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I have a Structure say MyStruct:-

#define SEC_DIMENSION 2
struct MyStruct
{
  char cChar;
  float fFloat;
  int   iInt;
};

struct MyStruct StructArray[SEC_DIMENSION][20];  //Two Dimensional Array of Structures.

Now I want to access this with Pointer.

struct MyStruct *StructPtr[SEC_DIMENSION];

I did assignment as follows:-

 StructPtr[0] = &StructArray[0][0];

Now, I want to access Members of Structure StructArray[0][1] i.e. StructArray[0][1].cChar or StructArray[0][1].fFloat

How can I access them by using StructPtr?

I tried using StuctPtr[0][1]->cChar then ((StructPtr[0])[1])->cChar Both returned an error.

With StructPtr[0]->cChar build was successful. But this is not what I want.

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4
  • "But this is not what I want." , what do you want exactly ? Commented Sep 7, 2011 at 5:46
  • I want to Access StructArray[0][1].cChar using Pointers. In This case It would be accessing StructArray[0][0].cChar not for Other Members of This Array. Commented Sep 7, 2011 at 5:49
  • you have yourself assigned the address of the structure at array location [0][0] into the 0th element of the array of pointers StructPtr. Therefore StructPtr[0] now holds the structure StructArray[0][0] 's base address. Therefore StructPtr[0]->cChar will give you the cChar component of StructArray[0][0] Commented Sep 7, 2011 at 5:52
  • @Swanand Purankar: Huh? Why is it StructArray[0][1] in the first case, and then suddenly StructArray[0][0]? Sorry, this is all still overly confusing. Provide a better descripotion of what you are trying to do. Commented Sep 7, 2011 at 5:52

2 Answers 2

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"Now I want to access this with Pointer" is not very descriptive. What you have in your example will not work. What you declared is not a pointer to an array but rather an array of pointers. Is that what you need? I don't know.

If you really need a pointer to an array, you can declare your pointer it as

struct MyStruct (*StructPtr)[20];

and then make it point to your array as

StructPtr = StructArray;

From this point on you can access the original array through this pointer as StructPtr[i][j]

StructPtr[i][j].cChar = 'a';

Alternatively, you can declare the pointer as

struct MyStruct (*StructPtr)[SEC_DIMENSION][20];

and then make it point to your array as

StructPtr = &StructArray;

From this point on you can access the original array through this pointer as (*StructPtr)[i][j]

(*StructPtr)[i][j].cChar = 'a';
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4 Comments

But struct MyStruct (*StructPtr)[SEC_DIMENSION][20] will declare (SEC_DIMENSION*20) pointers, Right? Instead of That 20, I want a Single Pointer with Which I could access All 20 structures and their members!
@Swanand Purankar: No. You need to get more familiar with C declarations first. struct MyStruct (*StructPtr)[SEC_DIMENSION][20] declares exactly one pointer. It will declare exactly one pointer that points to type struct MyStruct[SEC_DIMENSION][20].
If you do struct MyStruct *StructPtr[SEC_DIMENSION][20], then you'll get SEC_DIMENSION*20 pointers. But one you add those extra parentheses, the meaning of the declaration changes drastically. How you get just one pointer.
Okey! Thanks! And Thanks for Clearing a Concept!!
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I think you need a "pointer to an array of dimension [SEC_DIMENSION][20] of structures of type struct MyStruct":

struct MyStruct (*StructPtr)[SEC_DIMENSION][20];

StructPtr = StructArray;

StructPtr[i][j]->cChar;
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