Get to STL container's size_type using a variable

I want to write index-based for loops in a simple way. Since my code has to run in 32 and 64 bit, I get lots of warnings about implicit conversions because I usually use 32 bit integers but the STL uses 64 bit integers in 64 bit.

std::vector<MyClass> c;
for (FOO ui = 0, uiEnd = c.size(); ui < uiEnd; ++ui){...}

FOO = unsigned int works on 32 bit but generates warnings on 64 bit.
FOO = size_t works on 32 and 64 bit but is not entirely correct because the Standard does not require std::vector<MyClass>::size_type == size_t.
FOO = auto does not work (which is quite unfortunate) because it sets the type for each variable individually, so the type of the literal 0 (or 0UL or whatever) which differs from c.size() (in 32 or 64 bit, the type of c.size() can differ, so I cannot write the correct literal).
for (auto uiEnd = c.size(), ui = 0*uiEnd; ui < uiEnd; ++ui){...} works by expressing 0 with the correct type but is obscure.
FOO = decltype(c.size()) works but I find it obscure.
FOO = std::vector<MyClass>::size_type works but I find it tedious to repeat the type of the container.

I would prefer something like FOO = c::size_type, which clearly expresses what is happening. But is it possible to get the size_type of std::vector<MyClass> using only the variable c? I tried the following things without success:

c::size_type
typeid(c)::size_type
decltype(c)::size_type

Solution

Whereas you might use:

std::vector<MyClass> c;

for (decltype(c)::size_type ui = 0, uiEnd = c.size(); ui < uiEnd; ++ui){...}

I would use:

std::vector<MyClass> c;

std::size_t index = 0; // as you want index too
for (auto& myclass : c){...; ++index;}

And even in C++20:

std::vector<MyClass> c;

for (std::size_t index = 0; auto& myclass : c){...; ++index;}

ranges-v3 has enumerate view:

std::vector<MyClass> v;

for (const auto& [index, myclass] : v | ranges::v3::views::enumerate) {
    // ...
}

Demo