Unexpected behavior and compiler error when using pointers

Question

I'm new to C, and for a few months I've used pointers and tried to figure out how they work and learn syntax for using them, but i ran into a confusing part when compiler throws an error [1] 1473 segmentation fault (core dumped) ./a.out but i think code syntax is correct.

Since I am working with and learning about dynamic data structures, a lot of times everything seems fine, but occasionally i end up into some weird errors, so i provided this simple example of using pointers which confuses me.

#include <stdio.h>

int main ()
{
  int *b;
  printf ("Enter some int value -> ");
  scanf ("%d", b);

  printf ("Entered value is: %d", *b);


  return 0;
}

I am declaring a variable which is pointer to a integer in memory. Since its pointer in scanf() argument is just name of that variable, and when i print it to stdout(printf()) i use *b since pointer to pointer is actual value of that variable.

b is a uninitialized pointer. Derefencing it will cause undefined behavior — Thomas Sablik
– Thomas Sablik, Commented Oct 17, 2019 at 11:37
One thing you must always know about your pointers is where they point. In your program, the pointer b is uninitialized, so you have no idea where it points to (or whether it's even valid). Try making it point to the middle of an array: int arr[3]; int *b = &a[1];. Now you know where it points! You don't know what is in the place b points to (or whether it's a valid value), but b itself is valid and points to a definite and known place. — pmg
– pmg, Commented Oct 17, 2019 at 11:43
@ThomasSablik - it's not necessary to dereference it to get undefined behaviour. Accessing its value gives undefined behaviour. The simple act of passing it by value to a function (by definition) accesses it value. (A precursor to dereferencing a pointer is also to access its value). — Peter
– Peter, Commented Oct 17, 2019 at 12:36

pmg · Accepted Answer · 2019-10-17 11:47:49Z

4

*b is not pointing anywhere. Try this:

#include <stdio.h>
#include <stdlib.h>
int main ()
{
  int *b = malloc(sizeof(int)); // dynamically allocate memory
  printf ("Enter some int value -> ");
  scanf ("%d", b);

  printf ("Entered value is: %d", *b);
  free(b); // free the dynamically allocated memory
  return 0;
}

Or without dynamic memory allocation by pointing to a variable on the stack instead:

#include <stdio.h>
int main ()
{
  int a;
  int *b = &a;
  printf ("Enter some int value -> ");
  scanf ("%d", b);

  printf ("Entered value is: %d", *b); // or have a instead of *b
  return 0;
}

Make sure to consider the warnings that your compiler gives you. Your code will probably trigger a warning about using the uninitialized local variable 'b'. If you didn't get a warning, see if you can configure your compiler to be more strict.

edited Oct 17, 2019 at 11:47

pmg

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answered Oct 17, 2019 at 11:39

Blaze

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1 Comment

S.S. Anne Over a year ago

... by adding -Wall -Wextra -pedantic to your compiler flags.

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