记录一下mpl学习结果:
引用
1-1.
Turn vector_c<int,1,2,3> into a type sequence with elements (2,3,4) using transform.
对vector做一个加1就可以了
typedef mpl::vector_c<int,1,2,3> AAA;
typedef mpl::vector_c<int,2,3,4> CCC;
typedef mpl::transform<AAA,mpl::plus<_1, mpl::int_<1> > >::type BBB;
BOOST_STATIC_ASSERT((
mpl::equal<BBB,CCC>::type::value
));
引用
1-2.
Turn vector_c<int,1,2,3> into a type sequence with elements (1,4,9) using transform.
对两个AAA相乘
typedef mpl::vector_c<int,1,4,9> DDD;
typedef mpl::transform<AAA,AAA, mpl::multiplies<_1,_2> >::type EEE;
BOOST_STATIC_ASSERT((
mpl::equal<DDD,EEE>::type::value
));
引用
1-5.
There's still a problem with the dimensional analysis code in section 1.1. Hint: What happens when you do:
f = f + m * a;
Repair this example using techniques shown in this chapter.
加法操作支持两种类型,然后确保类型是一致的就可以了
template <class T, class D1, class D2>
quantity<T,D1>
operator+(quantity<T,D1> x, quantity<T,D2> y)
{
BOOST_STATIC_ASSERT((
mpl::equal<D1,D2>::type::value
));
return quantity<T,D1>(x.value() + y.value());
}