Pointer before Malloc

Question

what is different between this:

tElemPtr novyPrvok = (tElemPtr *)malloc(sizeof(tElemPtr));

and this:

tElemPtr novyPrvok = malloc(sizeof(tElemPtr));

I want to use it in InsertFirst function for inserting first element in the beginning of the List. Because, if I use without that pointer in front of malloc, Xcode tells me it is good, but I cant use "novyPrvok->data"

void InsertFirst (tList *L, int val) {  
    tElemPtr novyPrvok = (tElemPtr *)malloc(sizeof(tElemPtr));

    if(novyPrvok == NULL)
        Error();

    novyPrvok->data = val;
    novyPrvok->ptr = L->First;
    L->First = novyPrvok;              
}

you meant tElemPtr *novyPrvok = (tElemPtr *)malloc(sizeof(tElemPtr)); ? — OznOg, Oct 01 '18 at 18:54
typedefing pointers is a horrible habit, leading to difficult to read code. Do bot do it. The only exeption: function pointers — 0___________, Oct 01 '18 at 18:56
@P__J__ Function pointers are no exception. If you don't like the syntax, you can just typedef functions. — melpomene, Oct 01 '18 at 18:57
Casting malloc()'s return value is not something that needs to be, or should be, done. http://c-faq.com/malloc/mallocnocast.html — Shawn, Oct 01 '18 at 18:57
@melpomene function pointers are usually used another way then the "normal" pointers. — 0___________, Oct 01 '18 at 18:58
@Shawn it is a prehistoric opinion. C standard does not allow functions with no prototypes (it is the error now not the warning - and you cant compile the code the program with no included ), so that link is about 20years outdated. Obsolete opinion. Now it only question of taste — 0___________, Oct 01 '18 at 19:01
They are similar in a way that neither former nor the latter makes any sense. Check your indirection levels carefully. — AnT stands with Russia, Oct 01 '18 at 19:07
@P__J__ You'll find that many compilers, including gcc and clang, will complain about it but still compile programs that use malloc without including the appropriate header. — Shawn, Oct 01 '18 at 19:13
@Shawn just use the correct compile option to set the modern language standard. — 0___________, Oct 01 '18 at 19:38
@P__J__ Even with C11 (Which is the default in recent versions of gcc, not sure about clang), they will compile such code. Don't believe me? Try it yourself with a file with nothing but `int main(void) { int *x = malloc(sizeof *x); }` in it and compile with `gcc -std=c11`. You'll get some warnings, but it *will* compile (Tested with gcc 7 and 8, and clang 6). If that's not enough to demonstrate that, yes, compilers still let you use functions without declaring them first, well... I'm done. Peace. — Shawn, Oct 01 '18 at 20:37
hi, please can you tell me, how can I acces data, when I have it declared linke pointer? (*tElemPtr) typedef struct tElem { struct tElem *ptr; int data; } *tElemPtr; typedef struct { tElemPtr Act; tElemPtr First;} tList; — Timco Vanco, Oct 02 '18 at 17:33

OznOg · Answer 1 · 2018-10-01T19:13:45.120

1

no need to cast malloc return -> Do I cast the result of malloc?

BTW your code seems wrong; as tElemPtr seems to be a typedef on a pointer, I would expect the malloc to be:

tElemPtr novyPrvok = malloc(sizeof(*novyPrvok));

edited Oct 01 '18 at 19:13

answered Oct 01 '18 at 19:04

OznOg

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No need for `()`. Could use `sizeof *novyPrvok` instead of `sizeof(*novyPrvok)`. – chux - Reinstate Monica Oct 01 '18 at 19:14

score -1 · Answer 2 · answered Oct 01 '18 at 18:58

-1

your malloc is wrong, its only allocating space for a pointer. After that everything is Undefined Behavior. Should be

 tElemPtr novyPrvok = (tElemPtr *)malloc(sizeof(*tElemPtr));

answered Oct 01 '18 at 18:58

pm100

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You're right, but the code sample isn't. `tElemPtr novyPrvok = malloc(sizeof *novyPrvok);` – Shawn Oct 01 '18 at 19:00
You're still only allocating space for a pointer. Or at least you would be if `*tElemPtr` wasn't a syntax error. – melpomene Oct 01 '18 at 19:00

Pointer before Malloc

2 Answers2