A problem about template function overload

[miaohua1982]

the code as follows:
#include<iostream>
using namespace std;

template <int N>
void foo( const char (&str)[N])
{
cout<<"array"<<endl;
}
template <typename T>
void foo(const T& str);
template <>
void foo(char *const &str)
{
cout<<"char *"<<endl;
}

int main()
{
char arr[10];
foo(arr);
}

why the called function is the void foo(const char(&str)[N])? I test
the code by VC7, the output is
*array*, but according to my knowledge, the following code is also
OK:

char p[1];
char * const & str = p;

so why the output is not *char*? I mean the why the *exact match*
funcion is not the secnod "foo" with params (char *const &str)?

It has confused me so much. Is there anyone call tell me?
Thank you very much!

[=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?=]

On 2007-03-18 14:02, mi*********@gmail.com wrote:

the code as follows:
#include<iostream>
using namespace std;

template <int N>
void foo( const char (&str)[N])
{
cout<<"array"<<endl;
}
template <typename T>
void foo(const T& str);
template <>
void foo(char *const &str)
{
cout<<"char *"<<endl;
}

int main()
{
char arr[10];
foo(arr);
}

why the called function is the void foo(const char(&str)[N])? I test
the code by VC7, the output is
*array*, but according to my knowledge, the following code is also
OK:

char p[1];
char * const & str = p;

so why the output is not *char*? I mean the why the *exact match*
funcion is not the secnod "foo" with params (char *const &str)?

Because an array is not a pointer, it can however decay (is that the
correct word?) to a pointer, so the function taking an array is a better
match since no conversion is needed.

--
Erik Wikström

[miaohua1982]

On 3月18日, 下�10时28分, Erik Wikström <[e-mail]:

On 2007-03-18 14:02, [e-mail] wrote:

the code as follows:
#include<iostream>
using namespace std;

template <int N>
void foo( const char (&str)[N])
{
Â* Â* Â* cout<<"array"<<endl;
}
template <typename T>
void foo(const T& str);
template <>
void foo(char *const &str)
{
Â* Â*cout<<"char *"<<endl;
}

int main()
{
Â* Â* char arr[10];
Â* Â*foo(arr);
}

why Â*the called function is the Â*void foo(const char(&str)[N])? I test
the code by VC7, the output is
*array*, Â*but according to my knowledge, the following code is also
OK:

char p[1];
char * const & str = p;

so why the output is not *char*? I mean the why the *exact match*
funcion is not the secnod "foo" with params (char *const &str)?

Because an array is not a pointer, it can however decay (is that the
correct word?) to a pointer, so the function taking an array is a better
match since no conversion is needed.

--
Erik Wikström- éš�è—�被引用文å*— -

- 显示引用的文å*— -

well, I don't think so. Just have a look at the following code:
#include<iostream>
using namespace std;

template <int N>
void foo( const char (&str)[N])
{
cout<<"array"<<endl;
}

void foo(char * const &str)
{
cout<<"char *"<<endl;
}

int main()
{
char arr[10];
foo(arr);
}

the output in VC7 is "char*", so can you explain it?