Re: Bizarre : pointer as template parameter
On 8/27/2010 7:57 AM, persres@googlemail.com wrote:
On 27 Aug, 12:46, "Alf P. Steinbach /Usenet"<alf.p.steinbach
+use...@gmail.com> wrote:
* pers...@googlemail.com, on 27.08.2010 13:39:
On 27 Aug, 12:19, "pers...@googlemail.com"<pers...@googlemail.com>
wrote:
Hi,
I am using visual studio 2008.
I have the following code:
typedef int *PINT;
This is bad in two ways. First, don't typedef pointer types (except for the
purpuse of abstracting away the pointer nature). Second, reserve ALL UPPERCASE
for macros, don't use for anything else.
template<class T>
void foo(const T src)
{
}
template<>
void foo<int *> (const int * src)
{
} // line 43
int main()
{
int a;
foo(&a);
}
file.cpp (43) :error C2912: explicit specialization; 'void
foo<int*>(const int *)' is not a specialization of a function
template.
If I were to replace int * with PINT, it compiles.
Any explanations please.
'const PINT' is not the same as 'const int*'.
'const PINT' is the same as 'PINT const' is the same as 'int *const'.
If you adopt the convention of writing 'const' /after/ the type name then you
avoid most of these problems.
[snip extra example]
Cheers& hth.,
- Alf
--
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Do you know anything about the template errors?
Nah, he just pretends :-)...
The error, as Alf was trying to lead you, is due to your
misunderstanding of what typedef does and what 'const' qualifiies when
used. In your code
>>>> template<class T>
>>>> void foo(const T src)
>>>> {
>>
>>>> }
>>
>>>> template<>
>>>> void foo<int *> (const int * src)
The argument of the "specialization" cannot be matched to make that the
specialization. If you specialize your 'foo' template on 'int*', the
specialization would look like this:
template<>
void foo<int*> (int * const src ) ...
Can you spot a *single* significant difference? Once you find it,
reread Alf's reply. Do you get it now? If not, reread it again until
you do.
V
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