Monitoring UDP data on wireshark shows ARP packet - sockets

Question

I am trying to send UDP packet to my server 10.20.1.2 with port number 20000. I have implemented UDP client on PC and when i send data using sendto API , at the same time i monitor data on wireshark wireshark shows it as an ARP packet.
18967 5440.858646 PcsCompu_ef:b4:89 Broadcast ARP 42 Who has 10.20.1.2? Tell 192.168.1.70
192.168.1.70 is my machine ip where UDP client is running.
I am not sure how UDP packet is getting converted into ARP packet ?
I understand ARP is for finding MAC address of target node but here i already know MAC address of target device , How can i add it in my udp client so it directly starts UDP communication . My target device is one embedded camera , i am not expecting it to reply on ARP request so i want to prevent sending ARP request.
Below is my UDP client code :
Any inputs are highly appreciated. Thanks in advance.
/*
Simple udp client
*/
#include<stdio.h> //printf
#include<string.h> //memset
#include<stdlib.h> //exit(0);
#include<arpa/inet.h>
#include<sys/socket.h>
#define SERVER "10.20.1.2"
#define PORT 20000 //The port on which to send data
char message[3]={0x00, 0x00 , 0x24};
int main(void)
{
struct sockaddr_in si_other;
int s, i, slen=sizeof(si_other);
int ret;
if ( (s=socket(AF_INET, SOCK_DGRAM, IPPROTO_UDP)) == -1)
{
printf("socket failed");
}
memset((char *) &si_other, 0, sizeof(si_other));
si_other.sin_family = AF_INET;
si_other.sin_port = htons(PORT);
if (inet_aton(SERVER , &si_other.sin_addr) == 0)
{
fprintf(stderr, "inet_aton() failed\n");
exit(1);
}
ret = sendto(s, message, sizeof(message) , 0 , (struct sockaddr *) &si_other, slen);
close(s);
return 0;
}

Answer 1

Some clarifications regarding networking.
1. ARP must be sent and replied
Your camera has IP interface, which means it must handle ARP requests fine without any doubts. ARP is essential part of communicating via IP, camera without ARP support makes no sense. And ARP isn't a result of converting UDP, it's a preliminary step before sending actual UDP datagram. Once ARP reply is discovered with destination MAC-address, UDP packet is sent to that destination. The issue you see isn't about hardcoding MAC to avoid ARP.
2. Your code looks fine
Compiled it locally with minor corrections (missing #include <unistd.h> header with close() declaration), tested on several targets, client works as expected.
3. Something is wrong with your network topology
You are sending message from 192.168.1.70 to 10.20.1.2, which is weird. 192.168.0.0/24 and 10.0.0.0/8 are private IP addresses from different ranges, so they normally can't reach each other without black magic (like NAT traversal). And, what is much weirder, during your attempt ARP request is sent to strange destination. Let me illustrate different cases:
if both devices are in same subnet (e.g. 192.168.1.70 sends to 192.168.1.71), then message is sent directly, so client asks "who has 192.168.1.71" in ARP request.
if devices are in different subnets (e.g. 192.168.1.70 sends to 8.8.8.8), then message is sent through gateway, thus ARP request reads "who has 192.168.1.1" or whatever your gateway address is. Gateway MAC may be already in cache, in which case ARP isn't sent at all.
in your case subnets are obviously different, but ARP is asking about direct destination address rather than gateway MAC address.
It's a shot in the dark, but probably you have two network interfaces on your PC, one connected to 192.168.0.0 subnet, the other to 10.0.0.0 and ARP request is sent from both. If you sniff the wrong interface, you see weird ARP request and don't see UDP, which is actually sent after it. By the way, seeing single arp request is also confusing, because it should be repeated several times if noone answers.
Anyway, you need to check network topology and/or simplify it. Remove unnecessary network interfaces, configure PC and camera to be on the same subnet connected to the same switch/router and investigate further.