Thread: Numbers issue...

Hello all,

Basically, I've been scouring google and all the obscure sites therein.... I can't understand why this is happening. See, the below code code out and nabs values through snmp. I'm sorry the code isn't pretty, but I am not (yet?) a very good c programmer. So, after it has those values, it calls some sytem events that don't really matter. (It's rrdtool, if you'd like to know more, pm me. Way cool stuff.) Anyway, my problem is that when I get the values eth0in,eth0out, and loout, I have them as strings. And I can't seem to convert them to anything else, and I need to do math things with them. I've tried integer, long int, double, float, etc, using their respective conversion functions. But the number I get out is never the same. I'm thinking that maybe there's some funky memory values floating around near the string?

I really don't know. Again, sorry that this code isn't pretty.

UPDATED: alright, basically when one runs this code, they see the string version and the long int version of the numbers. You will see if you run it... but If you don't this is what the result is:

Code:

eth0in    625163572    625163572
eth0out 17025541       17025541
lo           3045189694   2147483647 //<<<WHAT IS THAT???

     >>>eth0 in 0 bps
     >>>eth0 out 0 bps
     >>>lo 0 bps

Notice how in the lo section, the string and the long int values are different. WHAT have I done? I need an exorcist!

Here is the code:

Code:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<strings.h>
#include<time.h>

char *rip (char *s)
{
   char *p;

   if ((p = strrchr(s,'\n')) != NULL) *p = '\0';
   if ((p = strrchr(s,'\r')) != NULL) *p = '\0';
   return (s);
}

int main()
{
   FILE *f;
   char buf[100];
   char tmp[16];
   char cmd[100];
   char host[50],commstr[30];
   long int eth0in,eth0out,lo,itmp,etime;
   long int peth0in,peth0out,plo,petime;//previous values
   short int first=0;

   strncpy(host,"localhost",sizeof(host));
   strncpy(commstr,"public",sizeof(commstr));

   sprintf(cmd,"snmpwalk %s %s interfaces.ifTable.ifEntry.ifInOctets.3 >/tmp/sysgrapheth0in",host,commstr);
   system(cmd);
   sprintf(cmd,"snmpwalk %s %s interfaces.ifTable.ifEntry.ifOutOctets.3 >/tmp/sysgrapheth0out",host,commstr);
   system(cmd);
   sprintf(cmd,"snmpwalk %s %s interfaces.ifTable.ifEntry.ifOutOctets.1 >/tmp/sysgraphlooutout",host,commstr);
   system(cmd);

   system("perl time.pl >/tmp/sysgraphtime");
   system("/sbin/clock >/tmp/sysgraphclock");

   f = fopen("/tmp/sysgraphtime","r");
   fgets(buf,sizeof(buf),f);
   fclose(f);

   etime = atoi(buf);

   f = fopen("/tmp/sysgrapheth0in","r");
       fgets(buf,sizeof(buf),f);
   fclose(f);

   tmp[0] = buf[53];
   tmp[1] = buf[54];
   tmp[2] = buf[55];
   tmp[3] = buf[56];
   tmp[4] = buf[57];
   tmp[5] = buf[58];
   tmp[6] = buf[59];
   tmp[7] = buf[60];
   tmp[8] = buf[61];
   tmp[9] = buf[62];
   tmp[10] = buf[63];

   eth0in=atol(rip(tmp));
   printf("eth0in %s %d\n",rip(tmp),eth0in);

   f = fopen("/tmp/sysgrapheth0out","r");
       fgets(buf,sizeof(buf),f);
   fclose(f);

   tmp[0] = buf[54];
   tmp[1] = buf[55];
   tmp[2] = buf[56];
   tmp[3] = buf[57];
   tmp[4] = buf[58];
   tmp[5] = buf[59];
   tmp[6] = buf[60];
   tmp[7] = buf[61];
   tmp[8] = buf[62];
   tmp[9] = buf[63];

   eth0out=atol(rip(tmp));
   printf("eth0out %s %d\n",rip(tmp),eth0out);

   f = fopen("/tmp/sysgraphlooutout","r");
       fgets(buf,sizeof(buf),f);
   fclose(f);

   tmp[0] = buf[54];
   tmp[1] = buf[55];
   tmp[2] = buf[56];
   tmp[3] = buf[57];
   tmp[4] = buf[58];
   tmp[5] = buf[59];
   tmp[6] = buf[60];
   tmp[7] = buf[61];
   tmp[8] = buf[62];
   tmp[9] = buf[63];

   lo=atol(rip(tmp));
   printf("lo %s %d\n",rip(tmp),lo);

   //time
   itmp=etime;
   etime=etime-petime;
   petime=itmp;

   //eth0in
   itmp=eth0in;
   eth0in=eth0in-peth0in;
   peth0in=itmp;
   eth0in=eth0in/etime; //bytes per second

   //eth0out
   itmp=eth0out;
   eth0out=eth0out-peth0out;
   peth0out=itmp;
   eth0out=eth0out/etime; //bytes per second

   //lo
   itmp=lo;
   lo=lo-plo;
   plo=itmp;
   lo=lo/etime; //bytes per second

   printf("\n     >>>eth0 in %d bps\n     >>>eth0 out %d bps\n     >>>lo %d bps\n\n",eth0in,eth0out,lo);

   return 0;
}

Thanks,

-Jack C
http://www.CREPinc.com

lol, along the lines of all the other thrown together peices of this code, I now have it system()'ing the problem peices to a perl script which puts it into another text file then I read that back.... there must be a better way!!

PLEASE?

Thanks,

I can't see why you don't do the whole thing in either perl or bash

because I don't know perl or bash

So, basically I'm screwed? I think its the variable type, but what's big enough????

> lo 3045189694 2147483647 //<<<WHAT IS THAT???
It means it's bigger than a signed integer on your system, so whatever conversion function you used returned the largest possible signed integer.

If you want, you could use strtoul(), but that won't buy you much if you're already well into 9-digit decimals.

Code:

   tmp[0] = buf[53];
   tmp[1] = buf[54];
   tmp[2] = buf[55];
   tmp[3] = buf[56];
   tmp[4] = buf[57];
   tmp[5] = buf[58];
   tmp[6] = buf[59];
   tmp[7] = buf[60];
   tmp[8] = buf[61];
   tmp[9] = buf[62];
   tmp[10] = buf[63];

Two things
1. Use strncpy, its much easier
2. you forgot the \0 to make it a proper string

Code:

   strncpy( tmp, &buf[53], 11 );
   tmp[11] = '\0';

Originally Posted by Salem

2. you forgot the \0 to make it a proper string

I figured that since I defined it as 16 chars, that there would still be several nulls left on the end. I was wrong.

Hm. Ok, well thank you, i will attempt to figure out a different way.