sizeof（void（））是一个合法的表达式吗？

Question

From [5.3.3/1] , I found that:

The sizeof operator shall not be applied to an expression that has function or incomplete type

From [3.9/5] I found that:

Incompletely-defined object types and cv void are incomplete types

Anyway, for sizeof does not evaluate it's operands, I would have said that sizeof(void()) was a legal expression (actually GCC compiles it and the result is 1).
On the other side, from here , void is not mentioned while discussing sizeof , neither when the types having size 1 are mentioned, nor in the list of the ones having an implementation defined size.

The question is thus: is sizeof(void()) a legal expression?
Is it guaranteed to have size equal to 1?
Or is it a legal expression resulting in an UB and that's all?

Answer 1

void()是一个函数类型（它是一个不带参数且不返回任何内容的函数），因此它不是sizeof()的有效类型。

Answer 2

From looking at CppReference.com - sizeof operator , the documentation literally states:

sizeof cannot be used with function types , incomplete types, or bit-field glvalues.

And since void() is a function type, then sizeof(void()) is not a legal expression.

In their usage example, we can see their error comment on this line:

std::cout << "size of function: " << sizeof(void()) << '\n'; // error

Answer 3

Also, if you compile the code, such as the example below:

#include <iostream>

int main()
{
   std::cout << sizeof(void());
}

The code compiles correctly and produces a value of 1, but if you look at the compilation, you see this:

main.cpp: In function 'int main()':

main.cpp:5:29: warning: invalid application of 'sizeof' to a function type [-Wpointer-arith]

std::cout << sizeof(void());

So, it is evident that sizeof() doesn't apply for function types, so the code produces a warning. It is invalid.

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Code here

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Answer 4

A little premise.

The question arose from a misinterpretation of the sizeof operator.
In fact the OP considered void() an expression that has incomplete type in the context of sizeof and the question itself can be read as - why sizeof accept the expression void() , that is an incomplete type and should not be accepted as mentioned in the working draft ?
That's why [3.9/5] is mentioned actually, otherwise it wouldn't have made sense.

That said, the fact is that the question contains actually two interesting questions:

• Why is sizeof(void()) not legal?
  This is the actual question as from the title itself.

• Why is sizeof((void())) not legal?
  This is the intended question of the OP.

Answers below:

• void() in sizeof(void()) is interpreted as a function type and it is ill-formed as for [5.3.3/1] (emphasis mine):

  The sizeof operator shall not be applied to an expression that has function or incomplete type, to the parenthesized name of such types , [...]

• (void()) in sizeof((void())) is an expression that has incomplete type void (note that sizeof is an unevaluated context) and it is ill-formed as for [5.3.3/1] (emphasis mine):

  The sizeof operator shall not be applied to an expression that has function or incomplete type , to the parenthesized name of such types, [...]

In both cases GCC compiles the code with a warning.

Answer 5

As already highlighted in the docs here http://en.cppreference.com/w/cpp/language/sizeof

Notes

sizeof() cannot be used with function types , incomplete types, or bit-field glvalues.

Since void() is a function type, so its not a valid type of sizeof()

Note:

void() is a function that takes no arguments and returns nothing

Quoting a Example from Docs:

//<< "size of function: " << sizeof(void()) << '\n'  // error

So Answers to your questions:

1)No it is not a legal Expression.

2)It will show as 1 , but will show a warning

3)same as 1).

Answer 6

Straigth from the C99 reference NO

Stated in the document under section :

The sizeof operator shall not be applied to an expression that has function type or an incomplete type, to the parenthesized name of such a type, or to an expression that designates a bit-field member.

According to item 19 and 20 of section :

19. The void type comprises an empty set of values; that cannot be completed.
20. ...A function type describes a function with specified return type. A function type is characterized by its return type and the number and types of its parameters. A function type is said to be derived from its return type, and if its return type is T, the function type is sometimes called ''function returning T''. The construction of a function type from a return type is called ''function type derivation''.

C99 Standard