Command Line Argument processing in c - c

Question

On running the below code, it stucks after displaying the argv[0], argv[1] and argv[2] line.
Further flow of code is blocked at this point, can any one help why it is stopping its execution or is it entering into an infinite loop.
#include <stdio.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <arpa/inet.h>
#include <string.h>
#include <stdlib.h>
#include "p8log.h"
#include <errno.h>
int main(int argc, char* argv[])
{
char* PORT;
char* IPADDR;
printf("Arg Count=%d\n",argc);
printf("Arguments are=%s,%s,%s\n",argv[0],argv[1],argv[2]);
printf("HELLO");
PORT=argv[1],
printf("WORLD");
IPADDR=argv[2];
printf("START");
printf("port num=%s",PORT);
printf("IP ADDR=%s",IPADDR);
printf("END");
/* some algorithm of calculation */
return 0;
}
Execution
./file-exe 11111 127.0.0.1
Output
Arg Count=3
Arguments are=./file-exe,11111,127.0.0.1

Answer 1

fflush(NULL); is good to do after any output, if you want to make sure it prints to screen. printf is buffered, so it can get lost.
./a.out 11111 127.0.0.1
Arg Count=3
Arguments are=./a.out,11111,127.0.0.1
HELLO
WORLD
START
port num=11111
IP ADDR=127.0.0.1
END
works fine, you needed some \n to break up lines, like so..
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char* argv[])
{
char* PORT;
char* IPADDR;
printf("Arg Count=%d\n",argc);
printf("Arguments are=%s,%s,%s\n",argv[0],argv[1],argv[2]);
printf("HELLO\n");
PORT=argv[1],
printf("WORLD\n");
IPADDR=argv[2];
printf("START\n");
printf("port num=%s\n",PORT);
printf("IP ADDR=%s\n",IPADDR);
printf("END\n");
fflush(NULL);
/* some algorithm of calculation */
return 0;
}