CODEWITHSUNDEEP
awk
user292832
user292832

Reputation: 1

fatal: division by zero attempted

when ever I am trying to calculate mean and standard deviation using awk i am getting "awk: fatal: division by zero attempted" error.

my command is

BEGIN{
recvs = 0;
routing_packets = 0;
}
{
state       =   $1;
layer       =   $3;
tipe_data   =   $6;
}
{
if (state == "r" && tipe_data == "tcp" && layer=="AGT") recvs++;
if (state == "s"  && layer == "AGT" && tipe_data =="AODV") routing_packets++;   
}

END{
printf("##################################################################################\n");

printf("\n");

printf("                       Normalized Routing Load = %.3f\n", routing_packets/recvs);

printf("\n");

printf("##################################################################################\n");
}

thanks

Upvotes: 0

Views: 3802

Answers (3)

user3442743
user3442743

Reputation:

Change this line

printf("                       Normalized Routing Load = %.3f\n", routing_packets/recvs);

to

if (recvs)printf("                       Normalized Routing Load = %.3f\n", routing_packets/recvs);

Upvotes: 0

bashrc
bashrc

Reputation: 4835

If this if check never succeeds the value of recvs will be equal to 0

if (state == "r" && tipe_data == "tcp" && layer=="AGT") recvs++;

Your program will go on nevertheless and hit this line routing_packets/recvs You should handle the scenario where recvs is equal to 0 depending on your requirement.

Upvotes: 0

Avinash Raj
Avinash Raj

Reputation: 174696

Yes, values assigned to recvs and routing_packets variables is zero.

recvs = 0;
routing_packets = 0;

And you are trying to to divide it. That is,

routing_packets/recvs

Which results in,

awk: fatal: division by zero attempted" error

Upvotes: 0

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