A problem about template function overload

[miaohua1982@gmail.com]

the code as follows:
#include<iostre am>
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=?=]

Re: A problem about template function overload

On 2007-03-18 14:02, miaohua1982@gma il.com wrote:

the code as follows:
#include<iostre am>
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@gmail.com]

Re: A problem about template function overload

On 3月18æ—¥, 下午10æ—¶28åˆ †, Erik Wikström <Erik-wikst...@telia. comwrote:

On 2007-03-18 14:02, miaohua1...@gma il.com wrote:
>
>
>
>
>

the code as follows:
#include<iostre am>
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- 隐藏被引用 æ–‡å— -
>
- 显示引用的 æ–‡å— -

well, I don't think so. Just have a look at the following code:
#include<iostre am>
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?