Doubt Related to Pointers usage in the function parameters?

I have a two programs they consists of main and a function.

In the first case.

I am sending the output parameters to the function like below.

In the function i am putting the values into the output parameters and then i am displaying

them in the main.

Here i need to access the output parameters with the '*' in function.

Code:

---

void getSumDifAndProduct(int a,int b,int *arr);

void getSumDifAndProduct(int a,int b,int *arr)
{
  *(arr+0) =  a+b;
  *(arr+1) =  (a-b);
  *(arr+2) = (a*b);
}

---

In the second case

Iam using the structure variable and p

Code:

---

typedef struct
{
  int sum;
  int diff;
  int product;

}OutStruct_t;


void getSumDifAndProduct(int a,int b,OutStruct_t *pSt);

void getSumDifAndProduct(int a,int b,OutStruct_t *pSt)
{
  pSt->sum =  a+b;
  pSt->diff =  (a-b);
  pSt->product = (a*b);
}

int main()
{
  int a=0,b=0;
  OutStruct_t *pSt;

  printf("Please Enter the a and b values \n");
  scanf("%d %d",&a,&b);

  pSt = (OutStruct_t *) malloc(sizeof(OutStruct_t));

  getSumDifAndProduct(a,b,pSt);

  printf("a and b is %d %d \n",a,b);
  printf("a and b sum is %d \n",pSt->sum);
  printf("a and b dif is %d \n",pSt->diff);
  printf("a and b product is %d \n",pSt->product);

  return 0;
}

---

Here if i use *pSt->sum why does it gives me error like

ex3.c:16: error: invalid type argument of `unary *'

Iam bit confused about this variation.

Yes, you may use (*pst).sum if you wish - but there's no benfit over pst->sum.

I also prefer

Code:

---

  arr[0] =  a+b;
  arr[1] =  (a-b);
  arr[2] = (a*b);

---

instead of

Code:

---

  *(arr+0) =  a+b;
  *(arr+1) =  (a-b);
  *(arr+2) = (a*b);

---

--
Mats

Because the -> operator alone 1) dereferences the pointer on the left hand to 2) access the struct member on the right hand. Why would you try to dereference pSt twice?

The a->b syntax essentially means (*a).b, so if you try to do *a->b, you get *(*a).b, and if a would be a pointer (not pointer to pointer), that would mean you first dereference and then try to dereference again, which can't be done.
Note also that in C, you should never cast the return of malloc.

In the first case if iam using the individual variable like below

Code:

---

void getSumDifAndProduct(int a,int b,int *sum,int *dif,int *product);

void getSumDifAndProduct(int a,int b,int *sum,int *dif,int *product)
{
        *sum =  a+b;
        *dif =  (a-b);
        *product = (a*b);
}

int main()
{
        int a=0,b=0,sum=0,dif=0,product=0;

        printf("Please Enter the a and b values \n");
        scanf("%d %d",&a,&b);

        getSumDifAndProduct(a,b,&sum,&dif,&product);

        printf("a and b is %d %d \n",a,b);
        printf("a and b sum is %d \n",sum);
        printf("a and b dif is %d \n",dif);
        printf("a and b product is %d \n",product);

        return 0;
}

---

i have to use the * before the variables .

else

sum = a+b;

i get error like below



ex1.c:7: warning: assignment makes pointer from integer without a cast


I understand here as we need to put the value in the memory location of sum .

But the structure variable should also be the same case .

Why should we avoid giving the * here .

You have to tell the compiler that it is a memory deference (that you want it to follow the pointer), but there are different ways to do that. You can use a pointer and an array in almost the same way, do x[0] and *x and *(x + 0) are all the same thing, and you can choose, when it comes to structures, to either use the (*x).y or x->y - both are identical when it comes to what the computer actually does when it executes the code.

Doing sum = a + b attempts to set the pointer itself to a + b, which will not achieve what you want.

--
Mats

Thanks a lot to all of you for the detailed replies