What is the difference between scanf("%d", *p) and scanf("%d", p)?

Question

Pointers are a new thing for me and I'm struggling to understand it, but I won't give in and hopefully learn it. What would be the difference between `scanf ("%d", *p)` and `scanf ("%d", p)`? In examples I saw that if I want to input some value in a variable, I should use `scanf ("%d", p)`. That doesn't make sense to me. Shouldn't it be `scanf ("%d", *p)`? I interpret it as: "put some integer value where the pointer is pointing" and for instance it is pointing on variable `x` and then it should be `x = 10`, but it isn't. And how then to use `scanf()` and pointers to set values in an array? Where and what am I getting wrong? I'm trying to learn this using C language, since it is the one which I'm supposed to use in my class. For example: ``` #include <stdio.h> int main () { float x[10], *p; int i; p = &x[0]; for (i = 0; i < 10; i++) { scanf("%d", p + i); } for (i = 0; i < 10; i++) { printf("%d", *(p + i)); } return 0; } ``` Why is only `p + i` in the first `for () {}` and `*(p + i)` in…

Accepted Answer

`*p` means go to the place `p` points to `&p` means take the address of `p`, or "get a pointer to" `p`

```
int i;
scanf("%d", &i); // gives scanf a pointer to i

int i;
int *p = &i;
scanf("%d", p); // a more contrived way to say the same thing
```

The obligatory visual explanation is Pointer Fun with Binky.

You can read the types from right to left:

`int *p` => "`p` has type `int *`" => `p` is a pointer to an `int` `int *p` => "`*p` has type `int`" => `*p` is the `int` pointed to by `p`

What is the difference between scanf("%d", *p) and scanf("%d", p)?

Solution

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