为什么 C++ 标准容器迭代器会忽略其底层容器的一些模板参数？

Question

Here is a specific example. In the code below, I would have expected a compilation error like "cannot assign value of type std::map< int, int, cmp >::iterator to variable of type std::map< int, int >::iterator". However, I am able to compile the code without any issues (g++ C++20 Wall). Thanks for help.

#include <map>

struct cmp {
    bool operator()(const int a, const int b) const { return a > b; }
};

int main(int argc, char *argv[]) {

    std::map<int, int> m1;
    std::map<int, int>::iterator m1_it = m1.begin();

    std::map<int, int, cmp> m2;
    std::map<int, int, cmp>::iterator m2_it = m2.begin();

    // Why do these work?
    m1_it = m2.begin();
    m2_it = m1.begin();

    return 0;
}

Answer 1

Standard sets few constraints on iterator types and just impose some behaviours.

In particular:

• they don't have to be in namespace std . So T* might be a valid iterator for std::vector<T> .

• they don't have to have same template parameters than their container. For above example T* might be an iterator for std::vector<T, Allocator> for any Allocator .

Whereas, indeed, it is less type safe, less template parameter allows less instantiations (faster compilation, less generated similar code, ...).

Notice that your code might not compile for compilers/libraries which uses all template parameters for their iterators.

Answer 2

One practical reason is that this might lead to smaller executables. Your two map types can share one iterator type, so all the iterator functions are also shared.