CODEWITHSUNDEEP
c++objecttypeidtypeinfo
fast-reflexes
fast-reflexes

Reputation: 5176

C++ Methods of store type_info objects doesn't work

I'm having some trouble understanding the correspondance between the return type of typeid and actual type_info objects which seems to work differently from usual objects. For example, I can do...

std::cout << typeid(int).name() << std::endl;

...and get a decent behaviour form the program... but this won't compile...

std::type_info a(int);
std::cout << a.name() << std::endl;

Compiler outputs:

type_info.cpp: In function 'int main()':
type_info.cpp:6:17: error: request for member 'name' in 'a', which is of non-class type 'std::type_info(int)'

...nor can I do...

if(a == typeid(int)) {/*Do something*/ } //Looong error message

What am I missing?

Upvotes: 7

Views: 6452

Answers (2)

dom0
dom0

Reputation: 7486

std::type_info is neither copy-constructible nor copy-assignable. Thus you can't do stuff like std::type_info a = typeid(...).

const std::type_info &someTypeId(typeid(someType));

You can also store a pointer to a std::type_info object, because typeid returns a lvalue.

There is also std::type_index in case you want to use std::type_info objects as a key in containers. It is essentially a thin wrapper around a pointer to a std::type_info. std::type_index is a C++11 feature.

Upvotes: 3

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171097

First off, vexing parse:

std::type_info a(int);

a is a function (taking int and returning std::type_info).

Second, std::type_info is not copyable, so you cannot store it by value. You can use a reference if it suits your needs:

const std::type_info &a(typeid(int));

If you need to actually store std::type_info objects (as if) "by value," use std::type_index instead; it was designed for this:

std::type_index a(typeid(int));
std::cout << a.name() << std::endl;

Upvotes: 12

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