array of structs problem - HELP

Posted on 1998-12-18
Last Modified: 2013-12-14
Can somebody tell me how to actually get this to work ??

 I have a DLL that returns a pointer to an array of structs (I think);
 I dont have a Lib file so I have had to load it explicitly (works fine).

 // header snippet
 typedef struct
 int i;
 char c[20];
 } X_Y;

 typedef X_Y  *X_Y;

 // DLL prototype
 typedef BOOL (WINAPI *LPGETLIST) (int iCount,  X_Y **ppUsers);
 // Documentation for the DLL function.
 // In args none.
 // Out, returns ..  iCount the number of users.
 // ppUsers ..  a pointer to the users array.
 // The data returned is in the order of the user list, top to bottom
 // The users array is initialized with int i and char c.

 X_Y *ppUsers;
 BOOL uReturnVal;
 int iCount, j;
 // main snippet

 hDLL = LoadLibrary("The.dll");
 // works fine

 lpGetList = (LPGETLIST) GetProcAddress(hDLL, "TheGetListFunc");
 uReturnVal = lpGetListDet (&iCount, &ppUsers);
 // works fine (uReturnVal TRUE)

// HOW do I access members of the pointed to structs
// I have tried this below DOES NOT WORK

 for(j = 0; j < iCount; j++) {
 printf("%d\n", ppUsers[j].i);      
 // print to screen      

// Free user structures and the array itself.
Question by:cosmicmike
  • 3

Author Comment

ID: 1255305
ignore this typo:
uReturnVal = lpGetListDet (&iCount, &ppUsers);
should be:
uReturnVal = lpGetList (&iCount, &ppUsers);


Expert Comment

ID: 1255306
printf( "%d\n", (*pUsers[j])->i )) ;

The trick is to read hungarian notation (on the front of variable names)

p = pointer to whatever, in your case a struct, therefore use -> to access a member
pp = pointer to pointer
The * dereferences a pointer, so *ppUsers is a pointer to a user struct


Author Comment

ID: 1255307
Tried this
printf( "%d\n", (*ppUsers[j])->i );  

Ad get this compile error
error C2232: '->i' : left operand has 'struct' type, use '.'

Author Comment

ID: 1255308
ppUsers is already a double pointer to a struct

Accepted Solution

lafanga earned 100 total points
ID: 1255309
X_Y *ppUser; // means that *ppUser is a Structure and ppUser is a pointer to the structure

For 1 given structure:
retrieve values by *ppUser.i or ppUser->i

Now, &ppUser is a pointer, to the pointer of the structures
Say &ppUser  = pppUser

So Your first member,  that is a pointer to structure 1 will be pppUser[0] or &ppUser[0]
Now, this pointer (&ppUser[0]) will access data as (&ppUser[0])->i

So the net result should be:
printf("%d\n", (&ppUsers[j])->i);


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