Python3 scapy using lambda (where it works in 2.7)

Question

This is my line of code:

ans,unans=srp(Ether(dst="ff:ff:ff:ff:ff:ff")/ARP(pdst="192.168.5.0/24"),timeout=2)
ans.summary(lambda (s,r): r.sprintf("IP: %ARP.psrc%   MAC: %Ether.src%")

using python 2.7, it scans and print the hosts using arp ping scan Using python3, it doesn't work and bring up this error:

  File "arp1.py", line 9
    ans.summary(lambda (s,r): r.sprintf("IP: %ARP.psrc%   MAC: %Ether.src%")
                       ^
SyntaxError: invalid syntax

@jonrsharpe to be honest, I still do not know how to fix the problem — Bob Ebert, Sep 30 '15 at 22:19
changing to lambda s_r: r.sprintf ---- says r is not defined — Bob Ebert, Sep 30 '15 at 22:24

score 0 · Answer 1 · answered Sep 30 '15 at 22:38

0

Alright, figured out this can replace the second line:

for snd,rcv in ans:
        print (rcv.sprintf(r"%ARP.psrc% & %Ether.src%"))

Since, python3 did "Removal of Tuple Parameter Unpacking"

answered Sep 30 '15 at 22:38

Bob Ebert

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Python3 scapy using lambda (where it works in 2.7)

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