std_text_routines.spinh

PUB str(s = string("")) | c
  REPEAT WHILE ((c := byte[s++]) <> 0)
    tx(c)


''
'' print an number with a given base
'' we do this by finding the remainder repeatedly
'' this gives us the digits in reverse order
'' so we store them in a buffer; the worst case
'' buffer size needed is 32 (for base 2)
''
''
'' signflag indicates how to handle the sign of the
'' number:
''   0 == treat number as unsigned
''   1 == print nothing before positive numbers
''   anything else: print before positive numbers
'' for signed negative numbers we always print a "-"
''
'' we will print at least prec digits
''
VAR
  byte buf[32]

PUB Num(val, base, signflag, digitsNeeded) | i, digit, r1, q1

  '' if signflag is nonzero, it indicates we should treat
  '' val as signed; if it is > 1, it is a character we should
  '' print for positive numbers (typically "+")
  
  if (signflag)
      if (val < 0)
        signflag := "-"
	val := -val
	
  '' make sure we will not overflow our buffer
  if (digitsNeeded > 32)
    digitsNeeded := 32

  '' accumulate the digits
  i := 0
  repeat
    if (val < 0)
      ' synthesize unsigned division from signed
      ' basically shift val right by 2 to make it positive
      ' then adjust the result afterwards by the bit we
      ' shifted out
      r1 := val&1  ' capture low bit
      q1 := val>>1 ' divide val by 2
      digit := r1 + 2*(q1 // base)
      val := 2*(q1 / base)
      if (digit => base)
        val++
	digit -= base
    else
      digit := val // base
      val := val / base

    if (digit => 0 and digit =< 9)
       digit += "0"
    else
       digit := (digit - 10) + "A"
    buf[i++] := digit
    --digitsNeeded
  while (val <> 0 or digitsNeeded > 0) and (i < 32)
  if (signflag > 1)
    tx(signflag)
    
  '' now print the digits in reverse order
  repeat while (i > 0)
    tx(buf[--i])

'' print a signed decimal number
PUB dec(val)
  num(val, 10, 1, 0)

'' print an unsigned decimal number with the specified
'' number of digits; 0 means just use as many as we need
PUB decuns(val, digits)
  num(val, 10, 0, digits)

'' print a hex number with the specified number
'' of digits; 0 means just use as many as we need
PUB hex(val, digits) | mask
  num(val, 16, 0, digits)

'' print a newline
PUB nl()
  tx(13)
  tx(10)