Accepting a pointer address?

Question

This is the source code

#include<stdio.h>
void main()
{
int *p,a=5;
p=&a;
scanf("%d",p);
printf("%d\t",*p);
printf("%d",a);
}

How can we accept an address of a pointer?.Cuz it has the address of variable 'a' already.There are no errors shown by the compiler. Also,i'm not able to understand the output.

output:(if my input is 45)

45   45

With `scanf("%d",p);` you are passing the address pointed by pointer `p`, which is the address of `a`. — LPs, Dec 15 '16 at 07:55
What are you trying to achieve? What do you mean by _accepting_ ? — moooeeeep, Dec 15 '16 at 07:55
Since `p=&a;`, `scanf("%d",p);` is the same as `scanf("%d",&a);`. — David Schwartz, Dec 15 '16 at 07:56
From `man scanf`: "*The scanf() family of functions scans input [...]; the results [...] are stored in the locations **pointed to by the pointer arguments** that follow format.*" — alk, Dec 15 '16 at 08:09

cokceken · Answer 1 · 2016-12-15T08:31:25.567

1

Your pointer is also a variable like a and has it's own address. You can access it by saying &p. But you say scanf("%d", p); so it is accessing pointer's pointing address.

EDIT: if you want to print pointer's address you can use printf("%p\n",(void *) &p);

edited Dec 15 '16 at 08:31

answered Dec 15 '16 at 07:54

cokceken

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but how can i accept the pointers pointing address?doesn't it have a fixed value? – bit_radiation_96 Dec 15 '16 at 07:57
when you define a pointer you allocate a memory space for that pointer. And you can't change it's address. – cokceken Dec 15 '16 at 07:59
1

You need to cast to `(void *)` before printing the value of a pointer: `printf("%p\n", (void *) &p)`. Failure to do so is undefined behavior. – ad absurdum Dec 15 '16 at 08:15
You can compile and run the code on some compilers i think. But it is safer so i will change my answer. Thank you. – cokceken Dec 15 '16 at 08:21
You are right that it works sometimes (often?), but it is in the Standard, and I, for one, am not always lucky! ;) – ad absurdum Dec 15 '16 at 08:23
That is a great link, @Lưu Vĩnh Phúc. Thanks for that – ad absurdum Dec 15 '16 at 09:08

e.jahandar · Answer 2 · 2016-12-15T08:26:20.153

0

You can access pointer itself by its address &p

int pa;
int *p;
scanf("%d", &pa);

p = (int*) pa;

printf("%p\t", p); // printing pointer stored address (got from input)
printf("%d\n", (void *) *p); // printing value stored in address stored in p, of course would be segfault in case of invalid address

Also use it with caution, its unportable and may resulting undefined behavior. it may work on 32bit machines because of int and int* have same length of bits but for other CPU's may not work.

edited Dec 15 '16 at 08:26

answered Dec 15 '16 at 07:58

e.jahandar

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4

What's the idea behind doing `scanf("%d", &p);`? In fact it provokes undefined behaviour, because `d` is defined to scan `int`s only. – alk Dec 15 '16 at 08:05
I agree that its not portable, but in fact on 32 bit machine int and int* have same length, so we can copy contents of int to int*. The int could store the address of something but could not dereference it. – e.jahandar Dec 15 '16 at 08:12
2

Cast to `(void *)` before printing an address. This is required by the Standard. – ad absurdum Dec 15 '16 at 08:21

Accepting a pointer address?

2 Answers2