Typedef problem

**ToniShinobi** · 01-25-2008

I have a problem with 2 structures in a project of mine. My llNode structure has an ArrayQueue structure in it, but the ArrayQueue structure contains pointers to other llNodes. For me to use llNode pointers in my ArrayQueue, i used a typedef at the beginning, but it won't work because llNode isn't defined at that point.

My code is below. Any help is greatly appreciated.

Code:

typedef llNode* arrEntry;

typedef struct arrayqueue {
  //Structure for a queue of a locations' connections
  int front;
  int back;
  int size;
  arrEntry* array;
} ArrayQueue;

typedef struct llnode { 
  //Structure for a linked list node containing location info
  string name;
  int x, y;
  arrayqueue roads;
  struct llnode* next;
} llNode;

**Dave_Sinkula** · 01-25-2008

Code:

typedef struct llnode* arrEntry;

typedef struct arrayqueue {
  //Structure for a queue of a locations' connections
  int front;
  int back;
  int size;
  arrEntry* array;
} ArrayQueue;

typedef struct llnode { 
  //Structure for a linked list node containing location info
  string name;
  int x, y;
  arrayqueue roads;
  struct llnode* next;
} llNode;

?

**Aia** · 01-25-2008

Code:

typedef struct llnode* arrEntry;

typedef struct arrayqueue {
  //Structure for a queue of a locations' connections
  int front;
  int back;
  int size;
  arrEntry* array;
} ArrayQueue;

typedef struct llnode { 
  //Structure for a linked list node containing location info
  char name[60];
  int x, y;
  ArrayQueue roads;
  struct llnode* next;
} llNode;

**ToniShinobi** · 01-25-2008

Thanks, i made the change in the first typedef line, and it works. Because i didn't have struct there it didn't work.

**Elysia** · 01-26-2008

string? Hmmm, are we using C/C++ mix here?

**ToniShinobi** · 01-26-2008

oh sorry, i had a typedef for string earlier in my code

Code:

typedef char* string;

which brings up another question.... my scanf's had problems when i tried to scan strings into char*'s, but when i changed them to char [5], the problem disappeared. why's that?

**matsp** · 01-26-2008

Probably because your char * hasn't got any memory actually assigned to it. Perhaps you need to malloc some memory or some such.

--
Mats

**Elysia** · 01-26-2008

It is typically a bad idea to typedef such things, such as char* because it's hard for any programmer besides you to know what the actual type is. A string can be char[] or char*, whatever works.
I recommend you avoid typenames such as that, when they're obscure.
Especially, I don't think it's a good idea to do away with *. So
typedef char string;
Would be better and you should do string* later.

**ToniShinobi** · 01-26-2008

oh, that makes sense... so you can have:

Code:

char * hello = "Hello world";

but you can't have

Code:

char * hello;
scanf(" &#37;s", &hello);

**Elysia** · 01-26-2008

Indeed. The second example is wrong on two occasions:
hello is just a variable pointing at some random memory location, to memory you haven't allocated, so you'll most likely get a crash when running that code. Secondly, scanf expects a char*, but since you are passing the address of a pointer, essentially you are passing a char**. Remove the & before the pointer to make it better. Though the code is far from ok...

And a note for the first code: it should be const char* hello = "Hello world", because the string literal is read-only and thus if you modify it you get an access violation.

**matsp** · 01-26-2008

Correct.

You can of course have something like:

Code:

char * hello;

hello = malloc(500);
scanf("%s", hello);

You should also be aware that scanf isn't the best function to read user input, particularly strings, as if the user types in too much input, you end up with a buffer overflow. Although there are some ways to counteract that, I would rather recommend the funciton fgets() - it is much more robust.

--
Mats

**Elysia** · 01-26-2008

Oh, right.
More about scanf: http://cboard.cprogramming.com/showp...37&postcount=9

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