richw@rosevax.Rosemount.COM (Rich Wagenknecht) (01/07/88)
Assuming I have a structure like:
struct junk {
type element1;
type element2;
}
(The elements in the structure are not important.)
And I have a pointer:
struct junk *junk_ptr;
How do I type cast my void pointer from malloc? The following doesn't seem
to work.
junk_ptr = (struct junk *)malloc( sizeof(struct junk));
Is the above statement possible? If not, any suggestions?
Thanks for your help,
--
Rich W
richw@rosevax.Rosemount.COM
Disclaimer: My views may not represent my own much less my employer's.
gwyn@brl-smoke.ARPA (Doug Gwyn ) (01/09/88)
In article <3834@rosevax.Rosemount.COM> richw@rosevax.Rosemount.COM (Rich Wagenknecht) writes: >junk_ptr = (struct junk *)malloc( sizeof(struct junk)); That's the right method. Two comments: (a) Make sure you properly declare malloc -- don't let it default to int-valued. (b) Ignore the "lint" warnings about possible pointer alignment problems; "lint" doesn't understand that malloc values are always pessimally aligned.