[General] CRC function in Modbus protocol

Hello,

I trying to make program for building and sending Modbus RTU packet. Program contain a function for CRC. Results of CRC must be two bytes placed in the end of packet. This is what I have yet, there is function for CRC and function send_packet(), which I call from main. But there is some error during compilation. I don´t know how to forward my packet[] array to CRC function. I don´t understand pointers very well.

Please help.

You forgot to mark the error line.

I guess you have an error in this line

You should try

Pointer is a just a phisical address of first element in array.
buf[pos] is equal to *but+pos

In the lines


and


is still error "expected an expression".

uplifting_mind said:

Hello,

I trying to make program for building and sending Modbus RTU packet. Program contain a function for CRC. Results of CRC must be two bytes placed in the end of packet. This is what I have yet, there is function for CRC and function send_packet(), which I call from main. But there is some error during compilation. I don´t know how to forward my packet[] array to CRC function. I don´t understand pointers very well.

Please help.

Code:

uint8_t packet[5];
int i;
int len;
uint16_t crc;
uint8_t crc_LByte;
uint8_t crc_HByte;
uint8_t *buf;
uint8_t pos;

// Compute the MODBUS RTU CRC
uint16_t ModRTU_CRC(uint8_t *buf, int len)
{
  uint16_t crc = 0xFFFF;
 
  for (int pos = 0; pos < len; pos++) {
    crc ^= (uint16_t)*buf[pos];          // XOR byte into least sig. byte of crc
 
    for (int i = 8; i != 0; i--) {    // Loop over each bit
      if ((crc & 0x0001) != 0) {      // If the LSB is set
        crc >>= 1;                    // Shift right and XOR 0xA001
        crc ^= 0xA001;
      }
      else                            // Else LSB is not set
        crc >>= 1;                    // Just shift right
    }
  }
  // Note, this number has low and high bytes swapped, so use it accordingly (or swap bytes)
  return crc;  
}

void send_packet(uint8_t address, uint8_t function, uint8_t lo_data, uint8_t hi_data) {

packet[0] = address;  // build packet
packet[1] = function;
packet[2] = lo_data;
packet[3] = hi_data;
	
    crc = ModRTU_CRC(packet, 4);
    crc_LByte = (crc & 0x00FF); // Low byte calculation
    crc_HByte = (crc & 0xFF00) / 256; // High byte calculation 

packet[4] = crc_LByte;
packet[5] = crc_HByte;
	
    for (i = 0; i < 6; i++) {
    printf("%d %02x\n", i, (int) packet[i]);    // send packet
    } 
}

Click to expand...

unsigned int modbuscrc(unsigned char *buf,int len)
{
unsigned int crc = 0xffff;
int pos;
for (pos = 0; pos < len; pos++)
{
crc ^= (unsigned int)buf[pos]; // XOR byte into least sig. byte of crc

for (int i = 8; i>0; i--) { // Loop over each bit
if ((crc & 0x0001) != 0) { // If the LSB is set
crc >>= 1; // Shift right and XOR 0xA001
crc ^= 0xA001;
}
else // Else LSB is not set
crc >>= 1; // Just shift right
}
}


return crc;
}


it is compiling like this.

- - - Updated - - -

I have a question also. maybe it will be a stupid question



void main(void)
{
setISR();
setRS();
buffer[0]=0x01; //slave address
buffer[1]=0b00000011; //function code
buffer[2]=0x00; //address low if i use 0x01 it is transmitting all data. if 0x00 it is not why? this is my qustion
buffer[3]=0x01; //address high
buffer[4]=0x02; //just for try crc generation changes
buffer[5]=0x02; //just for try crc generation changes
crc16(buffer,6); // creats crc and add it to the buffer 6 and 7.
__delay_ms(600);
writemdframe(buffer);
while (1)
{
}
}


void writemdframe(unsigned char *str)
{
while((*str)!=0)
{
writeRS(*str);
str++;
}
}
void writeRS(unsigned char c)
{
TXREG = c;
TXEN = 1;
while (!TRMT);
}

In the lines


Code:
for (int pos = 0; pos < len; pos++) {and


Code:
for (int i = 8; i != 0; i--) {is still error "expected an expression".

Click to expand...

I presume, the compiler doesn't support the c++ variable definition style inside for( ).

FvM said:

I presume, the compiler doesn't support the c++ variable definition style inside for( ).

Click to expand...

It´s work already. Thank you.