Can I have a ~destructor(void)?

Question

Looking at some old code of mine, I see that out of clumsiness I defined a destructor like so :

~ResourceManager(void);

This not only compiles, but works as expected. I changed it of course to

~ResourceManager();

but was I too quick to refactor? Is the first version correct and good C++ style?

EDIT

Since the question was closed and won't get any chance for proper disambiguation, I should put the related quote from the standard that answers this question, when destructors are put in perspective

12.4 Destructors

A special declarator syntax using an optional function-specifier (7.1.2) followed by ˜ followed by the destructor’s class name followed by an empty parameter list is used to declare the destructor in a class definition. In such a declaration, the ˜ followed by the destructor’s class name can be enclosed in optional parentheses; such parentheses are ignored. A typedef-name shall not be used as the class-name following the ∼ in the declarator for a destructor declaration.

So the standard mandates an empty parameter list. Maybe backwards compatibility with C practices for free functions (where f(void) is the way to declare an empty parameter list) took destructors along with them in implementations, but it certainly doesn't seem to be valid C++.

The `void` parameter makes no difference. It is a C-ism, where it actually makes a difference. You don't need it in C++. — juanchopanza, May 30 '14 at 14:23
Style-wise, leave `void` out. You'll almost never see it written that way. — Paul Roub, May 30 '14 at 14:24
@juanchopanza The question linked as duplicate does not put destructors in perpsective (nor any other member functions). — Nikos Athanasiou, May 30 '14 at 15:22
What "perspective" is needed? The duplicate explains the meaning of void as function parameter. — juanchopanza, May 30 '14 at 16:34

score 7 · Answer 1 · answered May 30 '14 at 14:38

7

This is a remnant from C. In C the empty parameter list does not mean a function that takes no arguments, but rather a function with unspecified number of arguments. In C++ foo(void) is deprecated, and foo() is preferred, as it specifically means a function that takes no parameters. The same goes with your destructor as well. Although, the two lines of code are equivalent you should prefer the ~ResourceManager() version.

answered May 30 '14 at 14:38

101010

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Hence in C: void f(); f(1, 2, 3) is ok, but void g(void); g(1, 2, 3), not ??? – May 30 '14 at 14:43
@DieterLücking correct. – 101010 May 30 '14 at 14:50

score 2 · Answer 2 · answered May 30 '14 at 14:23

2

There's no difference between these two lines of code. () is the same as (void). From a style perspective, () is what I've always seen for functions with no argument.

answered May 30 '14 at 14:23

Almo

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score 2 · Answer 3 · edited May 30 '14 at 14:37

2

Both lines are equivalent. Some developers like to explicit (void) in signature of parameterless functions to say to anyone reading it "It does not take any parameters"

edited May 30 '14 at 14:37

Oliver Matthews

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answered May 30 '14 at 14:24

quantdev

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Can I have a ~destructor(void)?

EDIT

3 Answers3