typeid(x).name() yields 'y', what does it mean?

August 5, 2023

In Code below, we get:
12 4
y
typeid().name should return data type,
but I cannot find which data type ‘y’ means 🤔

void alignment()
{
    struct Foo
    {
        int   i;
        float f;
        char  c;
    };

    std::cout << sizeof(Foo) << '\t' << alignof(Foo) << '\n';
    auto x = alignof(Foo);
    std::cout << typeid(x).name() << '\n';
}

I tried to find an association with data type names but nothing fits

>Solution :

I tried to find an association with data type names but nothing fits

Your compiler gave std::size_t name y, mine called it m, both are right and fully allowed to.
According to:
https://en.cppreference.com/w/cpp/types/type_info/name

Some implementations (such as MSVC, IBM, Oracle) produce a human-readable type name. Others, most notably gcc and clang, return the mangled name, which is specified by the Itanium C++ ABI. The mangled name can be converted to human-readable form using implementation-specific API such as abi::__cxa_demangle directly or through boost::core::demangle. It can also be piped through the command-line utility c++filt -t.

And alignof returns size_t, see here.

Example with demangling that using Boost:
https://godbolt.org/z/8959KdMhv