blob: 3cb279a747e416cb38d522d8c61e62751c5d6e7f (
plain) (
blame)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
|
/*
Quick and dirty program to make intel-hex from a binary.
Written by R.E.Wolff@BitWizard.nl
This file is in the public domain
Typing started:
Mon Jun 16 00:24:15 MET DST 1997
programming stopped:
Mon Jun 16 00:31:27 MET DST 1997
debugging finished (2 bugs found):
Mon Jun 16 00:32:52 MET DST 1997
---------------------------------------------------------
Doc written in timeout. Everything else in this file was done while
the timer was running.
I promised "Mark Kopecki" that writing the bin-to-intel-hex
converter would cost less than 15 minutes, and that it would be more
trouble to find a converter on the net than to write the converter
myself. I ended up spending over half an hour searching for
spec/converter/docs because of unreachable hosts on the internet. I
got a file with docs, after that it was 8 minutes.....
---------------------------------------------------------
*/
#include <stdio.h>
#include <unistd.h>
/* Intel Hex format:
ll aaaa tt dd....dd cc
ll = length
aaaa = address
tt = type
dd....dd = data
cc = checksum.
*/
int main (int argc, char **argv)
{
unsigned char buf[32];
int addr = 0;
int n,i;
while ((n = read (0, buf+4, 16)) > 0) {
buf[0] = n;
buf[1] = addr >> 8;
buf[2] = addr & 0xff;
buf[3] = 0x00;
buf[4+n] = 0x00;
for (i=0;i<4+n;i++)
buf[4+n] -= buf[i];
printf (":");
for (i=0;i<= 4+n;i++)
printf ("%02x", buf[i]);
printf ("\n");
addr += n;
}
printf (":0000000001ff\n");
exit (0);
}
|
