Thread: How will this variable assignment treat ...........

I Have a code snippet

Code:

#include <string.h>
#include <stdio.h>


typedef enum {
    e_NW_NFY_ID_UNKNOWN = 0,
    e_NW_NFY_ID_ETHERNET_PLUGIN,        	/*< Ethernet cable plugged in, pv_nfy_param: not used */
    e_NW_NFY_ID_ETHERNET_UNPLUG,        	/*< Ethernet cable unplugged, pv_nfy_param: not used */
    e_NW_NFY_ID_ADDRESS_CHANGED,        	/*< IP address changed */
    e_NW_NFY_ID_ADDRESS_EXPIRED,        	/*< IP address expired */
    e_NW_NFY_ID_DHCP_SUCCESS_DHCPv4,    	/*< DHCP started successfully and get IPV4 */
    /*e_NW_NFY_ID_DHCP_FAILURE_DHCPv4,*/  	/*NW_NFY_ID_NEW_ADDRESS_DHCPv4,*/
    e_NW_NFY_ID_DHCP_SUCCESS_LINKLOCAL,   /*< DHCP started successfully but get link local IP */
}E_NetworkStatus;


int minecall(void *ptr);


int minecall(void *ptr)
{
	E_NetworkStatus lNetworkStatus = (E_NetworkStatus)ptr ;
	printf("ptr:%x  ,lNetworkStatus = %x \n",(E_NetworkStatus)ptr,&lNetworkStatus);
	printf("Value : %d\n",lNetworkStatus);
	return 0;
}


int main()
{
	E_NetworkStatus NetworkStatus;
	NetworkStatus = e_NW_NFY_ID_ADDRESS_EXPIRED;
	printf("NetworkStatus:%x \n",NetworkStatus);
	minecall((void *)NetworkStatus);
	return 0;
}

Questions:

1. What will this assignment mean "E_NetworkStatus lNetworkStatus = (E_NetworkStatus)ptr ;"

Since the Output is a correct value.

2. Does the above assignment mean address is assigned to the variable "lNetworkStatus "?
Or is the value is assigned to the variable "lNetworkStatus "?
I am little confused with this assignment.....


Thanks in advance

void * is the closest you get to a generic type in C, and as such is often used when dealing with a group of functions that have to have the same signature (because they're being used via function pointer or similar). You don't have any of that going on, but the basic mechanics behind the idea are still there; any value that "fits" in a pointer (however large a pointer is on your system) can survive being casted to void* and back, and an enum with six things in it can definitely fit in a pointer.

Basic lesson: a void * does not necessarily imply "an address of something" like a "normal" pointer does.

Originally Posted by tabstop

void * is the closest you get to a generic type in C, and as such is often used when dealing with a group of functions that have to have the same signature (because they're being used via function pointer or similar). You don't have any of that going on, but the basic mechanics behind the idea are still there; any value that "fits" in a pointer (however large a pointer is on your system) can survive being casted to void* and back, and an enum with six things in it can definitely fit in a pointer.

Basic lesson: a void * does not necessarily imply "an address of something" like a "normal" pointer does.

Thanks for the reply

My confusion was with the "above highlighted line"

Why didn't we initialize the line with the below line:
E_NetworkStatus lNetworkStatus = *(E_NetworkStatus *)ptr ;