Pointer / dereference error

Question

I want to display "string pointer affected" but I get an error. Here is my code:

#include<stdio.h>

main()
{
  char* *p;

  char * s="string pointer affected";

  *p=s;

  printf("%s",*p);
}

An understanding of pointers & how to dereference them would avoid this problem. See [Evaluating the condition containing unitialized pointer - UB, but can it crash?](http://stackoverflow.com/questions/22465371/evaluating-the-condition-containing-unitialized-pointer-ub-but-can-it-crash/22466225#22466225) for an explanation. — Mogsdad, Dec 29 '15 at 16:05

score 6 · Accepted Answer · answered Dec 28 '15 at 17:30

6

p doesn't point to any known location, so writing to *p is a bad idea.

You mean to say:

p = &s;

answered Dec 28 '15 at 17:30

Paul Roub

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ameyCU · Answer 2 · 2015-12-28T17:36:46.970

5

You dereference a pointer which is not initialized , which will cause undefined behaviour . This is problem -

*p=s;

edited Dec 28 '15 at 17:36

answered Dec 28 '15 at 17:29

ameyCU

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@alk Sorry for my poor choice of words ,Sir . – ameyCU Dec 28 '15 at 17:36

score 4 · Answer 3 · edited Dec 29 '15 at 15:59

4

You are using an uninitialized variable in the line below and in the printf statement. If you replace

*p = s;

with

p = &s;

then it will work.

edited Dec 29 '15 at 15:59

Mogsdad

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answered Dec 28 '15 at 17:30

Weather Vane

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Ziffusion · Answer 4 · 2015-12-28T17:54:42.273

Try:

#include<stdio.h>

main()
{
    char *p; // <--------------------------

    char *s="string pointer affected";

    printf("===== s=%p\n", s);

    p=s;

    printf("===== p=%p\n", p);

    printf("%s\n", p);
}

The problem with the original code is that p is uninitialized. So you cannot dereference it.

If you do want to use a pointer to a pointer, allocate the pointer first, and then take its address.

#include<stdio.h>

main()
{
    char *q;

    char **p = &q;

    char *s="string pointer affected";

    *p=s;

    printf("%s\n", *p);
}

Pointer / dereference error

4 Answers4