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Referencing an array structure element with a pointer

Posted on 2003-10-26
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Last Modified: 2010-04-15
I would like to know how to reference an array in a function where the array of structures has been passed as a pointer.

If I have a structure:
typedef{
UInt32   A;
UInt16   B;
Char       C[5];
} Stuff

and I create an array:
  Stuff    myStuff[10];

and then pass a pointer to a routine:
Routine(&myStuff);

where the routine takes the pointer:
void Routine(Stuff *pStuff)
{

// do stuff

}

How do I reference say the 9th item A in the array myStuff from within the function Routine?

Thanks

Greg
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Question by:trifusion
5 Comments
 
LVL 15

Accepted Solution

by:
efn earned 250 total points
Comment Utility
The usual way to do this is to pass the address of the first element of the array, not the address of the array.  If you just pass the array as the argument, it automatically gets converted to a pointer to the first element.

Routine(myStuff);

Then the function can be declared either as you have it or like this:

void Routine(Stuff pStuff[])

The two forms are equivalent.

Inside the function, you can use the argument as an array name:

pStuff[8].A = 75;

--efn
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Expert Comment

by:dennis_george
Comment Utility
Hi Greg...

  you can access your array by using the following code....

typedef struct{
short   A;
int     B;
char    C[5];
} Stuff ;

 
Stuff myStuff[10];

void Routine(Stuff *pStuff)
{
     for(int i = 0; i < 10; i++){     // you can access a pointer as an array and vice-versa
          printf("%d",  pStuff[i].B );
      }
}


int main()
{
  int i ;

  for(i = 0; i < 10; i++){
        myStuff[i].B = i + 1;
  }

  Routine(myStuff);
  return 0 ;
}



Dennis
0
 
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Expert Comment

by:Ajar
Comment Utility
trifusion,

Because the array myStuff has been statically allocated memory ..i.e  
>> Stuff    myStuff[10];
instead of
>> Stuff   *myStuff  = (Stuff *) malloc(10* sizeof(Stuff));

it will be wise to declare the function Routine as
====================================
 void (Stuff  pStuff[] )
{
    // check if the size of pstuff is enough for accessing 9th element
    if( sizeof(pstuff)/sizeof(stuff)  >9)  
    { // ACCESS THE 9th element as
      pStuff[8].A = 75
    }
     else
    //   BE AFRAID THAT you will do illegal memory access and so quietly return
    return;
}
====================================
Basically    if you pass the pstuff   as    "Stuff  pStuff[]"   to a function then the function
can check the array size by using 'sizeof ()' .

where as  if you pass  'Stuff * pStuff' .. your function has no way to ensure that there wont be illegal access unless you
also pass the array length along with the pointer

void (Stuff  pStuff[] , int pStuff_len)
{
////Do something.
}


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Expert Comment

by:efn
Comment Utility
> Basically    if you pass the pstuff   as    "Stuff  pStuff[]"   to a function then the function
> can check the array size by using 'sizeof ()' .

I am inclined to doubt this assertion.  sizeof is a compile-time operation.  When the compiler compiles a function definition with an array argument of unspecified size, the compiler does not know what will be passed to the function at run time, so it can't give you the size of the array.  What it does give you is the size of the pointer that is actually passed.

I tested this with Microsoft Visual C++ 6.0 and Borland C++Builder 4.  With both compilers, when I passed an array of 17 integers, the called function reported the size as 4, the size of a pointer.

--efn
0
 

Author Comment

by:trifusion
Comment Utility
All the comments I am sure are valid, but I am self-taught C and I really was just looking for the correct syntax for describing the array once passed as a pointer to a function.

I appreciate the memory notes, I am programming in Palm OS and so I am very aware of the memory issues.

Thanks to all who replied.

Greg
0

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