Comparing two ways of initializing an int pointer

Question

Suppose I have a pointer

int *x;

Then I need to let the pointer point to an int of value, say 42.

Should I do this:

   *x = 42;

or this:

  int y = 42;
  x = &y;

? What is the usual practice?

I assume you are doing `*x=42` after a call to `x = malloc`, aren't you? — IrAM, Dec 08 '20 at 10:25
I was thinking that because as long as x is non-null, *x=42 should work even without malloc. Right? It would be like we just put 42 to a random memory case, but what would be wrong with it? — zell, Dec 08 '20 at 10:37
but random memory cannot guarantee it will be successfully stored, that would be **UB** (undefined behavior) — IrAM, Dec 08 '20 at 10:43
Regarding the common misconception that you can somehow store data inside pointers, see [Crash or “segmentation fault” when data is copied/scanned/read to an uninitialized pointer](https://stackoverflow.com/questions/37549594/crash-or-segmentation-fault-when-data-is-copied-scanned-read-to-an-uninitializ). — Lundin, Dec 08 '20 at 11:15

Vlad from Moscow · Accepted Answer · 2020-12-08T10:33:51.840

2

After this declaration

int *x;

the pointer x either is equal to NULL (if it is declared outside any function) or has an indeterminate value. So dereferencing the pointer like

*x = 42;

invokes undefined behavior.

You can write either

int y = 42;
x = &y;

or

x = malloc( sizeof( int ) );
*x = 42;

edited Dec 08 '20 at 10:33

answered Dec 08 '20 at 10:22

Vlad from Moscow

Suppose x is a local variable. Does "int *x; *x = 42;" is still undefined behavior? I was thinking that because x is guaranteed to be non-null, this one should work even without malloc, no? – zell Dec 08 '20 at 10:34
@zell The pointer is uninitialized and has indeterminate value. So you may not dereference it. – Vlad from Moscow Dec 08 '20 at 10:43
@zell: After `int *x;`, `x` is not guaranteed to be non-null. What made you think it was guaranteed to be non-null? – Eric Postpischil Dec 08 '20 at 11:19
Check the return value of `malloc()` before using it. – 12431234123412341234123 Dec 08 '20 at 12:23

score 0 · Answer 2 · answered Dec 08 '20 at 10:22

0

When you do:

int *x = 42;

you assigne the pointeur x to the memory case 42, that gonna make a segmentation fault. The right way to do what you wan't is:

int y = 42;
x = &y;

answered Dec 08 '20 at 10:22

DipStax

`int *x = 42;` isn't valid C and the OP isn't doing that either, so why did you bring that case up? – Lundin Dec 08 '20 at 11:15
It does not necessary cause a segmentation fault, many platforms do not even have the concept of segmentation fault. – 12431234123412341234123 Dec 08 '20 at 12:24

MED LDN · Answer 3 · 2020-12-08T12:30:56.890

0

x gets the adress of y (&y)

#include <stdio.h>

int main()
{
      int y=42;
      int x=&y;
      printf("\nY = %d is at adress X = %d \n",y,x);
}

edited Dec 08 '20 at 12:30

answered Dec 08 '20 at 10:23

MED LDN

Your code causes UB. `"%d"` expects a `int`, not a `int *` variable. – 12431234123412341234123 Dec 08 '20 at 12:25

3 Answers3