如何在 C++ 的模板函数中传递普通参数和模板参数？

Question

我在名为 myNamespace 的命名空间中有一个模板函数（如下所示）：

template <typename setX>
void getRandomItems(NaturalNumber size, setX &random, setX &items)
{
    assert(size <= items.size());

    //set of randomly selected indices for items
    set<NaturalNumber> index;
    NaturalNumber r, i;

    while(index.size() < size)
    {
        r = unifRand(0,items.size()-1);
        index.insert(r);
    }

    typename setX::iterator it, sit = items.begin();
    for(i = 0, it = index.begin(); it != index.end(); it ++)
    {
        //find the r-th elt in index
        r = *it;
        for(; i < r; i ++)
            sit++;

        random.insert(*sit);
    }
}

但是，每当我调用此函数时，我都会收到这些错误：

generic.h: In function ‘void myNamespace::getRandomItems(NaturalNumber, setX&, setX&) [with setX = std::set<std::basic_string<char> >, NaturalNumber = long unsigned int]’:
synthetic-graph.C:87:55:   instantiated from here
generic.h:74:32: error: no match for ‘operator=’ in ‘it = index.std::set::begin [with _Key = long unsigned int, _Compare = std::less<long unsigned int>, _Alloc = std::allocator<long unsigned int>, std::set<_Key, _Compare, _Alloc>::iterator = std::_Rb_tree_const_iterator<long unsigned int>]()’
/usr/include/c++/4.5/bits/stl_tree.h:224:5: note: candidate is: std::_Rb_tree_const_iterator<std::basic_string<char> >& std::_Rb_tree_const_iterator<std::basic_string<char> >::operator=(const std::_Rb_tree_const_iterator<std::basic_string<char> >&)
synthetic-graph.C:87:55:   instantiated from here
generic.h:74:32: error: no match for ‘operator!=’ in ‘it != index.std::set<_Key, _Compare, _Alloc>::end [with _Key = long unsigned int, _Compare = std::less<long unsigned int>, _Alloc = std::allocator<long unsigned int>, std::set<_Key, _Compare, _Alloc>::iterator = std::_Rb_tree_const_iterator<long unsigned int>]()’
/usr/include/c++/4.5/bits/stl_tree.h:291:7: note: candidate is: bool std::_Rb_tree_const_iterator<_Tp>::operator!=(const std::_Rb_tree_const_iterator<_Tp>::_Self&) const [with _Tp = std::basic_string<char>, std::_Rb_tree_const_iterator<_Tp>::_Self = std::_Rb_tree_const_iterator<std::basic_string<char> >]
generic.h:77:4: error: cannot convert ‘const std::basic_string<char>’ to ‘NaturalNumber’ in assignment

我已尝试了所有组合，但没有运气，请帮助我！

Answer 1

setX 不是一组 NaturalNumber，因此当您说 it = index.begin() 时，迭代器不兼容。您可以将 it 设为 set 的迭代器，但我不太清楚您真正想在这里做什么。

我还注意到，在您的内部循环中，您没有进行任何检查来确保 sit 不会超出其集合的末尾。

Answer 2

您正在尝试分配不兼容的迭代器。

也许你的意思是

set<NaturalNumber>::iterator it;
typename setX::iterator sit = items.begin();

而不是

typename setX::iterator it, sit = items.begin();