question on storage duration

[Mark]

Hello

given the very simple code:

#include <stdlib.h>
#include <stdio.h>

int g(void)
{
int i = 2;
return i;
}

int *f(void)
{
int *p;
p = malloc(sizeof *p);
/* skip result checking for the example's brevity */
*p = 2;
return p;
}

int main(void)
{
int i;
int *j;

i = g();
j = f();
printf("g()=%d, f()=%d\n", i, *j);
return 0;
}

Here is my reasoning.

In g() I declare automatic variable 'i' and initialize it with 2. The
storage, occupied by the variable, exists till the function returns. So we
can safely return the value. In function f() 'p' is as well automatic
variable, the value of which points to some storage, obtained by 'malloc()'
and this storage exists as long as the program runs. In main() local pointer
'j' is assigned to the returned value of g() and we can perfectly access the
value of that pointer.

Am I correct?

 

[Ben Bacarisse]

given the very simple code:

#include <stdlib.h>
#include <stdio.h>

int g(void)
{
int i = 2;
return i;
}

int *f(void)
{
int *p;
p = malloc(sizeof *p);
/* skip result checking for the example's brevity */
*p = 2;

The code that checks:

if (p)
*p = 2;

is shorter than the explanation that you want to be brief!

return p;
}

int main(void)
{
int i;
int *j;

i = g();
j = f();
printf("g()=%d, f()=%d\n", i, *j);
return 0;
}

Here is my reasoning.

In g() I declare automatic variable 'i' and initialize it with 2. The
storage, occupied by the variable, exists till the function
returns. So we can safely return the value. In function f() 'p' is as
well automatic variable, the value of which points to some storage,
obtained by 'malloc()' and this storage exists as long as the program
runs. In main() local pointer 'j' is assigned to the returned value of
g() and we can perfectly access the value of that pointer.

Am I correct?

Yes.

 

[Nick Keighley]

given the very simple code:

#include <stdlib.h>
#include <stdio.h>

int g(void)
{
    int i = 2;
    return i;
}

int *f(void)
{
    int *p;
    p = malloc(sizeof *p);
    /* skip result checking for the example's brevity */
    *p = 2;
    return p;
}

int main(void)
{
    int i;
    int *j;

    i = g();
    j = f();
    printf("g()=%d, f()=%d\n", i, *j);
    return 0;
}

Here is my reasoning.

In g() I declare automatic variable 'i' and initialize it with 2. The
storage, occupied by the variable, exists till the function returns. So we
can safely return the value. In function f() 'p' is as well automatic
variable, the value of which points to some storage, obtained by 'malloc()'
and this storage exists as long as the program runs.

or until free() is called for the memory

In main() local pointer
'j' is assigned to the returned value of g() and we can perfectly access the
value of that pointer.

Am I correct?

yes

 

[Richard Bos]

The code that checks:

if (p)
*p = 2;

is shorter than the explanation that you want to be brief!

No; the code that does _only_ that is completely useless, and indeed
dangerous, as a check for malloc() failure. You'd need to add code in
main(), as well, so as not to dereference a null pointer.

This is a mistake that is often made, in fact: check for null or error
returns, and then do nothing with the result of that check. The effect
is that the error, instead of being caught, is only propagated up a
level. It's even more dangerous than not checking at all.

Richard