CODEWITHSUNDEEP
c++c++11
HugoZou
HugoZou

Reputation: 13

How to get more detail about the identifier of auto in iterator of C++11

 void test()
 {
       vector <string> s={"hello world"};
       for(vector<string>::iterator it=s.begin();it!=s.end()&&!it->empty();it++)
      {
           for(auto it2=it->begin();it2!=it->end();it2++)//I don't konw  type of it2
           {
                *it2=toupper(*it2);
           }
       cout<<*it<<endl;
     }
}

In the first loop I can sure that iterator of type is vector<string>::iterator. I wonder what is type of the it2 (I have already tried use vector<string>::const). And how can I get the more detail about which type did auto equal.

Upvotes: 1

Views: 132

Answers (1)

Barry
Barry

Reputation: 303337

it is an iterator into a vector<string>. Which means that it "points to" a string. So when you write it->begin() and it->end(), you're iterating over the string, which would make it2 have type std::string::iterator. As smartly noted in the comments by Yakk and Mr. Wakely, this type does not matter. It is simply the type of something that "points to" a char with which we can iterator over our string. This is one of the main selling points of auto - if the name of the type doesn't matter, don't clutter your code with it.

Note that since you never actually need iterators in your code, you could simply avoid them with a range-based for loop:

for (auto& str : s) {
    for (auto& chr: str) {
        chr = toupper(chr);
    }
}

Alternatively, if you ever need to know the exact type of something, I'd recommend the following hack. Make a class template that takes a type but is never defined:

template <typename > struct TD; // for Type Description

And then just stick it in somewhere:

for (auto it2 = ... ) {
    TD<decltype(it2)> t;
    // ...
}

which would give the following error with gcc 5.2:

main.cpp:16:34: error: aggregate 'TD<__gnu_cxx::__normal_iterator<char*, std::__cxx11::basic_string<char> > > t' has incomplete type and cannot be defined
                TD<decltype(it2)> t;
                                  ^

Notice that the compile error has the complete type of it2 in it.

Upvotes: 5

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