Creating a pointer of a struct in C++

Question

i have a struct defined as

typedef struct  BSP_AREA {

    struct BSP_AREA  *otherBspArea;         /* Pointer to itself */
    long             BSPReturnCode;                
    short            sGroupNum;             /* default to 0 */
    char             filler4[6];            /* alignment bytes */
    unsigned char    byCobolWorkArea[20];   /* Cobol Work Area */
    // .... and lots of others fields

} * PBSP_PASS ;

So basically a BSP_AREA can hold a pointer reference to itself, to create a sort of link list.

In the code, i have a method createNode() which returns a BSP_AREA with method signature BSP_AREA createNode(). So in this method i just initialize a BSP_AREA and return it. Now after calling this method, to set the pointer it contains to itself i am doing this, but getting an error.

BSP_AREA cobolPassArea2 = createNode(data1);
cobolPassArea->otherBspArea = &cobolPassArea2;

How do i initialize the pointer otherBSP to the BSP_PASS_AREA returned from createNode()? Please dont advise to make createNode return a pointer, instead of a concrete BSP_AREA, as this is the requirement.

is this `C` or `C++`? If it's `C++` there's no need for the typedef. — OMGtechy, Jun 09 '14 at 17:35
It's hard to suggest an answer. However, in creating linked lists, using a pointer to an object that is not allocated in the heap sounds problematic. You can end with a dangling pointer very soon. — R Sahu, Jun 09 '14 at 17:40
"/vob/infra_dev/source/cppsrc/sourcefile.cpp", line 550.16: 1540-0202 (S) An expression of type "BSP_AREA" is not allowed on the left side of "->". — Saad Ahmed, Jun 09 '14 at 17:40
@OMGtechy it's a pointer typedef, so `PBSP_PASS` is different to `BSP_AREA`, but I agree that there is no need for it! — M.M, Jun 09 '14 at 23:38

score 1 · Answer 1 · edited Jun 09 '14 at 20:56

1

cobolPassArea->otherBspArea = &cobolPassArea2;

should be

cobolPassArea2.otherBspArea = &cobolPassArea2;

Notice that the -> has been changed to a ., in addition to adding the 2 to the variable name, on the left side.

edited Jun 09 '14 at 20:56

Andrew Barber

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answered Jun 09 '14 at 17:49

downhillFromHere

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I've edited in the points about what was changed. It would probably help to also explain the difference between the `->` and `.` operators there. – Andrew Barber Jun 09 '14 at 20:57
http://stackoverflow.com/questions/1238613/what-is-the-difference-between-the-dot-operator-and-in-c – macfij Jun 09 '14 at 21:24

vmp · Answer 2 · 2014-06-09T23:49:37.057

-1

#include <cstdio>
#include <cstdlib>

typedef struct  BSP_AREA 
{

  struct BSP_AREA  *otherBspArea;   
  int something;
} *PBSP_PASS ;


struct BSP_AREA create()
{
   BSP_AREA x;
   x.something = 5;
   return x;
}

// this main you need to force the compiler to accept (-fpermissive on g++)

int main()
{
   BSP_AREA s;
   s.otherBspArea = &(create());
   printf("%d", s.otherBspArea->something);
}

// this one should work fine... but still, those requirements are a little odd

 int main()
{
   BSP_AREA s, r;
   r = create();
   s.otherBspArea = &r;
   printf("%d", s.otherBspArea->something);
}

Is that what you want? This compiles with a warning: "taking address of temporary"

edited Jun 09 '14 at 23:49

answered Jun 09 '14 at 19:07

vmp

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It's not permitted to do `&(create())`, if your compiler generates an executable then it is a compiler extension. – M.M Jun 09 '14 at 23:39
i use g++... Well, c sort of let you do almost anything you want... It gave a warning tho (g++ -fpermissive file.cpp), and didn't compile without -fpermissive – vmp Jun 09 '14 at 23:42
@vmp: It didn't compile without permissive, because _it's not permitted to compile that code_. – Mooing Duck Jun 09 '14 at 23:52
yes I know that... that's why I posted 2 mains... And those requiriments are little bit weird... I wouldn't do either way – vmp Jun 10 '14 at 00:30

Creating a pointer of a struct in C++

2 Answers2