pointers

[RoSsIaCrIiLoIA]

If sizeof(unsigned)=4 is this allowed?

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

int main(void)
{unsigned *i;
i = malloc(1);
return 0;
}

and this

int main(void)
{unsigned *i;
i = malloc(1);
*i=98;
return 0;
}

 

[Emmanuel Delahaye]

RoSsIaCrIiLoIA wrote on 28/07/04 :

If sizeof(unsigned)=4 is this allowed?

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

int main(void)
{unsigned *i;
i = malloc(1);

Yes, as far as you don't use 'i'. (Is it useful is another story).

return 0;
}

and this

int main(void)
{unsigned *i;
i = malloc(1);
*i=98;

No. Using 'i' invokes an undefined behaviour.

return 0;
}

Two more things : malloc() could fail, and you are supposed to free the
allocated memory...

 

[Fred L. Kleinschmidt]

If sizeof(unsigned)=4 is this allowed?

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

int main(void)
{unsigned *i;
i = malloc(1);
return 0;
}

and this

int main(void)
{unsigned *i;
i = malloc(1);
*i=98;
return 0;
}

Why bother? Why not just use
i = malloc(sizeof(*i));
This is the same speed and much more useful - it works on all platforms,
i.e., when sizeof(unsigned) is 4 or 8 or anything else, and you can
change the type later and not worry about haviong to change all of the
malloc's.

 

[Keith Thompson]

RoSsIaCrIiLoIA wrote on 28/07/04 :

Yes, as far as you don't use 'i'. (Is it useful is another story).


No. Using 'i' invokes an undefined behaviour.

[...]

Using 'i' (e.g., by comparing it to NULL) should be ok; using *i, as
the quoted code fragment does, invokes undefined behavior.

 

[RoSsIaCrIiLoIA]

If sizeof(unsigned)=4 is this allowed?

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

int main(void)
{unsigned *i;
i = malloc(1); free(i);
return 0;
}

and this

int main(void)
{unsigned *i;
i = malloc(1);
*i=98; free(i);
return 0;
}