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From: David Abrahams (dave_at_[hidden])
Date: 2005-06-03 08:47:38


Vladimir Prus <ghost_at_[hidden]> writes:

> David Abrahams wrote:
>
>>> Below is the test program:
>>>
>>>
>>> char* cmdline3_[1] = {};
>>>
>>> template<class charT>
>>> void func(const charT* const argv[]) {}
>>>
>>> int main()
>>> {
>>> // P is 'const charT* const argv[]'cmdline3_
>>> // A is 'char* cmdline3_[]'
>>
>> I think A is 'char*[1]'
>
> Yes, but it still should be converted to char** for template argument
> deduction.
>
>>> // According to 14.8.2.1/2
>>> // If P is not reference type
>>> // -If A is an array type, the pointer type produced
>>> // by the array-to-pointer standard convertion (4.2)
>>> // is used in place of A for type deduction.
>>> //
>>> // So, A should become 'char**'
>>>
>>> // The following does not compile.
>>> func(cmdline3_);
>>>
>>> char** p = cmdline3_;
>>> // This does compile, even though the argument type used for
>>> // type deduction should be the same as in example above.
>>> func(p);
>>>
>>> }
>>>
>>> Anybody can explain what's going on?
>>
>> It's well known that converting a char** into a char const** is a
>> type-safety hole, so the language doesn't allow it (see D&E for the
>> details, I think).
>
> I know that. What's why the function argument is not (const charT* argv[])
> but (const charT* const argv[])
>
>> 'char*[1]' converts (by array-to-pointer) to 'char**',
>> but a 'const charT* const[]' parameter is equivalent to char const*
>> const*', so a similar logic applies.
>
> I don't think so. In the second call, passing "char**" to the same function
> works without any problems. It looks like in the first call, compiler
> "forgets" to promote array to pointer.

On second thought, I think you've found an EDG bug.

-- 
Dave Abrahams
Boost Consulting
www.boost-consulting.com

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