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Marc Boyer wrote:

> Here is a well defined program:
>
> #include <stdio.h>
> #include <string.h>
>
> int main(){
> void* p= &p;
> if ( strlen("abc") < sizeof(p) ){
> memcpy(p, "abc", strlen("abc")+1 );
> puts( (char*) p );

It's not defined and these issues have
already been gone over in this thread.

p is an indeterminate pointer after the memcpy call.

Accessing the value of p at that point,
as the cast operator does, is undefined.

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

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

int main(void)
{
void *p = &p;

if (strlen("abc") < sizeof(p)) {
strcpy(p, "abc");
puts((char *)&p);
} else {
puts("Pointers too small");
}
return 0;
}

--
pete

Le 07-10-2005, pete <> a écrit*:
> Marc Boyer wrote:
>
>> Here is a well defined program:
>>
>> #include <stdio.h>
>> #include <string.h>
>>
>> int main(){
>> void* p= &p;
>> if ( strlen("abc") < sizeof(p) ){
>> memcpy(p, "abc", strlen("abc")+1 );
>> puts( (char*) p );
>
> It's not defined and these issues have
> already been gone over in this thread.
>
> p is an indeterminate pointer after the memcpy call.

Yes, sorry. I ve not posted the right version.

> Accessing the value of p at that point,
> as the cast operator does, is undefined.

Of course.

> #include <stdio.h>
> #include <string.h>
>
> #include <stdio.h>
> #include <string.h>
>
> int main(void)
> {
> void *p = &p;
>
> if (strlen("abc") < sizeof(p)) {
> strcpy(p, "abc");
> puts((char *)&p);
> } else {
> puts("Pointers too small");
> }
> return 0;
> }

Yes


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À vélo, prendre une rue à contre-sens est moins dangeureux
que prendre un boulevard dans le sens légal. À qui la faute ?