The void type in C seems to be strange from various different situations. Sometimes it behaves like a normal object type, such as int or char, and sometimes it just means nothing (as it should).
Look at my snippet. First of all, it seems strange that you can declare a void object, meaning you just declare nothing.
Then I created an int variable and casted its result to void, discarding it:
If an expression of any other type is evaluated as a void
expression, its value or designator is discarded. (ISO/IEC 9899:201x, 6.3.2.2 void)
I tried to call my function with a void cast, but my compiler gave me (Clang 10.0):
error: too many arguments to function call, expected 0, have 1
So the void in a prototype means nothing, and not the type void.
But then, I created a pointer to void, dereferenced it, and assigning the “result” to my int variable. I got the “incompatible type” error. That means the void type does exist here.
extern void a; // Why is this authorised ???
void foo(void); // This function takes no argument. Not the 'void' type.
int main(void)
{
int a = 42;
void *p;
// Expression result casted to 'void' which discards it (per the C standard).
(void)a;
// Casting to 'void' should make the argument inexistant too...
foo((void)a);
// Assigning to 'int' from incompatible type 'void': so the 'void' type does exists...
a = *p;
// Am I not passing the 'void' type ?
foo(*p);
return 0;
}
Is void an actual type, or a keyword to means nothing ? Because sometimes it behaves like the instruction “nothing is allowed here”, and sometimes like an actual type.
EDIT: This questions is NOT a duplicate. It is a purely about the semantics of the void type. I do not want any explanation about how to use void, pointers to void or any other things. I want an answer per the C standard.
In C language the void type has been introduced with the meaning of 'don't care' more than 'null' or 'nothing', and it's used for different scopes.
The void keyword can reference a void type, a reference to void, a void expression, a void operand or a void function. It also explicitly defines a function having no parameters.
Let's have a look at some of them.
The void type
First of all void object exists and have some special properties, as stated in ISO/IEC 9899:2017, §6.2.5 Types:
The void type comprises an empty set of values; it is an incomplete object type that cannot be completed.
Pointers
The more useful reference to void, or void *, is a reference to an incomplete type, but itself is well defined, and then is a complete type, have a size, and can be used as any other standard variable as stated in ISO/IEC 9899:2017, §6.2.5 Types:
A pointer to void shall have the same representation and alignment requirements as a pointer to a character type.
Similarly, pointers to qualified or unqualified versions of compatible types shall have the same representation and alignment requirements.
All pointers to structure types shall have the same representation and alignment requirements as each other.
All pointers to union types shall have the same representation and alignment requirements as each other.
Pointers to other types need not have the same representation or alignment requirements.
Casting to void
It can be used as cast to nullify an expression, but allowing the completion of any side effect of such expression. This concept is explained in the standard at ISO/IEC 9899:2017, §6.3 Conversions, §6.3.2.2 void:
The (nonexistent) value of a void expression (an expression that has type void) shall not be used in any way, and implicit or explicit conversions (except to void) shall not be applied to such an expression.
If an expression of any other type is evaluated as a void expression, its value or designator is discarded. (A void expression is evaluated for its side effects.)
A practical example for the casting to void is its use to prevent warning for unused parameters in function definition:
int fn(int a, int b)
{
(void)b; //This will flag the parameter b as used
... //Your code is here
return 0;
}
The snippet above shows the standard practice used to mute compiler warnings. The cast to void of parameter b acts as an effective expression that don't generate code and marks b as used preventing compiler complains.
void Functions
The paragraph §6.3.2.2 void of the standard, covers also some explanation about void functions, that are such functions that don't return any value usable in an expression, but functions are called anyway to implement side effects.
void pointers properties
As we said before, pointers to void are much more useful because they allow to handle objects references in a generic way due to their property explained in ISO/IEC 9899:2017, §6.3.2.3 Pointers:
A pointer to void may be converted to or from a pointer to any object type.
A pointer to any object type may be converted to a pointer to void and back again; the result shall compare equal to the original pointer.
As practical example imagine a function returning a pointer to different objects depending on input parameters:
enum
{
FAMILY, //Software family as integer
VERSION, //Software version as float
NAME //Software release name as char string
} eRelease;
void *GetSoftwareInfo(eRelease par)
{
static const int iFamily = 1;
static const float fVersion = 2.0;
static const *char szName = "Rel2 Toaster";
switch(par)
{
case FAMILY:
return &iFamily;
case VERSION:
return &fVersion;
case NAME:
return szName;
}
return NULL;
}
In this snippet you can return a generic pointer that can be dependent on input par value.
void as functions parameter
The use of void parameter in functions definitions was introduced after the, so called, ANSI-Standard, to effectively disambiguate functions having variable number of arguments from functions having no arguments.
From standard ISO/IEC 9899:2017, 6.7.6.3 Function declarators (including prototypes):
The special case of an unnamed parameter of type void as the only item in the list specifies that the function has no parameters.
Actual compilers still support function declaration with empty parenthesis for backward compatibility, but this is an obsolete feature that will eventually be removed in future release of standard. See Future directions - §6.11.6 Function declarators:
The use of function declarators with empty parentheses (not prototype-format parameter type declarators) is an obsolescent
feature.
Consider the following example:
int foo(); //prototype of variable arguments function (backward compatibility)
int bar(void); //prototype of no arguments function
int a = foo(2); //Allowed
int b = foo(); //Allowed
int c = bar(); //Allowed
int d = bar(1); //Error!
Now resembling your test, if we call the function bar as follows:
int a = 1;
bar((void)a);
Triggers an error, because casting to void an object doesn't null it. So you are still trying to pass a void object as parameter to a function that don't have any.
Side effects
As requested this is a short explain for side effects concept.
A side effect is whichever alteration of objects and values derived from the execution of a statement, and which are not the direct expected effect.
int a = 0;
(void)b = ++a;
In the snippet above the void expression lose the direct effect, assigning b, but as side effect increase the value of a.
The only reference, explaining the meaning, in the standard can be found in 5.1.2.3 Program execution:
Accessing a volatile object, modifying an object, modifying a
file, or calling a function that does any of those operations are all
side effects, which are changes in the state of the execution
environment.
Evaluation of an expression in general includes both value
computations and initiation of side effects.
void is a type. Per C 2018 6.2.5 19, the type has no values (the set of values it can represent is empty), it is incomplete (its size is unknown), and it cannot be completed (its size cannot be known).
Regarding extern void a;, this does not define an object. It declares an identifier. If a were used in an expression (except as part of a sizeof or _Alignof operator), there would have to be a definition for it somewhere in the program. Since there cannot a definition of void object in strictly conforming C, a cannot be used in an expression. So I think this declaration is allowed in strictly conforming C but is not useful. It might be used in C implementations as an extension that allows getting the address of an object whose type is not known. (For example, define an actual object a in one module, then declare it as extern void a; in another module and use &a there to get its address.)
The declaration of functions with (void) as a parameter list is a kludge. Ideally, () might be used to indicate a function takes no parameters, as is the case in C++. However, due to the history of C, () was used to mean an unspecified parameter list, so something else had to be invented to mean no parameters. So (void) was adopted for that. Thus, (void) is an exception to the rules that would say (int) is for a function taking an int, (double) is for a function taking a double, and so on—(void) is a special case meaning that a function takes no parameters, not that it takes a void.
In foo((void) a), the cast does not make the value “not exist.” It converts a to the type void. The result is an expression of type void. That expression “exists,” but it has no value and cannot be used in an expression, so using it in foo((void) a) results in an error message.
From C Standard#6.2.5p19:
19 The void type comprises an empty set of values; it is an incomplete object type that cannot be completed.
This indicate that the void type exists.
Doubt 1:
void foo(void); // This function takes no argument. Not the 'void' type.
Correct.
From C Standard#6.7.6.3p10 [emphasis mine]:
10 The special case of an unnamed parameter of type void as the only item in the list specifies that the function has no parameters.
This is a special case they had to add to the language syntax because void foo(); already meant something different (void foo(); doesn't specify anything about foo's parameters). If it weren't for the old meaning of void foo();, void foo(); would have been the syntax to declare a no-argument function. You can't generalize anything from this. It's just a special case.
Doubt 2:
// Casting to 'void' should make the argument inexistant too...
foo((void)a);
No, it will not because void is also an object type though it is incomplete.
Doubt 3:
// Assigning to 'int' from incompatible type 'void': so the 'void' type does exists...
a = *p;
Yes, it does exist and hence the compiler is reporting error on this statement.
Doubt 4:
// Am I not passing the 'void' type ?
foo(*p);
Declaration of foo() function:
void foo(void);
^^^^
The void in parameter list indicates that function will not take any argument because it has been declared with no parameters.
Just for reference, check this from C Standard#5.1.2.2.1p1 [emphasis mine]:
1 The function called at program startup is named main. The implementation declares no prototype for this function. It shall be defined with a return type of int and with no parameters:
int main(void) { /* ... */ }
^^^^
Doubt 5:
extern void a; // Why is this authorised ???
This is authorized because void is a valid type and it is just a declaration. No storage will allocate to a.
In C, void can't be considered as a data type, it is a keyword used as a placeholder in place of a data type to show that actually there is no data. Hence this
void a;
is not valid.
while here
void foo(void);
void keyword is used to inform to the compiler that foo is not going to take any input argument nor it has return type.
In below case
int a = 42;
void *p;
a = *p; /* this causes error */
a = *p; is wrong because you can't dereference void pointer directly, you need to perform proper type casting first. for e.g
a = *(int*)p; /* first typecast and then do dereference */
Also this
foo(*p);
is wrong because of two reason,
firstly foo() doesn't expects any argument.
secondly you can't do *p as p is void pointer. Correct one is foo(*(int*)p); if foo() declaration is void foo(int);.
Note that this
(void)a;
doesn't do anything so your compiler might not giving any warning but when you do like
int b = (void)a;
compiler won't allow as void is not consider as data type.
Finally this
extern void a; // Why is this authorised ???
this is just a declaration not definition, a doesn't exist until you define it, since a is having extern storage class, you need to define somewhere & when you are going define like
a = 10;
compiler throws a error as
error: ‘a’ has an incomplete type
From C standard 6.2.5 Types
The void type comprises an empty set of values; it is an
incomplete object type that cannot be completed.
6.3.2.2 void
The (nonexistent) value of a void expression (an expression that has
type void) shall not be used in any way, and implicit or explicit
conversions (except to void) shall not be applied to such an
expression. If an expression of any other type is evaluated as a
void expression, its value or designator is discarded. (A void
expression is evaluated for its side effects.)
6.3.2.3 Pointers
A pointer to void may be converted to or from a pointer to any
object type. A pointer to any object type may be converted to a
pointer to void and back again; the result shall compare equal to the
original pointer.
A storage-class specifier or type qualifier modifies the keyword
void as a function parameter type list (6.7.6.3).
An attempt is made to use the value of a void expression, or an
implicit or explicit conversion (except to void) is applied to a
void expression (6.3.2.2).
First of all, it seems strange that you can declare a void object, meaning you just declare nothing.
void is an incomplete object type that cannot be completed. This mostly defines its uses in regular contexts, i.e. contexts that do not provide special treatment for void. Your extern declaration is one of such regular contexts. It is OK to use an incomplete data type in a non-defining declaration.
However, you will never be able to provide a matching definition for that declaration.
So the void in a prototype means nothing, and not the type void.
Correct. The parameter must be unnnamed. And the (void) combination is given special treatment: it is not one parameter of type void, but rather no parameters at all.
But then, I created a pointer to void, dereferenced it, and assigning the “result” to my int variable. I got the “incompatible type” error. That means the void type does exist here.
No. It is illegal to apply unary * operator to a void * pointer. Your code is invalid for that reason already. Your compiler issued a misleading diagnostic message. Formally, diagnostic messages are not required to properly describe the root of the problem. The compiler could've just said "Hi!".
Is void an actual type, or a keyword to means nothing ?
It is a type. It is an incomplete object type that cannot be completed.
Related
I am wondering if it is possible to include the restrict keyword only in the function definition and not in the function declaration like so:
void foo(char *bar);
void foo(char * restrict bar)
{
// do something
}
Since foo only takes one argument, any pointer aliasing would have to take place inside foo. There would be no need for the person calling the function to know about the restrict modifier. Would it be fine to omit the keyword in only the function declaration, just like with const?
You may use restrict on parameters in function declarations whether they are definitions or not, as it is allowed by the C grammar and there is no rule against it. However, they have no effect to the compiler in declarations that are not definitions. This is because 6.5.2.2 7 says qualifiers are removed when passing arguments to functions with prototypes:
… the arguments are implicitly converted, as if by assignment, to the types of the corresponding parameters, taking the type of each parameter to be the unqualified version of its declared type.
Thus, if a function declaration has a parameter of type int * restrict a, whatever argument you pass is converted to the unqualified type, int *.
Further, two otherwise identical function declarations are compatible even if the qualifiers on parameters are changed, because C 2018 6.7.6.3 15 says:
… (In the determination of type compatibility and of a composite type, … each parameter declared with qualified type is taken as having the unqualified version of its declared type.)
However, this applies only to the parameter itself. The parameter is not affected by a restrict that qualifies it. But it can point to a pointer that is restrict-qualified. For example, void foo(void * restrict *a); and void foo(void **a); declare different function types.
Although qualifiers on parameters in declarations have no effect to the compiler, they can signal to humans that the arguments are expected to conform to the restriction. Inside the function definition, the parameter is restrict-qualified, and anybody calling the function should respect that.
The next C99 source code can show you that the output of the program depends on restrict :
__attribute__((noinline))
int process(const int * restrict const a, int * const b) {
*b /= (*a + 1) ;
return *a + *b ;
}
int main(void) {
int data[2] = {1, 2};
return process(&data[0], &data[0]);
}
The software terminates with code 1 using restrict and 0 without restrict qualifier.
The compilation is done with gcc -std=c99 -Wall -pedantic -O3 main.c.
The flag -O1 do the job too.
And so on.
Is it legal to access pointers to functions with varying argument lists in via a void (*f)() pointer? The program below compiles without warnings with gcc and appears to run correctly, but is it legal C?
#include <stdio.h>
#include <stdlib.h>
typedef void funp();
static void funcall( funp* F, int args, double x)
{
switch( args)
{
case 0: F(); break;
case 1: F(x); break;
}
}
static void fun0( void)
{
printf( "zero\n");
}
static void fun1( double x)
{
printf( "one\t%f\n", x);
}
int main( )
{
funcall( (funp*)fun0, 0, 17.0);
funcall( (funp*)fun1, 1, 17.0);
return EXIT_SUCCESS;
}
I compiled this with
gcc -Wpedantic -Wall -Wextra -std=gnu11 -O2 -o ./funp funp.c
It would be undefined behavior if the nargs parameter didn't match the number of arguments the function took, but is it legal if there is a match?
In this particular case, the calls are legal.
Section 6.7.6.3p15 of the C standard spells out what makes two function type compatible (relevant part in bold):
For two function types to be compatible, both shall specify
compatible return types. Moreover, the parameter type lists, if
both are present, shall agree in the number of parameters and
in use of the ellipsis terminator; corresponding parameters
shall have compatible types. If one type has a parameter type list
and the other type is specified by a function declarator that is
not part of a function definition and that contains an empty
identifier list, the parameter list shall not have an ellipsis
terminator and the type of each parameter shall be compatible with
the type that results from the application of the default
argument promotions. If one type has a parameter type list and the
other type is specified by a function definition that contains a
(possibly empty) identifier list, both shall agree in the number
of parameters, and the type of each prototype parameter shall
be compatible with the type that results from the application
of the default argument promotions to the type of the
corresponding identifier. (In the determination of type
compatibility and of a composite type, each parameter declared
with function or array type is taken as having the adjusted type
and each parameter declared with qualified type is taken as having the
unqualified version of its declared type.)
So you have a typedef with type:
void()
And functions with type:
void(void)
void(double)
The two function definitions don't use ellipsis (...) so that satisfies the fist condition. For the second condition, let's look at what the default argument promotions are. Those are specified in section 6.5.2.2p6:
If the expression that denotes the called function has a
type that does not include a prototype, the integer promotions
are performed on each argument, and arguments that have type
float are promoted to double. These are called the default
argument promotions.
The first function has no arguments, so it is compatible. The second function has a single double argument, which matches the default argument promotions, so it is also compatible.
To give some more examples, the following functions would also be compatible:
void f1(long);
void f2(int);
But these would not:
void f3(float);
void f4(char);
void f5(short);
As another answer notes, the code you showed is valid C today. But that may change at any point in the future, due to the use of a function type with no parameter list.
6.11 Future language directions
6.11.6 Function declarators
1 The use of function declarators with empty parentheses (not
prototype-format parameter type declarators) is an obsolescent
feature.
An obsolescent feature is one that is subject to removal in future standard versions. So if you wish your code to be future proof, it's best to avoid it.
As mentioned in #StoryTeller's answer, the use of function declarators with empty parentheses is an obsolescent feature, but it can be avoided:
#include <stdio.h>
#include <stdlib.h>
typedef void funp(void);
static void funcall( funp* F, int args, double x)
{
switch( args)
{
case 0:
F();
break;
case 1:
{
typedef void fn(double);
((fn *)F)(x);
}
break;
}
}
static void fun0( void)
{
printf( "zero\n");
}
static void fun1( double x)
{
printf( "one\t%f\n", x);
}
int main( void )
{
funcall( (funp*)fun0, 0, 17.0);
funcall( (funp*)fun1, 1, 17.0);
return EXIT_SUCCESS;
}
EDIT: Changed parameter list of main to void for compliance.
In answer to the query:
"Moreover, the parameter type lists, if both are present, shall agree in the number of parameters" would seem to mean that the types of funp and of fun1 are incompatible. Is it ok to cast?
The answer is yes, it is OK to cast. From C11 draft 6.3.2.3 para 8:
A pointer to a function of one type may be converted to a pointer to a function of another type and back again; the result shall compare equal to the original pointer. If a converted pointer is used to call a function whose type is not compatible with the referenced type, the behavior is undefined.
In the code, the pointer to fun1 has been converted to a different function pointer type in the call to funcall, and converted back to the original type within funcall so can be used to call fun1.
in the following example I make a CAST of a function without arguments in a pointer to a function that should receive an argument. Assuming that it gives the desired result, is it possible that this procedure causes some malfunction?
online test: https://onlinegdb.com/SJ6QzzOKI
typedef void (*Callback)(const char*);
Callback cb;
void inserisce_cb(void* c) {
cb=c;
}
void esegue_cb(){
cb("pippo");
}
void scriveTitolo(const char* titolo) {
Uart_Println(titolo);
}
void scriveTitolo2() {
Uart_Println("pluto");
}
void main(){
inserisce_cb(scriveTitolo);
esegue_cb();
inserisce_cb(scriveTitolo2);
esegue_cb();
}
Converting a pointer to a function to another pointer to a function is defined by the c standard, but using the resulting pointer to call a function with an incompatible type is not, per C 6.3.2.3 8:
A pointer to a function of one type may be converted to a pointer to a function of another type and back again; the result shall compare equal to the original pointer. If a converted pointer is used to call a function whose type is not compatible with the referenced type, the behavior is undefined.
The declaration void scriveTitolo2() { … } defines a function that does not have a parameter type list (it uses the old C style of an identifier list, with that list being empty) and that takes no arguments. A Callback pointer points to a function that has a parameter type list and takes a const char * argument. These are incompatible per C 2018 6.7.6.3 15:
For two function types to be compatible,… If one type has a parameter type list and the other type is specified by a function definition that contains a (possibly empty) identifier list, both shall agree in the number of parameters,…
Since they do not agree in the number of parameters, they are incompatible.
The above speaks only to the issue of converting from void (*)() to void (*){const char *) and using the result to call the function. There is a separate issue in that the function pointer is passed to inserisce_cb, which takes an argument of type void *, which is a pointer to an object type. The C standard does not define the behavior of converting a pointer to a function type to a pointer to an object type. To remedy this, inserisce_cb should be declared to take a pointer to a function type, such as void inserisce_cb(Callback c).
If scriveTitolo2 can be changed, then the compatibility issue can be resolved by changing it to take a const char * parameter that is unused, changing its definition to void scriveTitolo2(const char *).
(Note that it is preferable to declare scriveTitolo2 with the modern C style, as void scriveTitolo2(void) { … }, rather than without the void. This is unrelated to the question, as it would not make the function types compatible, but this format of declaration provides more information to the compiler in many circumstances.)
Additional thoughts to Eric's answer, which holds true for C99 as well:
If you call a function with an argument list not compatible to the function's parameter list, this is according to C99 §6.5.2.2 (6) undefined behavior.
It may work, depending on your compiler's ABI. There are compilers that let the called function clean up the stack, other compilers let the caller clean up. The former case will most likely crash, the latter ... who knows.
You can declare your scriveTitolo2 with an ignored parameter:
void scriveTitolo2(const char*) {
/* ... */
}
And everyone is happy: you and the compiler.
Is adding additional const specifiers to function arguments allowed by the standard, like in the following?
foo.h:
int foo(int x, char * data);
foo.c:
// does this match the prototype?
int foo(const int x, char * const data) {
// this implementation promises not to change x or move data inside the function
}
GCC accepts it with -std=c99 -Wpedantic -Wall -Werror, but that's not necessarily the same as standard-compliant.
This answer shows that the C++ standard allows this - does the C (99) standard allow this too?
There's another question here and a good answer here for C++
This is explicitly allowed by a special case in the rules for function parameter lists. N1570 §6.7.6.3p131 says:
In the determination of type compatibility and of a composite type, each parameter declared with function or array type is taken as having the adjusted type and each parameter declared with qualified type is taken as having the unqualified version of its declared type.
But you must also understand that the "unqualified version" of a type like const char * is still const char *, because the type const char * is derived from the type const char, and §6.2.5p26 says
A derived type is not qualified by the qualifiers (if any) of the type from which it is derived.
That means that the declaration
void foo (const int x);
is compatible with the definition
void foo (int x) { ... }
but the declaration
void bar (const char *x)
is not compatible with the definition
void foo (char *x) { ... }
You might be wondering why these rules are the way they are. The short version is that in C, all arguments are always passed by copying the value (but not any data pointed to by the value, if there are pointers involved), so it doesn't matter whether an actual argument is const T; the callee receives it as a regular old T regardless. But if you copy a pointer to constant data, the copy still points to constant data, so it does matter and that qualifier should be preserved.
1 Document N1570 is the closest approximation to the 2011 ISO C standard that is publicly available at no charge.
To the best of my knowledge, these rules have not changed significantly since the original 1989 standard. Pre-C89 "K&R" C didn't have prototypes, nor did it have const, so the entire question would be moot.
From the C99 spec, 6.7.5.3.15 Function declarators, this is legal:
If one type has a parameter type list and the other type is specified by a function definition that contains a (possibly empty) identifier list, both shall agree in the number of parameters, and the type of each prototype parameter shall be compatible with the type that results from the application of the default argument promotions to the type of the corresponding identifier. (In the determination of type compatibility and of a composite type, each parameter declared with function or array type is taken as having the adjusted type and each parameter declared with qualified type is taken as having the unqualified version of its declared type.)
cv-qualifiers on the parameter (and not on the type of the parameter) do not affect the type of the parameter, so do not affect the prototype. Things that don't affect the prototype like this can be different between the declaration and definition with no problem. Similarly, the name of the parameter does not affect the prototype, so can also be different between the declaration and definition.
Now what is confusing is that a const appearing here may be on the parameter or may be on the type of the parameter, depending on exactly where it is. If it is part of the type of the parameter, then it does affect the prototype, so must be consistent between the declaration and definition:
int foo(const int x); // const on the parameter
int foo(int * const x); // also const on the parater
int foo(const int *x); // const in the type, not on the parameter
int foo(int const *x); // also const in the type
Is void a data type in the C programming language? If so, what type of values can it store? If we have int, float, char, etc., to store values, why is void needed? And what is the range of void?
Void is considered a data type (for organizational purposes), but it is basically a keyword to use as a placeholder where you would put a data type, to represent "no data".
Hence, you can declare a routine which does not return a value as:
void MyRoutine();
But, you cannot declare a variable like this:
void bad_variable;
However, when used as a pointer, then it has a different meaning:
void* vague_pointer;
This declares a pointer, but without specifying which data type it is pointing to.
Yes, void is a type. Whether it's a data type depends on how you define that term; the C standard doesn't.
The standard does define the term "object type". In C99 and earlier; void is not an object type; in C11, it is. In all versions of the standard, void is an incomplete type. What changed in C11 is that incomplete types are now a subset of object types; this is just a change in terminology. (The other kind of type is a function type.)
C99 6.2.6 paragraph 19 says:
The void type comprises an empty set of values; it is an incomplete
type that cannot be completed.
The C11 standard changes the wording slightly:
The void type comprises an empty set of values; it is an incomplete object type that
cannot be completed.
This reflects C11's change in the definition of "object type" to include incomplete types; it doesn't really change anything about the nature of type void.
The void keyword can also be used in some other contexts:
As the only parameter type in a function prototype, as in int func(void), it indicates that the function has no parameters. (C++ uses empty parentheses for this, but they mean something else in C.)
As the return type of a function, as in void func(int n), it indicates that the function returns no result.
void* is a pointer type that doesn't specify what it points to.
In principle, all of these uses refer to the type void, but you can also think of them as just special syntax that happens to use the same keyword.
The C Standard says that void is an incomplete type that cannot be completed (unlike other incomplete types that can be completed). This means you cannot apply the sizeof operator to void, but you can have a pointer to an incomplete type.