What is the difference in the use of structure pointer and structure variable?

Question

struct Book {
    int i;
} variable, *ptr;

while accessing struct memebers we use variable.i or ptr->i my question is what is the difference between/use of variable and *ptr

`ptr` is uninitialized, therefore any access through it (`*` or `->`) will have UB. Is that what you meant? — edmz, May 12 '16 at 19:04
Possible duplicate of [What are the differences between a pointer variable and a reference variable in C++?](http://stackoverflow.com/questions/57483/what-are-the-differences-between-a-pointer-variable-and-a-reference-variable-in) — Turtle, May 12 '16 at 19:07
Thank you for your response. Sorry i didn't get you. Can you explain it little clearly, if possible — Lokesh, May 12 '16 at 19:12
More specifically, imagine that the structure is big, say, 1000 bytes. A structure variable allocates 1000 bytes of memory. A pointer to the structure allocates probably 4 or 8 bytes (the size of a memory address). — Lee Daniel Crocker, May 12 '16 at 19:27
It's the same as the difference between an int pointer and an int variable. The variable contains an integer, the pointer contains the location of another variable. — Barmar, May 12 '16 at 20:03

score 7 · Accepted Answer · answered May 12 '16 at 19:07

7

Imagine a dog.

Now imagine a dog leash.

Now imagine the dog, clipped to the leash.

The leash represents a pointer to the dog. If you create a pointer (leash) and don't also have a structure to point to (dog), then you can't go to the park and play frisbee.

If you have a structure, but don't have a pointer, you can still do lots of things.

Using a pointer requires having a structure to point to. You can either declare a structure, and then point to it using the & operator, or you can call a function like malloc or calloc which will return dynamically allocated memory that you can use as the structure:

void demo() {
    struct Book b1;
    struct Book b2;

    typedef struct Book * Bookptr;

    Bookptr p;

    // Assign pointer to existing object using address operator:

    p = &b1;
    p->i = 10;
    p = &b2;
    p->i = 12;
    printf("Book 1 has i: %d, while Book 2 has i: %d\n", b1.i, b2.i);

    // Use dynamically allocated memory
    p = calloc(1, sizeof(struct Book));
    p->i = 3;
    printf("Dynamic book has i: %d\n", p->i);
    free(p);
}

answered May 12 '16 at 19:07

aghast

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Hmmm - So one can have 1) a dog on a leash 2) a leash and no dog (dangling pointer) 3) a leash tied to the [south pole](https://en.wikipedia.org/wiki/South_Pole#Ceremonial_South_Pole) (NULL pointer) or 4) a dog not on a leash (memory leak). ;-) – chux - Reinstate Monica May 12 '16 at 19:19
:-) That was cool... – Lokesh May 12 '16 at 19:27
Thank you @Austin... I got it – Lokesh May 12 '16 at 19:28
@chux (4) is also little precious Fifi, that runs around the house crapping on stuff. You don't have to use pointers to use a struct. ;-) – aghast May 12 '16 at 19:35
Explain please: [link](http://aestheticsofjoy.com/wp-content/uploads/2009/09/invisible_dogs3.jpg) – clickMe May 12 '16 at 20:01
2

@SebTu - those would be smart pointers. – Michael Dorgan May 12 '16 at 20:26
@SebNag Null pointers. – aghast Feb 20 '19 at 23:45

score 2 · Answer 2 · answered May 12 '16 at 19:07

variable will have memory associated with it and therefore can be directly accessed when created. Because the memory is given at compile time, the . means the compiler can directly lookup the values in the structure without having to do any sort of in-direct jumping

ptr will only be a pointer to memory and cannot be used until pointed at something that has memory (or given memory via a dynamic memory allocation.) The -> means the compiler must read the memory first and then jump to the location.

Thanks Michael for your clarification – Lokesh May 12 '16 at 19:30 — Lokesh, May 12 '16 at 19:30

What is the difference in the use of structure pointer and structure variable?

2 Answers2