Bus Error asprintf. OS X 10.3.6. gcc 3.3

[Neil Dunn]

Hi,

I don't seem to be able to get asprintf to work when compiling under
OS X.

Using the following code:

#include <string.h>
#include <stdio.h>

int main() {

char **w;

asprintf(w,"foo o");

printf("%s\n",* w);

return 0;

}

and compiling using gcc with the -lc and -Wall flags, I get no erros.

The program runs spitting out a Bus Error. The same code on a Linux
machine gcc 3.2.2 compiles and runs as expected.

Any help would be great.

Neil

[Robert Gamble]

On Sun, 21 Nov 2004 16:44:47 -0600, Merrill & Michele wrote:

Thank you for your response. I take objection that I don't make serious
efforts to communicate and that the indentifiers are inane. Only MANGO has
no pedigree in my posts, and what would I have called it?
ARBITRARY_IDENT IFIER_
FOR_PEDAGOGICAL _PURPOSES_HAVIN G_NO_GREAT_SIGN IFICANCE?
You are free to take objection but if you could reasonably format your
code when you post you will be more likely to receive positive responses.
As far as identifier names, you are notorious for using wildly ungermane
names.
[code snipped] There's at least a couple fatal flaws you've discerned here.

As you can see from my comments, there are numerious things wrong with
your code. Fixing the above issues and making some other modification
based on your implied intentions yields the following:

#include <stdio.h>
#define FAMSIZ 7
#define MANGO 3

static int *increase(int);

int buysfor[FAMSIZ];

int main(void) {
int i;
int *ptr;
/* initialize */
for (i = 0; i < FAMSIZ; ++i) buysfor[i] = i;
ptr = increase(MANGO) ;
printf("value returned from increase: %3d\n", *ptr);
return 0;
}

static int *increase(int mango) {
int i;
for (i = 0; i < FAMSIZ; ++ i) buysfor[i] += mango;
return &buysfor[0];
}
Your program builds and behaves as I would think, but does it not fall under
the same criticism that buysfor[] would not be visible in the *increase
function call? Does this not fall under the worst type of undefined
behavior, namely, the type that DOESN'T cause Scott Nudds to fly out of your
hard drive?

Um...no. Notice that I declared buysfor outside of a function definition
or block. The scope for buysfor is until the end of the translation unit
(file scope).
#include <stdio.h>
#define FAMSIZ 7
#define MANGO 3

static int *increase(int);

int buysfor[FAMSIZ];

int main(void) {
int i;
int *ptr;
/* initialize */
for (i = 0; i < FAMSIZ; ++i) buysfor[i] = i;
ptr = increase(MANGO) ;
printf("value returned from increase: %8d\n", ptr);
return 0;
}

static int *increase(int mango) {
int i;
for (i = 0; i < FAMSIZ; ++ i) buysfor[i] += mango;
return &buysfor[2];
Furthermore, it was my intention to actually see the address in memory of
some pointer. For that, I'm uncertain what the syntax would be on a printf:
printf("address is %d\n ", ptr); ? MPJ

printf("address is %p\n",(void *)ptr);

Rob Gamble

[Merrill & Michele]

"Robert Gamble"

MPJ:

As you can see from my comments, there are numerious things wrong with
your code. Fixing the above issues and making some other modification
based on your implied intentions yields the following:

#include <stdio.h>
#define FAMSIZ 7
#define MANGO 3

static int *increase(int);

int buysfor[FAMSIZ];

I missed that this declaration was at file scope.

Um...no. Notice that I declared buysfor outside of a function definition
or block. The scope for buysfor is until the end of the translation unit
(file scope). Furthermore, it was my intention to actually see the address in memory of some pointer. For that, I'm uncertain what the syntax would be on a printf: printf("address is %d\n ", ptr); ?

printf("address is %p\n",(void *)ptr);

I can well understand if you've had enough of this problem. This printf
certainly works and table B-1 shows that the character p after % means that
the compiler's looking for a void *. Doesn't it seem like ptr is already a
pointer and that therefore void * ptr would be a creature of type void ** ?
MPJ

[Robert Gamble]

On Sun, 21 Nov 2004 21:55:43 -0600, Merrill & Michele wrote:

"Robert Gamble" [snip] printf("address is %p\n",(void *)ptr);

I can well understand if you've had enough of this problem. This printf
certainly works and table B-1 shows that the character p after % means that
the compiler's looking for a void *. Doesn't it seem like ptr is already a
pointer and that therefore void * ptr would be a creature of type void ** ?
MPJ

No.

Robert Gamble