Why does C++ allow us to surround the variable name in parentheses when declaring a variable?

问题

For example a declaration such as that:

int (x) = 0;

Or even that:

int (((x))) = 0;

I stumbled upon this because in my code I happened to have a fragment similar to the following one:

struct B
{
};

struct C
{
  C (B *) {}
  void f () {};
};

int main()
{
  B *y;
  C (y);
}

Obviously I wanted to construct object C which then would do something useful in its destructor. However as it happens compiler treats C (y); as a declaration of variable y with type C and thus it prints an error about y redefinition. Interesting thing is that if I write it as C (y).f () or as something like C (static_cast<B*> (y)) it will compile as intended. The best modern workaround is to use {} in constructor call, of course.

So as I figured out after that, it\'s possible to declare variables like int (x) = 0; or even int (((x))) = 0; but I\'ve never seen anyone actually using declarations like this. So I\'m interested -what\'s the purpose of such possibility because for now I see that it only creates the case similar to the notorious \"most vexing parse\" and doesn\'t add anything useful?

回答1:

Grouping.

As a particular example, consider that you can declare a variable of function type such as

int f(int);

Now, how would you declare a pointer to such a thing?

int *f(int);

Nope, doesn't work! This is interpreted as a function returning int*. You need to add in the parentheses to make it parse the right way:

int (*f)(int);

The same deal with arrays:

int *x[5];   // array of five int*
int (*x)[5]; // pointer to array of five int

回答2:

There's generally allowed to use parentheses in such declarations because the declaration, from the syntactical point of view looks always like this:

<front type> <specification>;

For example, in the following declaration:

int* p[2];

The "front type" is int (not int*) and the "specification" is * p[2].

The rule is that you can use any number of parentheses as needed in the "specification" part because they are sometimes inevitable to disambiguate. For example:

int* p[2]; // array of 2 pointers to int; same as int (*p[2]);
int (*p)[2]; // pointer to an array of 2 ints

The pointer to an array is a rare case, however the same situation you have with a pointer to function:

int (*func(int)); // declares a function returning int*
int (*func)(int); // declares a pointer to function returning int

This is the direct answer to your question. If your question is about the statement like C(y), then:

Put parentheses around the whole expression - (C(y)) and you'll get what you wanted
This statement does nothing but creating a temporary object, which ceases to living after this instruction ends (I hope this is what you intended to do).

来源：https://stackoverflow.com/questions/29675601/why-does-c-allow-us-to-surround-the-variable-name-in-parentheses-when-declarin

标签

c++

most-vexing-parse