Thread: Assigning value to structure members using -> operator

I appreciate that you're trying to figure out the syntax, but I suggest that you don't do things that way. Let's look at your print loop:

Code:

for(i =0; i < num;++i)
   printf("%s\t%d\t%.2f\n",(ptr+i)->name,(ptr+i)->age,(ptr+i)->weight);

This is fine, but because you are iterating by index, we would more commonly use array index syntax:

Code:

for (i = 0; i < num; ++i)
   printf("%s\t%d\t%.2f\n", ptr[i].name, ptr[i].age, ptr[i].weight);

As you can see, this makes it a bit easier to reason: ptr[i] is the "current element" in each iteration.

If you want to use pointer syntax, then work with pointers, e.g.,

Code:

struct person *end;
for (end = ptr + num; ptr != end; ++ptr)
   printf("%s\t%d\t%.2f\n", ptr->name, ptr->age, ptr->weight);

As you can see, by iterating with a pointer instead of an index, we don't need to do pointer arithmetic beyond incrementing the pointer, as long as we aren't say, trying to access other elements in the array via the pointer. This is because now ptr is the "pointer to the current element" in each iteration.

So, back to your question. It looks like &(ptr+i)->name is actually wrong. It should indeed have been just (ptr+i)->name. Again, if you had used array index notation to match your iteration by index:

Code:

for (i = 0; i < num; ++i)
{
   printf("Enter name, age and weight of the person respectively:\n");
   scanf("%s%d%f", ptr[i].name, &ptr[i].age, &ptr[i].weight);
}

whereas if you actually want to work with pointers proper:

Code:

struct person *end;
for (end = ptr + num; ptr != end; ++ptr)
{
   printf("Enter name, age and weight of the person respectively:\n");
   scanf("%s%d%f", ptr->name, &ptr->age, &ptr->weight);
}

By the way, you should check the return value of scanf to see if the read was successful.

changing this

Code:

for(i =0; i < num;++i)
   {
       printf("Enter name, age and weight of the person respectively:\n");
       scanf("%s%d%f",&(ptr+i)->name,&(ptr+i)->age,&(ptr+i)->weight);
   }

to this

Code:

for(i =0; i < num;++i)
   {
       printf("Enter name, age and weight of the person respectively:\n");
       scanf("%s%d%f",(ptr+i)->name,(ptr+i)->age,(ptr+i)->weight);
   }

give Segmentation fault

You removed some necessary & from your scanf call.

Use a better compiler to tell you when you get this wrong.

Code:

$ gcc -Wall foo.c
foo.c: In function ‘main’:
foo.c:23:14: warning: format ‘%d’ expects argument of type ‘int *’, but argument 3 has type ‘int’ [-Wformat=]
        scanf("%s%d%f",(ptr+i)->name,(ptr+i)->age,(ptr+i)->weight);
              ^
foo.c:23:14: warning: format ‘%f’ expects argument of type ‘float *’, but argument 4 has type ‘double’ [-Wformat=]

Exactly. Notice that I didn't say anything about &(ptr+i)->age or &(ptr+i)->weight, because you wrote that "Since Pointer ptr is already an address, what is the need of putting & in front of it?", so I assumed that you understood that this was for the name member only, as name is an array thus it will be converted to a pointer to its first element, whereas age and weight are neither arrays nor pointers.

EDIT:
Oh! Sorry. I realised that I misread your statement. Unfortunately, you are wrong in what you were looking at. Yes, ptr, being a pointer, contains an address, but you are not passing ptr as an argument. You are passing &(ptr+i)->name as the argument.

Put it another way: before you start trying to learn about how to use this syntax, you should learn how to use scanf with arrays and structures, e.g., you should understand why this is wrong:

Code:

struct person x;
scanf("%s%d%f", x.name, x.age, x.weight);

and why this is technically wrong, even though it may work:

Code:

struct person x;
scanf("%s%d%f", &x.name, &x.age, &x.weight);

but this is correct (disregarding the possible buffer overflow):

Code:

struct person x;
scanf("%s%d%f", x.name, &x.age, &x.weight);

Thanks for clarifying the difference

I understand how this works

Code:

struct person x;
scanf("%s%d%f", x.name, &x.age, &x.weight);

or

Code:

scanf("%s%d%f",&(*ptr).name,&(*ptr).age,&(*ptr).weight);

Here user input is assigned to age and weight variable by referring their address like a normal variable

i.e

Code:

int a;
scanf("%d", &a);

But when we use pointer, consider the above example of int a

Code:

int a;
int *ptr = &a;
scanf("%d", ptr);

Here I don't have to write

Code:

scanf("%d", &ptr);

since pointer ptr itself represent the address

But in structure

I've to use &

Code:

for(i =0; i < num;++i)
   {
       printf("Enter name, age and weight of the person respectively:\n");
       scanf("%s%d%f",(ptr+i)->name,&(ptr+i)->age,&(ptr+i)->weight);
   }

For me it seems like for a same situation, two rules (But I know there's a difference between both, that's what I'm looking for)

Originally Posted by Athul

Here user input is assigned to age and weight variable by referring their address like a normal variable

Other than spelling, (*ptr).age is exactly the same as ptr->age. They are the same thing, so &(*ptr).age is the same thing as &ptr->age.

Originally Posted by Athul

Here I don't have to write

Code:

scanf("%d", &ptr);

since pointer ptr itself represent the address

ptr contains the address of the int object named a. This is critical, because that's exactly the same thing as what you are doing to read the age member of the struct person object.

Originally Posted by Athul

But in structure

I've to use &

Code:

for(i =0; i < num;++i)
   {
       printf("Enter name, age and weight of the person respectively:\n");
       scanf("%s%d%f",(ptr+i)->name,&(ptr+i)->age,&(ptr+i)->weight);
   }

For me it seems like for a same situation, two rules (But I know there's a difference between both, that's what I'm looking for)

It is the same situation, except that you confused yourself by forgetting that your aim is to read into the members of the struct person object, not the struct person object itself. You are not writing &ptr, which is the address of ptr. You are writing &(ptr + i)->age, which is the address of the age member of the ith struct person object in the array.

To break it down:
ptr is a pointer to the first element of the struct person array
(ptr + i) is a pointer to the ith element of the struct person array
(ptr + i)->age is the age member of the ith element of the struct person array; this is equivalent to (*(ptr + i)).age and ptr[i].age
&(ptr + i)->age is the address of the age member of the ith element of the struct person array

Notice that at no point do you write &ptr and leave it as that, therefore at no point do you pass the address of ptr as an argument to scanf.