type_info doesn't account for cv qualifiers: is this right?

问题

Is it the correct behaviour or is it a quirk of g++4.5 that this code prints 1?

#include <iostream>
#include <typeinfo>
using namespace std;

int main(){
    struct A{};
    cout<<(typeid(A)==typeid(const A)&&typeid(A)==typeid(const volatile A)&&typeid(A)==typeid(volatile A));
}

I thought that types differing for cv-qualifiers were threated as very distinct types, even though less cv-qualified types could be implicitly cast to more cv-qualified types.

回答1:

typeid ignores cv qualifiers, as per the C++ standard (taken from §5.2.8 from ISO/IEC 14882:2003) :

The top-level cv-qualifiers of the lvalue expression or the type-id that is the operand of typeid are always ignored. [Example:

class D { ... };
D d1;
const D d2;

typeid(d1) == typeid(d2);       // yields true
typeid(D) == typeid(const D);   // yields true
typeid(D) == typeid(d2);        // yields true
typeid(D) == typeid(const D&);  // yields true

—end example]

So, the result you're seeing is expected.

回答2:

As @SanderDeDycker mentioned cv qualifiers are ignored. But not for pointers!

example:

#include <iostream>
#include <typeinfo>

int main()
{
    int bar = 1;
    const int cbar = 10;
    volatile int vbar = 10;
    const volatile int cvbar = 1000;
    int &rbar = bar;
    const int &crbar = bar;
    volatile int &vrbar = bar;
    const volatile int &cvrbar = bar;
    int *pbar = &bar;
    const int *cpbar = &bar;
    volatile int *vpbar = &bar;
    const volatile int *cvpbar = &bar;
    const volatile int * const cvpcbar = &bar;
    const volatile int * const volatile cvpcvbar = &bar;
    int&& rrbar = 894354;
    const int&& rrcbar = 894354;

    std::cout << typeid(bar).name() << '\n';
    std::cout << typeid(cbar).name() << '\n';
    std::cout << typeid(vbar).name() << '\n';
    std::cout << typeid(cvbar).name() << '\n';
    std::cout << typeid(rbar).name() << '\n';
    std::cout << typeid(crbar).name() << '\n';
    std::cout << typeid(vrbar).name() << '\n';
    std::cout << typeid(cvrbar).name() << '\n';
    std::cout << typeid(pbar).name() << '\n';
    std::cout << typeid(cpbar).name() << '\n';
    std::cout << typeid(vpbar).name() << '\n';
    std::cout << typeid(cvpbar).name() << '\n';
    std::cout << typeid(cvpcbar).name() << '\n';
    std::cout << typeid(cvpcvbar).name() << '\n';
    std::cout << typeid(rrbar).name() << '\n';
    std::cout << typeid(rrcbar).name() << '\n';
    std::cout << "\n\n";
}

Output

int
int
int
int
int
int
int
int
int * __ptr64
int const * __ptr64
int volatile * __ptr64
int const volatile * __ptr64
int const volatile * __ptr64
int const volatile * __ptr64
int
int

As you can see cv qualifiers are not ignored in the 5 cases involving pointers. Output is the same for MSVS or g++ and C++11, C++14, C++17.

来源：https://stackoverflow.com/questions/10007596/type-info-doesnt-account-for-cv-qualifiers-is-this-right

标签

c++

g++

const

volatile

rtti