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. - Volodya

On Fri, Jun 03, 2005 at 07:38:13AM -0400, David Abrahams wrote:

Vladimir Prus <ghost@cs.msu.su> writes:

...

Hello, the program_options library fails one tests on EDG compilers (Intel, Como), which a failure to deduce template arguments for a certain call. After reading the standard for some time, I failed to understand the reason.

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]'

...

// 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).

Or the example in [conv.qual] in the standard: int main() { const char c = 'c'; char* pc; const char** pcc = &pc; //1: not allowed *pcc = &c; *pc = 'C'; //2: modifies a const object }

'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.

That was what I thought the problem was, but why does the second call succeed? jon