Template argument deduction

[Vladimir Grigoriev]

Let consider the following template function

template <typename T>
const T & f( const T &x, const T &y )
{
std::cout << typeid( x ).name() << std::endl;
std::cout << typeid( y ).name() << std::endl;

return ( x );
}

When I run this function using Borland C++ compiler adding the statements

char s1[] = "ABCDE";
char s2[] = "GFHIJ";

std::cout << typeid( f( d1, s2 ) ).name() << std::endl;

I get the result

const char *
conts char *
const char *

When I run the code using Microsoft C++ 2005 EE I get

char const [6]
char const [6]
char const [6]

So it is interesting what should be displayed?

And if Microsoft C++ shows that the return value of f() is of type char
const [6] when how does look the corresponding instantiated function?

Vladimir Grigoriev


std::cout << typeid( f(

[Igor Tandetnik]

Post by Vladimir Grigoriev
Let consider the following template function
template <typename T>
const T & f( const T &x, const T &y )
{
std::cout << typeid( x ).name() << std::endl;
std::cout << typeid( y ).name() << std::endl;
return ( x );
}
When I run this function using Borland C++ compiler adding the
statements
char s1[] = "ABCDE";
char s2[] = "GFHIJ";
std::cout << typeid( f( d1, s2 ) ).name() << std::endl;
I get the result
const char *
conts char *
const char *
When I run the code using Microsoft C++ 2005 EE I get
char const [6]
char const [6]
char const [6]
So it is interesting what should be displayed?

VC is correct, I believe.

Post by Vladimir Grigoriev
And if Microsoft C++ shows that the return value of f() is of type
char const [6] when how does look the corresponding instantiated
function?

Not char const[6] - char const (&)[6]. A function can't return an array, but it can return a reference to an array. However, when the argument of typeid operator is a reference, it produces the type_info object for the referent.

const char(&f(const char(&x)[6], const char(&y)[6]) )[6];

[Vladimir Grigoriev]

Thank you, Igor, very much. I was trying to place [6] before the list of
parameters and not at the end.:)

Vkadimir Grigoriev

Post by Vladimir Grigoriev
Let consider the following template function
template <typename T>
const T & f( const T &x, const T &y )
{
std::cout << typeid( x ).name() << std::endl;
std::cout << typeid( y ).name() << std::endl;
return ( x );
}
When I run this function using Borland C++ compiler adding the
statements
char s1[] = "ABCDE";
char s2[] = "GFHIJ";
std::cout << typeid( f( d1, s2 ) ).name() << std::endl;
I get the result
const char *
conts char *
const char *
When I run the code using Microsoft C++ 2005 EE I get
char const [6]
char const [6]
char const [6]
So it is interesting what should be displayed?

VC is correct, I believe.

Post by Vladimir Grigoriev
And if Microsoft C++ shows that the return value of f() is of type
char const [6] when how does look the corresponding instantiated
function?

Not char const[6] - char const (&)[6]. A function can't return an array, but
it can return a reference to an array. However, when the argument of typeid
operator is a reference, it produces the type_info object for the referent.

const char(&f(const char(&x)[6], const char(&y)[6]) )[6];

[Alex Blekhman]

Post by Vladimir Grigoriev
Thank you, Igor, very much. I was trying to place [6] before the
list of parameters and not at the end.:)

The common way to cope with such nasty declarations is to use
`typedef' keyword:

typedef char my_buff_t[6];

const my_buff_t& f(const my_buff_t& x, const my_buff_t& y);

This way you can pack the syntactic ugliness into nice descriptive
type names.

HTH
Alex

[Vladimir Grigoriev]

Thanks Alex. I was trying to do the same but without using typedefs. So I
got errors.

Vladimir Grigoriev

Post by Vladimir Grigoriev
Thank you, Igor, very much. I was trying to place [6] before the list of
parameters and not at the end.:)

The common way to cope with such nasty declarations is to use `typedef'
typedef char my_buff_t[6];
const my_buff_t& f(const my_buff_t& x, const my_buff_t& y);
This way you can pack the syntactic ugliness into nice descriptive type
names.
HTH
Alex