Why type cast a void pointer?

Question

Being new to C, the only practical usage I have gotten out of void pointers is for versatile functions that may store different data types in a given pointer. Therefore I did not type-cast my pointer when doing memory allocation.

I have seen some code examples that sometimes use void pointers, but they get type-cast. Why is this useful? Why not directly create desired type of pointer instead of a void?

You don't need to cast from `void*` because the language defines an automatic conversion from `void*` to any other type of pointer, but not all programmers are aware of this (not even all C book authors, unfortunately). — Fred Foo, Jun 07 '13 at 14:14
You have to cast a void pointer if you want to access its structure.(struct) — , Jun 07 '13 at 14:15
What exactly is your question? Why use `void` pointers or why to cast them? — alk, Jun 07 '13 at 14:25
@larsmans More precisely, the C standard defines implicit compatibility between `void *` all **data** pointer types. Function pointers are out of the game (if, however, your system is POSIX, and why would any sane programmer use anything non-POSIX, then function pointers are compatible with `void *` too). — , Jun 07 '13 at 14:33
[This may be relevant here.](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) — milleniumbug, Jun 07 '13 at 14:41
[It is “code”, not “codes.”](http://english.stackexchange.com/questions/20455/is-it-wrong-to-use-the-word-codes-in-a-programming-context) — , Jun 07 '13 at 14:46

score 22 · Answer 1 · edited May 23 '17 at 12:10

22

There are two reasons for casting a void pointer to another type in C.

If you want to access something being pointed to by the pointer ( *(int*)p = 42 )
If you are actually writing code in the common subset of C and C++, rather than "real" C. See also Do I cast the result of malloc?

The reason for 1 should be obvious. Number two is because C++ disallows the implicit conversion from void* to other types, while C allows it.

edited May 23 '17 at 12:10

Community

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answered Jun 07 '13 at 14:43

wolfgang

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But argument 2 is invalid: one should not compile C code with a C++ compiler. – Jun 07 '13 at 14:45
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@H2Co3: True, but sometimes you have code defined in header files in macros or inline functions which is meant to be usable in both C and C++ code which includes those headers. – Adam Rosenfield Jun 07 '13 at 14:47
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@H2CO3 Nevertheless, some people do it. I was trying to explain the phenomenon, not justify it. – wolfgang Jun 07 '13 at 14:49
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3rd reason: You are forced by a brain-dead coding guideline. s/implicit cast/implicit conversion/ – undur_gongor Jun 07 '13 at 14:57
@undur: "I was just following orders" doesn't count. Good catch on the cast/conversion thing, though. – wolfgang Jun 07 '13 at 14:59
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Reason 3: To tell people looking at your code that this area of memory is of that type, since `a->b = malloc(sizeof(*a->b))` doesn't tell anything while `a->b = (struct Monster*)malloc(sizeof(*a->b))` does. – xryl669 Jan 13 '22 at 11:34
@xryl669 Completely disagree! Record like **a->b = malloc(sizeof(*a->b))** tell you: allocating memory of size needed for **b** and assigning it to **b**. It is perfect and clear as for me. ;) – tvorez Jul 10 '23 at 20:50

score 16 · Answer 2 · answered Jun 07 '13 at 14:35

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You need to cast void pointers to something else if you want to dereference them, for instance you get a void pointer as a function parameter and you know for sure this is an integer:

void some_function(void * some_param) {
    int some_value = *some_param; /* won't work, you can't dereference a void pointer */
}

void some_function(void * some_param) {
    int some_value = *((int *) some_param); /* ok */
}

answered Jun 07 '13 at 14:35

Guillaume

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You don't *need to;* a more readable (preferred by me) approach is `int *tmp = param; int val = *tmp;`. – Jun 07 '13 at 14:38
@H2CO3 You have to if the pointer is a void type. As is in Guillaume's example. – Jun 07 '13 at 14:42
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@Armin You don't **have to,** I repeat - the cast can be avoided if you assign the pointer to a temporary. – Jun 07 '13 at 14:43
@H2CO3 But you **have to** if you don't use a temp , as is shown in his example. – Jun 07 '13 at 14:44
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@Armin That's right, but that's not what I'm saying - I rather suggest that it's not globally inevitable. – Jun 07 '13 at 14:45

Why type cast a void pointer?

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