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C++ 模板参数更改对指针的引用

转载 作者:太空狗 更新时间:2023-10-29 23:09:24 25 4
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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
//...

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(S1S2)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/

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