Function to check if a value is in an array

Question

I can get the code working without using a function, but as it is I get this error message:

error: expected expression before 'int'
is_authorized(x,int *authPorts[N]);

By researching this I think it might be something to do with passing an array to a function but everywhere I found something that seemed like it might be relevant had too much information that I didn't understand for me to use it.

Here's my code:

#include <stdio.h>
#define N 8

int is_authorized(int port,int *auth_ports[]);

int main(void) {  
  int x = 73;

  /* authorised ports: */
  int authPorts[N] = {20,73,60,80,212,434,2211,434};

  is_authorized(x,int *authPorts[N]);

  return 0;
}

int is_authorized(int port, int *auth_ports[]){
  int i;
  for (i = 0; i < N; i++){                    
      if (port == *auth_ports[i])
         printf("1");
      else
          printf("0");}
}

Answer 1

You are trying to define an array, inside an argument.

You can learn more on how to pass array to function here

To answer your question, change to

is_authorized(x, authPorts);

Answer 2

you do not need the asterix(*) for your array , whenever you want to pass an array to a function the compiler copy the address of the array and send it to the function without needing to tell him that it's called passing by refrence .

Answer 3

fixed code

#include <stdio.h>
#define N 8

int is_authorized(int port,int *auth_ports);



int main(void) {

  int x = 73;

  /* authorised ports: */
  int authPorts[N] = {20,73,60,80,212,434,2211,434};


   is_authorized(x, authPorts);

  return 0;
}

int is_authorized(int port, int *auth_ports){
  int i;
  for (i = 0; i < N; i++){                    
      if (port == auth_ports[i])
         printf("1");
      else
          printf("0");}
  return 0;
}