial.println();
}

void loop() 
{
  
}

smat.png

#8

Did you test it with numbers as specified below? Did it work? If so, then OK.

gfvalvo:
Try different numbers that have leading zeros in data.myData[1] and data.myData[0] meaning the 16 bit value is less than 0x1000. It's not a very good solution if it only works for some numbers.

#9

The sketch of Post#6 is now checking leading 0 of every byte of the result. Therefore, the print() method would show the leading 0s.

Tested for:
0x0102*0x0102 = 0x10404
smny.png

How to discard all those meaningless and odd looking 0s that are appearing at the very left of the result?

smny.png

#10

Don't print anything until you find the first non-zero byte. Then print all bytes (with added leading zeros as necessary).

#11

Thanks for all the replay, The reason I was asking this was that I am using below code to encrypt message using xtea(below code)

but I do not think the encryption is correct because the multiplication result was wrong.

void encipher(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]) {
  unsigned int i;
  uint32_t v0 = v[0], v1 = v[1], sum = 0, delta = 0x9E3779B9;
  for (i = 0; i < num_rounds; i++) {
    v0 += (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);
         
              sum += delta;

    v1 += (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);
   
  }

  v[0] = v0; v[1] = v1;

}

as you can see sum is equal numround which 32 times by delta

#12

This might be a case of misdiagnosing a problem (assuming there is a problem in the first place). It's pretty common for bit-twiddling algorithms to discard the upper bits of arithmetic operations, since they're not needed.

What leads you to believe that that encryption function is not working correctly?

#13

There’s another active thread by the OP on this subject.

#14

Can you provide a link for that thread

I know is wrong since when I encrypt it different language like php, Java it different.

Thank you

#15