Re: [PATCH 1/3] ipc: avoid dereference of null pointer and quiet the GCC warning about uninitialized variable

[Ivan Shmakov]

>>>>> Marin Ramesa <mpr@hi.t-com.hr> writes:
>>>>> On 18.12.2013 10:46:40, Richard Braun wrote:

[...]

 >> No, that's wrong.  The && and || operators are guaranteed to be
 >> evaluated left-to-right, and yield if the first operand compares
 >> equal to 0.  And that's exactly why this check against NULL is done
 >> first.

 > In the expression (!a && !b), if !a equals 0, the compiler must check
 > !b == 0 in order to return [FALSE].  If !a equals 0, that means the
 > entry is a null pointer, and evaluation of !b is a dereference of a
 > null pointer.

        Well, when explaining these operators to the students, I’d
        rather call them “conditionals” (along with A ? B : C and ‘if’.)
        Specifically:

        • the A && B operator evaluates the A expression; if the result
          is zero, zero is returned; if the result is non-zero, the B
          expression is evaluated, and its result is returned;

        • the A || B operator evaluates the A expression; if the result
          is non-zero, it is returned; if the result is zero, the B
          expression is evaluated, and its result is returned.

        Essentially, A && B is equivalent to (A ? B : 0), while A || B
        is equivalent to (A ? A : B), except that in latter case, A is
        evaluated only once.  (Thus A++ || B doesn’t result in A being
        incremented twice, as would be in the case of A++ ? A++ : B.)

        One may wish to check the following simplictic C code.

#include <stdio.h>

int
main ()
{
  int a = 0, b = 42;

  int c
    = a && puts ("a is true");
  int d
    = a || puts ("a is false");
  int e
    = b && puts ("b is true");
  int f
    = b || puts ("b is false");

  /* . */
  return 0;
}

        Note that only some of the puts () calls are in fact evaluated.

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