Thread: How can we demonstrate interchangeability of Arrays and Pointers in a structure?

We know that arrays and pointers in C are interchangeably. Namely;

Code:

//Pseudocode for giving instance.
int* bPtr;
int b;

bPtr=&b;

*(bPtr+i)  | bPtr[i]

*(b+i)      | b[i]

So I have tried to perform this knowledge to the structures. But I've failed. How can I overcome this problem? Or can't we do this in the structures? Or is there a special case in the structures?


Here the programme which runs:

Code:

#include <stdio.h>
#include <stdlib.h>


struct card{
    char *face;
    char *suit;
};


int main()
{
    struct card aCard;
    struct card *cardPtr;


    aCard.face="Ace";
    aCard.suit="Spade";


    cardPtr = &aCard;


    printf("%s%s%s\n%s%s%s\n%s%s%s\n",aCard.face," of ",aCard.suit,
                                      cardPtr->face," of ",cardPtr->suit,
                                      (*cardPtr).face," of ",(*cardPtr).suit);


    return 0;
}

Here the programme which doesn't run.

Code:

#include <stdio.h>
#include <stdlib.h>


struct card{
    char face[50];
    char suit[50];
};


int main()
{
    struct card aCard;
    struct card *cardPtr;


    aCard.(face[])="Ace";
    aCard.(suit[])="Spade";


    cardPtr = &aCard;


    printf("%s%s%s\n%s%s%s\n%s%s%s\n",aCard.face," of ",aCard.suit,
                                      cardPtr->face," of ",cardPtr->suit,
                                      (*cardPtr).(face[])," of ",(*cardPtr).(suit[]));
                                    //(*cardPtr).face[]," of ",(*cardPtr).suit[]);
    return 0;                       //(*cardPtr).face," of ",(*cardPtr).suit);
}

This should solve your problem :

Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
  
  
struct card{
    char face[50];
    char suit[50];
};
  
  
int main()
{
    struct card aCard;
    struct card *cardPtr;
 
 
    strcpy(aCard.face, "Ace");
    strcpy(aCard.suit, "Spade");
 
  
    cardPtr = &aCard;
  
  
    printf("%s%s%s\n%s%s%s\n%s%s%s\n",aCard.face," of ",aCard.suit,
                                      cardPtr->face," of ",cardPtr->suit,
                                      (*cardPtr).face," of ",(*cardPtr).suit);
 
 
    return 0;
}

I hope that I have helped .

For these declarations:

Code:

int array[10];
int index = 5;

array[index] is the same as *(array + index), which is the same as *(index + array), which is the same as index[array] .

Originally Posted by iMalc

For these declarations:

Code:

int array[10];
int index = 5;

array[index] is the same as *(array + index), which is the same as *(index + array), which is the same as index[array] .

*I thought that it was a joke. But then, I just stared with mouth open at the window output like .
*Alright. I understand that I didn't know wrong this subject. I just knew nothing.
*Thank you for sharing that something confusing. It's opened a new door for me about my study.

Originally Posted by iMalc

array[index] is the same as *(array + index), which is the same as *(index + array), which is the same as index[array] .

Just to be pedantic, array[index] and *(array + index) are not necessarily the same. They do give equivalent results, but the way they do that (say at the level of machine instructions) is often different. And that difference can be detected in various ways (performance measures, number of instructions to achieve an effect, etc).

index[array] and array[index] are equivalent because addition is commutative for pointers and integral values (i.e. x + y and y + x are equivalent)