请考虑以下代码:
#include <iostream>
#include <typeinfo>
template< typename Type >
void func( Type var )
{
std::cout << __FUNCTION__ << ": var = " << var << " [" << typeid( var ).name( ) << "]." << std::endl;
std::cout << "-> var is SCALAR. Size = " << sizeof( Type ) << std::endl;
}
#if 1
template< typename Type >
void func( Type * var )
{
std::cout << __FUNCTION__ << ": var = " << var << " [" << typeid( var ).name( ) << "]." << std::endl;
std::cout << "-> var is ARRAY. Size = " << sizeof( Type * ) << std::endl;
}
#endif
int main( )
{
typedef char char16[ 16 ];
char16 c16 = "16 bytes chars.";
std::cout << "Size of char16 = " << sizeof( char16 ) << std::endl;
func( c16 );
return 0;
}
如果我编译并运行它,我会看到以下内容:
> g++ -Wall -g3 spec_f_pointer.cpp -o spec_f_pointer
> ./spec_f_pointer
Size of char16 = 16
func: var = 16 bytes chars. [Pc].
-> var is ARRAY. Size = 8
显然sizeof
里面印有func
指的是指针的大小,而不是指针的大小typedef
数组,如给出的main()
.
现在我想知道如何正确地完成我的技巧func
专注于正确了解我的 typedef 及其大小的方式。
请问这里有人可以帮助我吗?
真的感谢。
EDIT
实施专业化:
template< typename Type >
void func( Type * const &var )
{
std::cout << __FUNCTION__ << ": var = " << var << " [" << typeid( var ).name( ) << "]." << std::endl;
std::cout << "-> var is ARRAY. Size = " << sizeof( Type * ) << std::endl;
}
输出是:
Size of char16 = 16
func: var = 16 bytes chars. [A16_c].
-> var is SCALAR. Size = 16
我注意到类型发生了变化Pc
to A16_c
。
有帮助吗?
如果您想将函数专门用于数组,请执行以下操作:
template<typename T, int N>
void func(T(&var)[N])
{
typedef T Type[N];
std::cout << __FUNCTION__ << " [" << typeid( var ).name( ) << "]." << std::endl;
std::cout << "-> var is ARRAY. Size = " << sizeof( Type ) << std::endl;
std::cout << "Number of elements: " << N << std::endl;
std::cout << "Size of each element: " << sizeof(T) << std::endl;
}
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