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`Passing an array of a structure as argument

I would like know how to pass by reference an array of a structure as argument to a function and how I can use this array in the function.  Could somebody tell me how to do this please ?  Thanks.

My code looks like this:

struct TEST
{
  int a;
  int b;
  int c;
  bool d;
}

void main()
{
  TEST TstArray[10];
  F1(???????);
]

Void F1(?????)
{
   //Here , how could I use the array ?
}
0
tam031198
Asked:
tam031198
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1 Solution
 
NorbertCommented:
Answer is coming
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NorbertCommented:
void F1(TEST& *Array, int Size)
{
    for(int i=0;i<Size;i++)
    {
       Array[i].a=Something
.
    }
}
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nietodCommented:
Small problem there.  Typo?  You want.

void F1(TEST *Array, int Size)
     {
         for(int i=0;i<Size;i++)
         {
            Array[i].a=Something
     .
         }
     }

or, I beleive this works, but I never use it, so I'm not sure.

void F1(TEST Array[], int Size)
     {
         for(int i=0;i<Size;i++)
         {
            Array[i].a=Something
     .
         }
     }





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alexoCommented:
Nietod's version is the correct one.

>> I beleive this works, but I never use it, so I'm not sure.
It works and also conveys more information to the reader.

And the call will be:

F1(TstArray, 10);
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nietodCommented:
>>  conveys more information to the reader

What information? Just, the fact that it is an array and not an pointer to 1 item.  Or is there something else?  
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tam031198Author Commented:
Here is my code:

#include <stdio.h>
struct TEST
{
      int a;
      int b;
      int c;
      bool d;
};

void main()
{
      void F1(TEST);  

        TEST TstArray[10];
      F1(TstArray);
        printf("Element 0 = %d\nElement 1 = %d\n",TstArray[0].a,TstArray[1].a);
}    

void F1(TEST *TstArray)
{
      TstArray[0].a = 1;
      TstArray[1].a = 10;
}

and I got this error while compiling:
error C2664: 'F1' : cannot convert parameter 1 from 'struct TEST [10]' to 'struct TEST'.

Could you tell me why please ?
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nietodCommented:
You have declared two versions (overloads) of F1.  One was supposed to be a forward reference for the second, but it doesn't have the same parameters. It takes a signel TEST argument.  The other takes a pointer to an array of TEST  arguments.

Change the

void main()
     {
           void F1(TEST);  

       TEST TstArray[10];
           F1(TstArray);
       printf("Element 0 = %d\nElement 1 = %d\n",TstArray[0].a,TstArray[1].a);
     }    

to

void main()
     {
           void F1(TEST *);  

       TEST TstArray[10];
           F1(TstArray);
       printf("Element 0 = %d\nElement 1 = %d\n",TstArray[0].a,TstArray[1].a);
     }    

Note the "*" in the F1 declaration.
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tam031198Author Commented:
Thank you very much Nietod, you are the King !!!!!!
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alexoCommented:
>> What information? Just, the fact that it is an array and not an pointer to 1 item.
Yep... King :-)
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NorbertCommented:
Your function works but using your definition it does not know anything about the size of the array therefore it is better to pass the size also because the function then can handle arrays of different size

nietod: the '&' only was a typo - sometimes I should read what i have written before I hit the Submit button :-)
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nietodCommented:
Alex, I prefer "your highness."  
Nordbert, I figured it was a typo, the rest of the code suggested that.
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alexoCommented:
Nordbert, any relation to Dilbert?
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NorbertCommented:
Who is Dilbert ???
Nietod: Typos hapens to everyone :-)
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nietodCommented:
Mostly they happen to em.
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alexoCommented:
>> Who is Dilbert ???
Man, what planet did you live on in the last 4 years?
www.unitedmedia.com/comics/dilbert
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