#!/bin/bash
# Code supplied by Donald W. Cragun in response to post on www.unix.com:
# https://www.unix.com/shell-programming-and-scripting/250310-replace-special-character-next-present-byte-4.html
/usr/xpg4/bin/awk -F 'ú' '
BEGIN {	x="0123456789ABCDEF" }
#{	printf(" Input:%s\nOutput:", $0) }
{	printf("%s", $1)
	for(i = 2; i <= NF; i++) {
		# Print 3 spaces.
		printf("   ")

		# Get repetition count.  (This seems likely to be error prone,
		# Dnd therefore dangerous, but this is what was specified by
		# Dineshkumar Nakka.
		if((cnt = substr($i, 1, 2)) ~ "[0-9][0-9]") {
			# Use decimal if both characters are decimal digits.
			base = cnt
			cnt = cnt + 0
		} else if (cnt ~ "[0-9A-Fa-f][[0-9A-Fa-f]") {
			# Use hexadecimal.
			base = "0x" cnt
			cnt = (index(x, toupper(substr(cnt, 1, 1))) - 1) * 16 +\
				index(x, toupper(substr(cnt, 2))) - 1
		} else {
			# Bad input.
			printf("\n\n\n*** Invalid count \"%s\" on line %d\n\n",
				cnt, NR)
			ec = 1
		}

		# Print the character to be repeated cnt times.
		chr = substr($i, 3, 1)
#		printf("|cnt=%d(from %s),chr=%s|", cnt, base, chr)
		for(j = 1; j <= cnt; j++)
			printf("%s", chr)

		# Print the remainder of the field.
		printf("%s", substr($i, 4))
	}
	# Add trailing <newline>.
	print ""
}
END {	exit ec }' sample.dat > output.dat

302619ú1A? 
ABCD172 2 B10001ú0E Fú08 DineshGú14 KumarNakkaú0E ú3C?INú14?EFGH340
ABC196 7 MIB513AMMODú07 MIByx66mcp00ú06 302619ú1A 00005014072605331600ú0A 980ú32 201407260533160ú14 2TRG212 7 98ú09 3P PAUTO FMGBN  MIB513AMMOAMERICAN MODER
N INSURANCE302619ú40 97871407259787    140724 NBSú19 A2014072520140724

302619   ?????????????????????????? 
ABCD172 2 B10001                 F           DineshG                 KumarNakka                    ????????????????????????????????????????????????????????????IN   ??????????????EFGH340
ABC196 7 MIB513AMMOD          MIByx66mcp00         302619                             00005014072605331600             980                                   201407260533160                 2TRG212 7 98            3P PAUTO FMGBN  MIB513AMMOAMERICAN MODER
N INSURANCE302619                                           97871407259787    140724 NBS                      A2014072520140724

 Input:302619ú1A? 
Output:302619   |cnt=26(from 0x1A),chr=?|?????????????????????????? 
 Input:ABCD172 2 B10001ú0E Fú08 DineshGú14 KumarNakkaú0E ú3C?INú14?EFGH340
Output:ABCD172 2 B10001   |cnt=14(from 0x0E),chr= |              F   |cnt=8(from 08),chr= |        DineshG   |cnt=14(from 14),chr= |              KumarNakka   |cnt=14(from 0x0E),chr= |                 |cnt=60(from 0x3C),chr=?|????????????????????????????????????????????????????????????IN   |cnt=14(from 14),chr=?|??????????????EFGH340
 Input:ABC196 7 MIB513AMMODú07 MIByx66mcp00ú06 302619ú1A 00005014072605331600ú0A 980ú32 201407260533160ú14 2TRG212 7 98ú09 3P PAUTO FMGBN  MIB513AMMOAMERICAN MODER
Output:ABC196 7 MIB513AMMOD   |cnt=7(from 07),chr= |       MIByx66mcp00   |cnt=6(from 06),chr= |      302619   |cnt=26(from 0x1A),chr= |                          00005014072605331600   |cnt=10(from 0x0A),chr= |          980   |cnt=32(from 32),chr= |                                201407260533160   |cnt=14(from 14),chr= |              2TRG212 7 98   |cnt=9(from 09),chr= |         3P PAUTO FMGBN  MIB513AMMOAMERICAN MODER
 Input:N INSURANCE302619ú40 97871407259787    140724 NBSú19 A2014072520140724
Output:N INSURANCE302619   |cnt=40(from 40),chr= |                                        97871407259787    140724 NBS   |cnt=19(from 19),chr= |                   A2014072520140724