Template function with std::map gives error: variable or field declared void

Question

Trying to create template function what couts the std::map elements, get this errors. I know what the map have four template arguments, but two have defaut values, cant understand what I have to do.

template<typename key, typename val> void arr_out (std::map<key, val>::iterator begin, std::map<key, val>::iterator end)
{
    std::cout << "map: " << std::endl;
    while(begin != end)
    {
        std::cout << (*begin).first << ": " << (*begin).second << std::endl ;
        begin++;
    }
    std::cout << std::endl;
}

error: variable or field ‘arr_out’ declared void
error: expected ‘)’ before ‘begin’
error: expected ‘)’ before ‘begin’

@AndyG There are still errors [even if you do that](http://coliru.stacked-crooked.com/a/8281c1698a8bb346) — Borgleader, Apr 19 '18 at 12:11

score 5 · Answer 1 · answered Apr 19 '18 at 12:11

5

You should add typename keyword before each of template function arguments:

template<typename key, typename val> 
void arr_out (typename std::map<key, val>::iterator begin, 
              typename std::map<key, val>::iterator end)

answered Apr 19 '18 at 12:11

Pustovalov Dmitry

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score 2 · Answer 2 · answered Apr 19 '18 at 12:12

Add typename before iterator, it indicates that its nested value type of template.

template<typename key, typename val> 
void arr_out(typename std::map<key, val>::iterator begin, typename std::map<key, val>::iterator end)
{
    std::cout << "map: " << std::endl;
    while(begin != end)
    {
        std::cout << (*begin).first << ": " << (*begin).second << std::endl ;
        begin++;
    }
    std::cout << std::endl;
}

Jarod42 · Answer 3 · 2018-04-20T15:59:11.160

2

to make code valid, you have to add typename,

template<typename key, typename val>
void arr_out (typename std::map<key, val>::iterator begin,
              typename std::map<key, val>::iterator end);

and as parameters are not deducible, call it:

std::map<Key, Value> m;
arr_out<Key, Value>(m.begin(), m.end());

An alternative to simplify call site is:

template <typename It>
void arr_out (It begin, It end);

or with some SFINAE:

template <typename It>
auto arr_out (It begin, It end)
-> decltype(void(std::cout << (*begin).first << (*begin).second));

edited Apr 20 '18 at 15:59

answered Apr 19 '18 at 22:02

Jarod42

203,559
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Thx for tip with typename It ! I had to guess. – EzR1d3r Apr 20 '18 at 11:35

score 1 · Answer 4 · answered Apr 19 '18 at 12:40

The rules of template deduction don't allow you to infer key or val from std::map<key, val>::iterator.

The other answers tell you how to correct the definition, but you have to specify the type parameters when you use it.

int main()
{
    std::map<int, std::string> m;
    // arr_out(m.begin(), m.end()); // errors relating to template argument deduction
    arr_out<int, std::string>(m.begin(), m.end()); // Ok
    return 0;
}

Template function with std::map gives error: variable or field declared void

4 Answers4