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Use err(3). Rewrote man page in mdoc format.

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This commit is contained in:
Philippe Charnier 1997-07-16 06:51:12 +00:00
parent 15fc002be8
commit ca22ec7d08
Notes: svn2git 2020-12-20 02:59:44 +00:00 
svn path=/head/; revision=27444
2 changed files with 46 additions and 56 deletions
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59
usr.bin/key/key.1
View File

@ -1,24 +1,26 @@
.ll 6i
.pl 10.5i
.\" @(#)key.1 1.0 (Bellcore) 12/2/91
.\" from: @(#)key.1 1.0 (Bellcore) 12/2/91
.\" $Id$
.\"
.lt 6.0i
.TH KEY 1 "2 December 1991"
.AT 3
.SH NAME
key \- Stand\-alone program for computing responses to S/Key challenges.
.SH SYNOPSIS
.B key [\-n <count>] <Sequence> <key>
.SH DESCRIPTION
.I key
Takes the optional count of the number of one time access
.Dd December 2, 1991
.Dt KEY 1
.Os
.Sh NAME
.Nm key
.Nd stand\-alone program for computing responses to S/Key challenges
.Sh SYNOPSIS
.Nm
.Op Fl n Ar count
.Ar sequence
.Ar key
.Sh DESCRIPTION
.Nm Key
takes the optional count of the number of one time access
passwords to print
along with a (maximum) sequence number and key as command line args,
it prompts for the user's secret password, and produces both word
and hex format responses.
.SH EXAMPLE
.sh
Usage example:
.Sh EXAMPLE
Usage example:
.sp 0
>key \-n 5 99 th91334
.sp 0
@ -29,21 +31,18 @@ and hex format responses.
.... 4 more passwords.
.sp 0
>
.LP
.SH OPTIONS
.LP
.B \-n <count>
the number of one time access passwords to print.
.Sh OPTIONS
.Bl -tag -width indent
.It Fl n Ar count
The number of one time access passwords to print.
The default is one.
.SH DIAGNOSTICS
.SH BUGS
.LP
.SH SEE ALSO
.BR skey(1),
.BR keyinit(1),
.El
.Sh SEE ALSO
.Xr keyinfo 1 ,
.Xr keyinit 1 ,
.Xr skey 1
.\" .BR keysu(1),
.BR keyinfo(1)
.SH AUTHOR
.Sh AUTHOR
Command by Phil Karn, Neil M. Haller, John S. Walden
.SH CONTACT
.Sh CONTACT
staff@thumper.bellcore.com

43
usr.bin/key/skey.c
View File

@ -9,9 +9,12 @@
* C848 666B 6435 0A93
* >
*/
#include <err.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#ifdef __MSDOS__
#include <dos.h>
@ -21,11 +24,7 @@
#include <skey.h>
char *readpass();
void usage();
int getopt();
extern int optind;
extern char *optarg;
static void usage __P((void));
int
main(argc,argv)
@ -33,7 +32,7 @@ int argc;
char *argv[];
{
int n,cnt,i;
char passwd[256],passwd2[256];
char passwd[256] /* ,passwd2[256] */;
char key[8];
char *seed;
char buf[33];
@ -50,31 +49,25 @@ char *argv[];
/* could be in the form <number>/<seed> */
if(argc <= optind + 1){
/*look for / in it */
if(argc <= optind){
usage(argv[0]);
return 1;
}
if(argc <= optind)
usage();
slash = strchr(argv[optind], '/');
if(slash == NULL){
usage(argv[0]);
return 1;
}
if(slash == NULL)
usage();
*slash++ = '\0';
seed = slash;
if((n = atoi(argv[optind])) < 0){
fprintf(stderr,"%s not positive\n",argv[optind]);
usage(argv[0]);
return 1;
usage();
}
}
else {
if((n = atoi(argv[optind])) < 0){
fprintf(stderr,"%s not positive\n",argv[optind]);
usage(argv[0]);
return 1;
usage();
}
seed = argv[++optind];
}
@ -95,10 +88,8 @@ char *argv[];
}
/* Crunch seed and password into starting key */
if(keycrunch(key,seed,passwd) != 0){
fprintf(stderr,"%s: key crunch failed\n",argv[0]);
return 1;
}
if(keycrunch(key,seed,passwd) != 0)
errx(1, "key crunch failed");
if(cnt == 1){
while(n-- != 0)
f(key);
@ -121,9 +112,9 @@ char *argv[];
return 0;
}
void
usage(s)
char *s;
static void
usage()
{
fprintf(stderr,"Usage: %s [-n count] <sequence #>[/] <key> \n",s);
fprintf(stderr,"usage: key [-n count] <sequence #>[/] <key>\n");
exit(1);
}