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Cracow, 13.03.2008

Hi,

Assuming that I want to pass a reference to a function, and
make sure that the function will not make any changes to the object
passed, is it necessary to declare two "const's" or only one, i.e.
for example:

void fun(const Type& const x);

or

void fun(const Type& x);

Also, is there a difference between

const Type& const x

and

const Type const & x


L. Bieniasz

Leslaw Bieniasz wrote:

>
> Cracow, 13.03.2008
>
> Hi,
>
> Assuming that I want to pass a reference to a function, and
> make sure that the function will not make any changes to the object
> passed, is it necessary to declare two "const's" or only one, i.e.
> for example:
>
> void fun(const Type& const x);

should not compile.

>
> or
>
> void fun(const Type& x);

should compile

> Also, is there a difference between
>
> const Type& const x
>
> and
>
> const Type const & x

not really: both are ill-formed, though in different ways. The first applies
a cv-qualifier to a reference type and the second violates [7.1.5/1].


Best

Kai-Uwe Bux

Leslaw Bieniasz <> wrote in
news:

>
> Cracow, 13.03.2008
>
> Hi,
>
> Assuming that I want to pass a reference to a function, and
> make sure that the function will not make any changes to the object
> passed, is it necessary to declare two "const's" or only one, i.e.
> for example:
>
> void fun(const Type& const x);

This is illegal because references aren't objects and therefore can't be
const by definition (can't modify a non-existant thing in the first
place).

>
> or
>
> void fun(const Type& x);

This is generally sufficient.

>
> Also, is there a difference between
>
> const Type& const x
>
> and
>
> const Type const & x
>

Yes, they are both illegal in different ways. The first tries to make
something which isnt an object const and the second has two const
specifications for Type.

joe