Accessing values in array of pointer inside of a struct

Question

I am trying to access the values of struct members inside an array of struct pointers inside of a struct. So I have a struct Room that contains an array of pointers to other Rooms called outboundConnection[]. I keep getting the error

error: request for member ‘name’ in something not a structure or union printf("Connection %i: %s", i, x.outboundConnections[i].name);

My struct is set up like this:

typedef struct
{
    char* name; //Name of the room
    char type; //Type of room
    int numOutboundConnections; //Number of rooms connected
    int isSelected;; // 0 means not selected, 1 means selected 
    struct Room *outboundConnections[6]; //Pointers to rooms connected
} Room;
// Room constructor used for filling roomBank
Room room_init(char* name, int s, int c)
{
    Room temp;
    temp.name = calloc(16, sizeof(char));
    strcpy(temp.name, name);
    temp.isSelected = s;
    temp.numOutboundConnections = c;
    return temp;
}

I am adding connections to the outboundConnections array using this function:

void ConnectRoom(Room *x, Room *y)
{
    (*x).outboundConnections[(*x).numOutboundConnections] = malloc(sizeof(Room));
    (*x).outboundConnections[(*x).numOutboundConnections] = y;
    (*x).numOutboundConnections++;
    (*y).outboundConnections[(*y).numOutboundConnections] = malloc(sizeof(Room));
    (*y).outboundConnections[(*y).numOutboundConnections] = x;
    (*y).numOutboundConnections++;
}

I am having issues with getting the name struct member in the outboundConnections array.

printf("Connection %i: %s", i, x.outboundConnections[i].name);

I have tried using ->name and (*x.outboundConnections[i]).name. I am wondering if I am even correctly assigning Rooms to the outboundConnections array or if my issue is how I am trying to access the member variable.

Kon · Accepted Answer · 2018-10-24 19:55:41Z

1

outboundConnections is an array of pointers, so outboundConnections[i] is a Room *. Also, x is a pointer as well. You should use -> (as opposed to .) to access its members:

x->outboundConnections[i]->name

edited Oct 24, 2018 at 19:55

answered Oct 24, 2018 at 19:53

Kon

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3 Comments

Teastruction Over a year ago

I tried this an got another error of error: invalid type argument of ‘->’ (have ‘Room’)

Kon Over a year ago

where does this printf appear. In that context, what is x?

Teastruction Over a year ago

It in in a function where I was trying to print out the name member variable of the Room structs in outboundConnections. Here is the function:

void testConnections(Room x) { 	int i; 	for(i=0; i<6; i++) 	{ 		printf("Connection %i: %s", i, x->outboundConnections[i]->name); 	} }

Carey Gregory · Accepted Answer · 2018-10-24 20:16:16Z

1

First off, replace the (*x). notation with x-> as others have pointed out.

But more importantly, when you do this:

(*x).outboundConnections[(*x).numOutboundConnections] = malloc(sizeof(Room));
(*x).outboundConnections[(*x).numOutboundConnections] = y;

You just leaked sizeof(Room) bytes. The malloc call is unnecessary if you're storing pointers to Room. You would only need it if you were storing a copy of Room. So your code should look like this if you're storing pointers:

x->outboundConnections[x->numOutboundConnections] = y;

Or this if you're storing a copy of Room:

x->outboundConnections[x->numOutboundConnections] = malloc(sizeof(Room));
*x->outboundConnections[x->numOutboundConnections] = *y;

answered Oct 24, 2018 at 20:16

Carey Gregory

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Comments

Barmar · Accepted Answer · 2018-10-24 20:50:42Z

0

outboundConnections is an array of pointers, so you have to dereference the pointer to get to the name member.

printf("Connection %i: %s", i, x->outboundConnections[i]->name);

In all cases where you write (*x).something you can simplify that to x->something.

Another problem is in the declaration of the structure itself. You have:

struct Room *outboundConnections[6];

but you've never declared struct Room. You declared an anonymous structure and then gave it a typedef Room. You need to change the declaration to:

typedef struct Room
{
    char* name; //Name of the room
    char type; //Type of room
    int numOutboundConnections; //Number of rooms connected
    int isSelected;; // 0 means not selected, 1 means selected 
    struct Room *outboundConnections[6]; //Pointers to rooms connected
} Room;

edited Oct 24, 2018 at 20:50

answered Oct 24, 2018 at 19:53

Barmar

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13 Comments

Teastruction Over a year ago

I tried this an got another error of error: invalid type argument of ‘->’ (have ‘Room’)

Barmar Over a year ago

I don't understand that. x is Room*, not Room.

Barmar Over a year ago

You didn't do temp->, did you? temp is a structure, not a pointer.

Barmar Over a year ago

You need to go back to your textbook or tutorial and review the sections on structures and pointers, so you know when to use . or ->.

Barmar Over a year ago

You don't have struct Room declared anywhere.

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