My Program not printing array by pointer arithmatic

Question

#include<iostream>
using namespace std;
int main()
{
    int a,count=0;
    cin>>a;
    int* arr;
    while(a)
    {
    int x= a%10;
    a=a/10;
    *(arr+count)=x;
    count++;
    }
    for(int i=0 ;i< count-1; i++)
    {
        cout<<*(arr+i);
    }
}
/* o/p :segementation error
// i/o :5*/

** array with pointers

why segmentation error

partitioning int in int array **

segmentation error

@tkausl i thought we can create array using pointers ...thats why array — shubhs_001, Nov 06 '17 at 13:23
@ShubhamJangam - You were misguided. Don't worry, we all were at some point. It's the legacy of C and C++. — StoryTeller - Unslander Monica, Nov 06 '17 at 13:24
`arr` is never initialised. So the usages of `*(arr + count)` and `*(arr + i)` both give undefined behaviour. — Peter, Nov 06 '17 at 13:24
@BLUEPIXY actually that doesnt give compile time error which may then could have produced. — shubhs_001, Nov 06 '17 at 13:25
@ShubhamJangam - Nowhere is it specified that such a thing should give a compile time error. C++ doesn't hold your hand, and will gladly let you shoot yourself in both feet. — StoryTeller - Unslander Monica, Nov 06 '17 at 13:26
@ShubhamJangam Why should it give you compile error? Writing a program which compiles fine, but crashes immediately, is easy. — Algirdas Preidžius, Nov 06 '17 at 13:26
@ShubhamJangam - accessing the value of uninitialised variables gives undefined behaviour. A compiler is not required to diagnose undefined behaviour. That said, most compilers have an option to give warnings about such things - but that option is also disabled by default. Read your compiler documentation to work out how to get such warnings. — Peter, Nov 06 '17 at 13:27
@StoryTeller - actually, according to Bjarne Stroustrup, "C makes it easy to shoot yourself in the foot; C++ makes it harder, but when you do it blows your whole leg off". — Peter, Nov 06 '17 at 13:30
@Peter - That may be the vision Bjarne had in mind, but I don't think it's quite the reality. — StoryTeller - Unslander Monica, Nov 06 '17 at 13:31
@Peter Thanks for help... but following solution by me also seems helpful .... If any suggestion... always welcome !! — shubhs_001, Nov 06 '17 at 15:40

score 3 · Answer 1 · answered Nov 06 '17 at 13:25

3

Accessing an uninitialized value has undefined behaviour (there are exceptions, but none that apply to your case).

arr is uninitialized:

int* arr;

The value of arr is used here:

*(arr+count)=x;

Therefore the behaviour of your program is undefined.

answered Nov 06 '17 at 13:25

eerorika

232,697
12
197
326

how to initialise it then ... I am noobie here – shubhs_001 Nov 06 '17 at 13:27
@ShubhamJangam Like this: `arr = address_of_an_array;`. Replace `address_of_an_array` with the address of the array that you wish to point to. – eerorika Nov 06 '17 at 13:28
1

@ShubhamJangam use new and delete for dynamically allocating memory – Albin Paul Nov 06 '17 at 13:29
3

@ShubhamJangam - Use `std::vector` – StoryTeller - Unslander Monica Nov 06 '17 at 13:32
@ShubhamJangam The correct syntax for declaring an array of int in C and C++ is: `int arr[SIZE];` where SIZE is non-zero. – Michaël Roy Nov 06 '17 at 14:08
@ShubhamJangam to expand on Michaels comment: `SIZE` must be known at compile time in C++. If you need dynamic size, use `std::vector` like StoryTeller suggested. – eerorika Nov 06 '17 at 14:24
@user2079303 Following code also seems helpfull .... if any suggestions... welcome :) – shubhs_001 Nov 06 '17 at 15:32
Thanks people for help :) – shubhs_001 Nov 06 '17 at 15:32

score -1 · Answer 2 · answered Nov 06 '17 at 15:30

-1

#include<stdio.h>
int main()
{
    int a,count=0;
    int i;
    scanf("%d",&a);
    int* arr=&i;
    while(a)
    {
    int x= a%10;
    a=a/10;
    *(arr+count)=x;
    count++;
    }
    for(int i=0 ;i< count; i++)
    {
        printf("%d",*(arr+i));
    }
    return 0;
}

**YES we can create dynamic array also like this may be not a good practice but really helpful to save memory **

answered Nov 06 '17 at 15:30

shubhs_001

5
4

`int i;` is a single integer. It is not an array. `*(arr+count)=x;` has undefined behaviour for all other values than `count == 0`. – eerorika Nov 06 '17 at 15:47
@user2079303 above program works fine... i just want to collect separate integers from 'a' . no undefined behavious seems to be present also...... *(arr+count) is just getting contagious memory – shubhs_001 Nov 06 '17 at 16:10
the program may *appear* to "work fine", but that's just because you were unlucky. Just because the program "works" doesn't mean that the behaviour isn't undefined. There is no guarantee that your program would continue working if you change compilation options, swap the compiler, run on different cpu, or use different input ... – eerorika Nov 06 '17 at 19:27
`*(arr+count)` is "just" getting *contiguous* memory **that you never allocated and therefore may be clobbered by other parts of your program ... or you may be clobbering the the memory that was used for something else**. If you "want to collect separate integers from 'a'", then what you need is an array (or some other data structure that represents a collection). – eerorika Nov 06 '17 at 19:34

My Program not printing array by pointer arithmatic

2 Answers2