i got the following:
I am a bit confused with the malloc. Why do you have to do this, (frac *) , before it??Code:typedef struct { int n, d; } frac; int main(void) { frac *f1; f1 = (frac *) malloc(sizeof(frac));
Pointers to structures
i got the following:
I am a bit confused with the malloc. Why do you have to do this, (frac *) , before it??Code:typedef struct { int n, d; } frac; int main(void) { frac *f1; f1 = (frac *) malloc(sizeof(frac));
You don't. In fact, you shouldn't.
(If your compiler is telling you you should, that's because you're using a C++ compiler.)
Or perhaps you forgot to #include <stdlib.h>.(If your compiler is telling you you should, that's because you're using a C++ compiler.)
dwk
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Even that is questionable since most modern compilers will implicitly accept malloc() without including stdlib.h. Though I applaud dwks for being so thorough and properOriginally Posted by dwks
Someone once told me that it did not matter if you did, because there are always tools to spot the bugs in your code.
So to answer the question: In C, you do not have to do it. You can, but you do not have to.
Is it good practice to do it? Debatable. I am sure many would have their own opinions.
Explicitly casting the return of malloc() is like saying, "I don't trust the language." The whole point (in C, not C++) of void * is that it can be silently converted to any other pointer type -- and any pointer type can silently convert to void *. Using an explicit cast is like denying this aspect of the language, and IMHO indicates a misunderstanding of the point of void * in the first place.
Calling functions without prototypes is just wrong. Including stdlib.h isn't optional.
Code://try //{ if (a) do { f( b); } while(1); else do { f(!b); } while(1); //}
I am just saying that IMHO, since void* can be anything (as you say, any pointer type can be implicitly converted to void*), so that does not necessarily mean that void* is the type you are assigning to. Maybe it is merely my C++ background.
That casting the return can mask the call without prototypes is just wrong and it is bad compiler, but I concede that that is not the aspect I am worried about.
Anyhow. I am just pointing out that, no, it is not wrong to cast the return, and no, it is not wrong to not do it either. In C, it is a taste thing.
200% agreed.Calling functions without prototypes is just wrong. Including stdlib.h isn't optional.
I never once said the word "optional" I said I appreciate his use of proper coding.
