Boost :: typeof编译器问题:mangling typeof,改为使用decltype

时间:2009-11-05 15:54:09

标签: c++ templates boost typeof

简短的例子:

#include <boost/typeof/typeof.hpp>
#include <boost/proto/core.hpp>
using namespace boost;


template<class T, class U>
BOOST_TYPEOF_TPL(T() + U()) add2(const T& t, const U& u)
{
    return t + u;
};

int main(){

     typedef BOOST_TYPEOF(add2(2.5, 1.5)) type; // get type -> works

     BOOST_STATIC_ASSERT((is_same<type, double>::value)); // check if double -> no error -> double

     double xxx = add2(1.5,1.5); // cause the problems
        return 0;
 }

当我尝试编译这个时,我变成了一个错误:

g ++ - 4.3:抱歉,未实现:mangling typeof,改为使用decltype

g ++ - 4.2:internal compiler error: in write_type, at cp/mangle.c:1648 Please submit a full bug report, with preprocessed source if appropriate. See <URL:http://gcc.gnu.org/bugs.html> for instructions. For Debian GNU/Linux specific bug reporting instructions, see <URL:file:///usr/share/doc/gcc-4.2/README.Bugs>.

gcc version 4.3.2(Debian 4.3.2-1.1) gcc版本4.2.4(Debian 4.2.4-6)

问题出在哪里或我做错了什么?a

1 个答案:

答案 0 :(得分:3)

The example in the typeof documentation首先将BOOST_TYPEOF_TPL宏的结果包装在模板化结构中,然后在声明函数时使用它。这对你有用吗?

template<class T, class U>
struct result_of_add2
{
    typedef BOOST_TYPEOF_TPL(T() + U()) type;
};

template<class T, class U>
typename result_of_add2<T, U>::type add2(const T& t, const U& u)
{
    return t + u;
};