[英]hide function template, declare specializations
This is a followup to C++ templates: prevent instantiation of base template 这是C ++模板的后续内容:防止实例化基本模板
I use templates to achieve function overloading without the mess of implicit type conversions: declare the function template, define desired specializations (overloads). 我使用模板来实现函数重载,而不会造成隐式类型转换的麻烦: 声明函数模板,定义所需的专业化(重载)。 all is well except wrong code does not produce errors until the link phase:
一切正常,除非错误的代码在链接阶段之前不会产生错误:
lib.hpp: lib.hpp:
template<class T> T f(T v);
lib.cpp: lib.cpp:
#include "lib.hpp"
template<> long f(long v) { return -v; }
template<> bool f(bool v) { return !v; }
main.cpp: main.cpp中:
#include <iostream>
#include "lib.hpp"
int main()
{
std::cout
<< f(123L) << ", "
<< f(true) << ", "
<< f(234) << "\n"
;
}
gcc output: gcc输出:
c++ -O2 -pipe -c main.cpp
c++ -O2 -pipe -c lib.cpp
c++ main.o lib.o -o main
main.o(.text+0x94): In function `main':
: undefined reference to `int get<int>(int)'
I'd like to have it fail during compilation of main.cpp. 我想在main.cpp编译期间使它失败。 Can I somehow declare only specializations actually implemented?
我可以以某种方式声明仅实际实施的专业化课程吗?
What are my options? 我有什么选择? The target is C++03, and I'm mainly interested in gcc-4.x and VC9.
目标是C ++ 03,我主要对gcc-4.x和VC9感兴趣。
It seems to produce a linker error even if you don't put it in the separate file. 即使不将其放在单独的文件中,它似乎也会产生链接器错误。
However, to produce a compiler error for other instantiations, implement the function and use a compile-time assertion, eg 但是,要为其他实例产生编译器错误,请实现该函数并使用编译时断言,例如
#include <boost/static_assert.hpp>
template <class T> T f(T)
{
//assert some type-dependent "always-false" condition,
//so it won't be triggered unless this function is instantiated
BOOST_STATIC_ASSERT(sizeof(T) == 0 && "Only long or bool are available");
}
template<> long f(long v) { return -v; }
template<> bool f(bool v) { return !v; }
int main()
{
//f(100);
f(100L);
f(false);
}
And just for general information, C++0x has a much more elegant way to deal with it: 仅作为一般信息,C ++ 0x拥有一种更为优雅的处理方式:
template <class T> T f(T) = delete;
template<> long f(long v) { return -v; }
template<> bool f(bool v) { return !v; }
The best way is to implement that basic template with something invalid (not illegal) C++ code. 最好的方法是使用无效的(不是非法的)C ++代码来实现该基本模板。 For example,
例如,
template<class T> T f(T v) { return v.Default_Implementation_Not_Available; }
This error will be compile time; 该错误将是编译时; and it's generated only when you instantiate any version other than 'long' & 'bool'.
并且仅当您实例化“ long”和“ bool”以外的任何版本时才生成。 If you don't instantiate 'int' version, compilation will go fine.
如果您不实例化“ int”版本,则编译会很好。
I don't believe it's possible to do what you want. 我不相信您可以做自己想做的事情。 See these FAQs for more info:
有关更多信息,请参见以下常见问题解答:
How can I avoid linker errors with my template functions? 如何避免模板函数出现链接器错误?
How can I avoid linker errors with my template classes? 如何避免模板类出现链接器错误?
编译main.cpp时,编译器无法知道其他某个编译单元中可能存在哪些模板专业化-因此无法在编译时标记此错误,您必须等待链接时间。
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