Typedef in template class not working

Question

That might be a silly question, but I'm staring at this code for a while and can't think of what's wrong. On compiling:

#include <string>
#include <map>

template <typename T>
struct my_string_map {
    typedef std::map<std::string, T> type;
};
template <typename T> bool add_to_string_map(my_string_map<T>::type map,
        std::string str, T x) {
    if (map.find(str) != map.end()) 
        return false;
    map[str] = x;
    return true;
}

I get:

foo.cpp:8: error: template declaration of ‘bool add_to_string_map’
foo.cpp:8: error: expected ‘)’ before ‘map’
foo.cpp:8: error: expected primary-expression before ‘str’
foo.cpp:8: error: expected primary-expression before ‘x’

(Definition of my_string_map class was taken from another thread of this forum)

When i specialize what template type I'm using, like in

bool add_to_string_int_map(my_string_map<int>::type map,
        std::string str, int x) {
    if (map.find(str) != map.end()) 
        return false;
    map[str] = x;
    return true;
}

everything works fine. Why it's not working and/or how to get it working?

Thanks in advance for your help

Define "not working". Presumably you get a compiler error message. What is it? — Oliver Charlesworth, Jul 03 '11 at 13:31

score 7 · Answer 1 · edited Jun 09 '16 at 14:46

7

Try putting typename before the my_string_map<T>::type argument.

For more info, see: https://isocpp.org/wiki/faq/templates#nondependent-name-lookup-types.

edited Jun 09 '16 at 14:46

Gunther Struyf

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answered Jul 03 '11 at 13:37

Oliver Charlesworth

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1

Whoa, thanks. It works. The more i use templates, the more weird they seem... – karoluch Jul 03 '11 at 13:39

score 4 · Answer 2 · answered Jul 03 '11 at 13:38

4

my_string_map<T>::type is a dependent name, as it depends on the template parameter T. For dependent names, you need to use typename:

template <typename T>
bool add_to_string_map(typename my_string_map<T>::type map, std::string str, T x) {
    ...
}

answered Jul 03 '11 at 13:38

hammar

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score 1 · Answer 3 · answered Jul 03 '11 at 13:35

1

The compiler will not be able to deduce T. It seems obvious to a human but from the point of view of the compiler, this kind of deduction is impossible. The specialization works because you're no longer asking the compiler to deduce T.

answered Jul 03 '11 at 13:35

Puppy

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Why wouldn't it be able to? There's a parameter (the third) of type `T`. – Ben Voigt Jul 03 '11 at 14:07
@Ben: Last time I checked, it won't do that. – Puppy Jul 03 '11 at 21:16

Typedef in template class not working

3 Answers3