也许我不知道如何搜索，但我找不到任何人谈论这个是事实。
我的结构有一个非类型参数，它依赖于一个类型参数。
template<
typename SpecType,
SpecType NonType >
struct Struct
//...
当 SpecType 是对指针的引用时(例如 const char *&)，NonType 的行为就好像它是实际的专门论点，而不是引用。
更令人惊讶的是，如果我将 NonType 显式转换为 SpecType，一切都会按预期进行!
IBM说了一些关于转换为数组和函数指针的内容，但我不理解它与我的疑问有关。
当我创建没有嵌入模板类型(S1 和 S2)的结构时，不会发生同样的事情。
当然我可以改成:
template<
typename SpecType,
SpecType &NonType >
但它无法解释我所看到的。 任何人都可以给出一个深刻的(或愚蠢的，如果是我的愚蠢)解释吗？
下面的例子有点冗长，但看看它的输出我想我的问题会更清楚:
#include <iostream>
#include <typeinfo>
using namespace std;
void f1( const char **p )
{
cout << "---------------------------------------------" << endl;
cout << "f1( const char **p ): p = \"" << p << "\"" << endl;
}
void f1( const char *p )
{
cout << "---------------------------------------------" << endl;
cout << "f1( const char *p ): p = \"" << p << "\"" << endl;
}
void f1( const int **p )
{
cout << "---------------------------------------------" << endl;
cout << "f1( const int **p ): p = \"" << p << "\"" << endl;
}
void f1( const int *p )
{
cout << "---------------------------------------------" << endl;
cout << "f1( const int *p ): p = \"" << p << "\"" << endl;
}
template<
typename SpecType,
SpecType NonType >
struct Struct
{
void f( )
{
cout << "---------------------------------------------" << endl;
cout << "SpecType is " << typeid( SpecType ).name( ) << endl;
cout << "NonType is " << typeid( NonType ).name( ) << endl;
cout << "NonType = \"" << NonType << "\"" << endl;
cout << "( SpecType )NonType = \"" << ( SpecType )NonType << "\"" << endl;
cout << "*NonType = \"" << *NonType << "\"" << endl;
cout << "*NonType[ 0 ] = \"" << **NonType << "\"" << endl;
f1( NonType );
}
};
template< const char *&P >
struct S1
{
void f( )
{
cout << "---------------------------------------------" << endl;
cout << "&P = \"" << &P << "\"" << endl;
cout << "P = \"" << P << "\"" << endl;
cout << "*P = \"" << *P << "\"" << endl;
f1( P );
}
};
template< const char **P >
struct S2
{
void f( )
{
cout << "---------------------------------------------" << endl;
cout << "P = \"" << P << "\"" << endl;
cout << "*P = \"" << *P << "\"" << endl;
cout << "*P[ 0 ] = \"" << **P << "\"" << endl;
f1( P );
}
};
const char * const_pname = "name";
const int pint[] = { 42, 51 };
const int *const_pint = pint;
int main( )
{
cout << "=============================================" << endl;
cout << "const_pname = " << const_pname << endl;
cout << "@const_pname = 0x" << hex << ( unsigned long )const_pname << dec << endl;
cout << "&const_pname = 0x" << hex << ( unsigned long )&const_pname << dec << endl;
cout << "=============================================" << endl;
cout << "Struct< const char *&, const_pname > constpTtname" << endl;
Struct< const char *&, const_pname > constpTtname;
constpTtname.f( );
cout << "=============================================" << endl;
cout << "Struct< const int *&, const_pint > constpTtint" << endl;
Struct< const int *&, const_pint > constpTtint;
constpTtint.f( );
cout << "=============================================" << endl;
cout << "S1< const_pname > s1" << endl;
S1< const_pname > s1;
s1.f( );
cout << "=============================================" << endl;
cout << "S2< &const_pname > s2" << endl;
S2< &const_pname > s2;
s2.f( );
return 0;
}
输出是:
$ ./nontype_mutant
=============================================
const_pname = name
@const_pname = x401624
&const_pname = 0x601e18
=============================================
Struct< const char *&, const_pname > constpTtname
---------------------------------------------
SpecType is PKc
NonType is PKc
NonType = "$@"
( SpecType )NonType = "name"
*NonType = "name"
*NonType[ 0 ] = "n"
---------------------------------------------
f1( const char *p ): p = "$@"
=============================================
Struct< const int *&, const_pint > constpTtint
---------------------------------------------
SpecType is PKi
NonType is PKi
NonType = "0x601e20"
( SpecType )NonType = "0x4017a8"
*NonType = "0x4017a8"
*NonType[ 0 ] = "42"
---------------------------------------------
f1( const int *p ): p = "0x601e20"
=============================================
S1< const_pname > s1
---------------------------------------------
&P = "0x601e18"
P = "name"
*P = "n"
---------------------------------------------
f1( const char *p ): p = "name"
=============================================
S2< &const_pname > s2
---------------------------------------------
P = "0x601e18"
*P = "name"
*P[ 0 ] = "n"
---------------------------------------------
f1( const char **p ): p = "0x601e18"
最佳答案
我尝试使用三个编译器编译您的代码，其中两个具有非常相似的行为，给出以下消息(大约):
test.cpp:44:41: error: indirection requires pointer operand ('int' invalid)
cout << "*NonType[ 0 ] = \"" << **NonType << "\"" << endl;
^~~~~~~~~
test.cpp:93:18: note: in instantiation of member function 'Struct<const char *&, const_pname>::f' requested here
constpTtname.f( );
^
test.cpp:44:41: error: indirection requires pointer operand ('int' invalid)
cout << "*NonType[ 0 ] = \"" << **NonType << "\"" << endl;
^~~~~~~~~
test.cpp:98:17: note: in instantiation of member function 'Struct<const int *&, const_pint>::f' requested here
constpTtint.f( );
^
2 errors generated.
错误消息对我来说似乎是正确且不言而喻的。这是使用 clang 的结果。 Comeau 的基于 EDG 的编译器是另一个给出与此非常相似的消息的编译器。
g++ 编译了它(我认为是错误的)并给出了类似于您报告的输出。
关于C++ 模板参数更改对指针的引用，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/5316262/