How to map 6 bytes of raw data to long long type?

Question

I'm trying to solve some problems for a contest, and I have a problem with it. I have to implement a function that has 3 arguments like this:

void foo(unsigned char* A, unsigned char* B, unsigned char* C);

And what comes in, is 6 bytes of data. I have to convert these 6 bytes of data to integer data (maybe long long). The problem is, I don't know how to map the bytes in the right order.

I'm using an Arduino Uno board.

Please help me with it.

• chrisl: What is the right order of the six bytes? And how are they contained in the char arrays from the functions parameters?
• qqq1ppp: @chrisl 0x123412341234 to {0x12,0x34,0x12,0x34,0x12,0x34}
• chrisl: I'm not quite sure, what exactly you want to do. As I understand you get 3 char arrays with each 2 bytes in the correct order as parameters into your function, and then you want to declare a long long and fill the bytes from the arrays into this variable. Is this correct?
• qqq1ppp: @chrisl nope. each of the parameters inputs 6 byte of data. and I want to convert it like what I said/

Answer 1

The trick is to use a union data type and get full control of the mapping. The issue of data representation and endian is now under your control:

uint64_t foo(uint8_t* A, uint8_t* B, uint8_t* C)
{
 union {
 uint64_t X;
 struct {
 uint8_t A[2];
 uint8_t B[2];
 uint8_t C[2];
 uint8_t D[2];
 };
 } map;
#if defined(MSB_ORDER)
 for (int i = 0, j = 1; i < 2; i++, j--) {
 map.A[i] = 0;
 map.B[i] = C[j];
 map.C[i] = B[j];
 map.D[i] = A[j];
 }
#else
 for (int i = 0; i < 2; i++) {
 map.A[i] = A[i];
 map.B[i] = B[i];
 map.C[i] = C[i];
 map.D[i] = 0;
 }
#endif
 return map.X;
}
void setup()
{
 Serial.begin(9600);
 while (!Serial);
 uint8_t a[] = { 0x12, 0x34 };
 uint8_t b[] = { 0x56, 0x78 };
 uint8_t c[] = { 0x9a, 0xbc };
 uint64_t res = foo(a, b, c);
 Serial.print((uint32_t) (res >> 32), HEX);
 Serial.print((uint32_t) res, HEX);
}
void loop()
{
}

Cheers!

Answer 2

uint64_t foo(uint8_t* A, uint8_t* B, uint8_t* C)
{
 uint64_t aux = 0;
 aux = A[0];
 aux <<= 8;
 aux |= A[1];
 aux <<= 8;
 aux |= B[0];
 aux <<= 8;
 aux |= B[1];
 aux <<= 8;
 aux |= C[0];
 aux <<= 8;
 aux |= C[1];
 return aux;
}

following sample code:

 uint8_t a[] = { 0x12, 0x34 };
 uint8_t b[] = { 0x56, 0x78 };
 uint8_t c[] = { 0x9a, 0xbc };
 uint64_t result = foo(a, b, c);
 std::cout << std::hex << result;

returns 123456789abc, try the code here, excluding the std::out part, the same code will run on arduino without any problem

Answer 3

When passing data to functions you do this...

foo(a,b,c);

You then extract the variables or data passed to the function like..

void foo(unsigned char* A, unsigned char* B, unsigned char* C) {
 A = whatever;
 B = whatever;
 C = whatever;
}

• qqq1ppp: wow. I'm not that noob ;; I wanna map the raw 6 byte data to "long long" type variable. memcpy won't work as I want, cause of the endian rules..
• Brian Moreau: Well your function was written wrong. So is each VAR or CHAR 2 bytes then?