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Home/ Questions/Q 709695
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T04:31:54+00:00 2026-05-14T04:31:54+00:00

I’m writing a program to check to see if a port is open in

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I’m writing a program to check to see if a port is open in C. One line in particular copies one of the arguments to a char array. However, when I try to compile, it says:

error: incompatible types in
assignment

Heres the code. The error is on the assignment of addr

#include <sys/socket.h>
#include <sys/time.h>
#include <sys/types.h>
#include <arpa/inet.h>
#include <netinet/in.h>
#include <errno.h>
#include <fcntl.h>
#include <stdio.h>
#include <netdb.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>

int main(int argc, char **argv) {
  u_short port;                /* user specified port number */
  char addr[1023];             /* will be a copy of the address entered by u */
  struct sockaddr_in address;  /* the libc network address data structure */
  short int sock = -1;         /* file descriptor for the network socket */

  port = atoi(argv[1]);
  addr = strncpy(addr, argv[2], 1023);
  bzero((char *)&address, sizeof(address));  /* init addr struct */
  address.sin_addr.s_addr = inet_addr(addr); /* assign the address */
  address.sin_port = htons(port);            /* translate int2port num */

  sock = socket(AF_INET, SOCK_STREAM, 0);
  if (connect(sock,(struct sockaddr *)&address,sizeof(address)) == 0) {
    printf("%i is open\n", port);
  }
  if (errno == 113) {
    fprintf(stderr, "Port not open!\n");
  }
  close(sock);
  return 0;
}

I’m new to C, so I’m not sure why it would do this.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T04:31:54+00:00Added an answer on May 14, 2026 at 4:31 am

    addr is an array so you can’t assign to it directly.

    Change addr = strncpy(addr, argv[2], 1023); to strncpy(addr, argv[2], 1023);

    A pointer to what you passed in is returned, but this value isn’t needed. The call to strncpy alone will copy the string from argv[2] to addr.


    Note: I notice sometimes you pass in the address of your array and sometimes you pass in the array itself without the address of operator.

    When the parameter only asks for char*…

    Although both will work passing in addr instead of &addr is more correct. &addr gives a pointer to a char array char (*)[1023] whereas addr gives you a char* which is the address of the first element. It usually doesn’t matter but if you do pointer arithmetic then it will make a big difference.

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