CCS function for Integer-to-string conversion for UART output

Thanks, I saw that code when reviewing various related posts.  But it is to expert, too hard for me to understand.  Don't know why the long string is there.  Anyway, I went ahead and wrote my own, as you suggested, using my beginner-type coding, and it works good.  Maybe this will be useful for someone in the early stages like myself.  Hey, I finally contributed something to the Forum!

int string_to_int (char *std)  //std is a pointer to a string
{
 int aa;     //length of string excluding trailing \0
 int bb;     //intermediate value
 int cc = 0;    //intermediate addend
 int ee;     //10s value for multiplier
 aa = strlen(std);  //returns
 for (i = 0; i < aa; i++)
 {
  switch (i)        //assumes string is 4 or less digits
  {
  case 3: {ee = 1000; break;}
  case 2: {ee = 100; break;}
  case 1: {ee = 10; break;}
  case 0: {ee = 1; break;}
  }
  bb = transint(std[aa-1-i]) * ee;
  cc = cc + bb;
 }
 return cc;
}

int transint(char c)
{
 switch (c)
 {
 case '0': return 0;
 case '1': return 1;
 case '2': return 2;
 case '3': return 3;
 case '4': return 4;
 case '5': return 5;
 case '6': return 6;
 case '7': return 7;
 case '8': return 8;
 case '9': return 9;
 case '\r': return 10;
 default: return -1;
 }
}

The long string in that itoa() implementation is a lookup table the maps an integer digit to an ASCII character. It is one of more unusual implementations because that table handles digits from -35 to 35. That implementation is attributed to Lukas Chmela.

Your string_to_int() is essentially the same as the standard atoi() function. Going from string to integer is easier to code than integer to string. Here is an implementation of atoi. I have tried avoid pointers and the more compact C language constructs. It is limited to positive numbers. A full atoi() implementation would handle the '-' character.

int atoi(const char *s)
{
  int a; // Accumulator
  char c; // Current character
  int  d; // Current digit
  int i; // Array index

  i = 0; // Init array index
  a = 0; // Init accumulator to zero.
  c = s[i]; // Get first character
  i++; //Advance the index

  // Convert while not null terminator and is a numeric character
  while((c!=0)&&(c>='0')&&(c<='9'))
  {
    d = c - '0'; // Convert character to digit
    a = a * 10;  // Make room for digit
    a = a + d; // Add digit to accumulator
    c = s[i]; // Get next character
    i++; //Advance the index
  }
  return(a);
}

There are a lot of itoa() implementations out there.

http://www.jb.man.ac.uk/~slowe/cpp/itoa.html

The simple and straight forward implementations of itoa() use division and multiplication that on the MSP430 can be very slow. The repeated subtraction method is faster but the code can be difficult to understand.

Yes thank you Norman, I studied your code and it is very understandable.  The subtraction of '0'  to yield the integer value is now obvious to me, having reviewed the ascii table of character representations.  Also, your shifting of the accumulator is much more efficient than the 'switch' statement I used.  Finally, with these hints, I was able to understand the code with the lookup table.