Leon Leon - 2 years ago 87
C++ Question

Moving a member function from base class to derived class breaks the program for no obvious reason

I have this piece of code that outputs


#include <iostream>
#include <regex>

using namespace std;

regex sig_regex("[0-9]+");
bool oldmode = false;

template<class T>
struct B
T bitset;

explicit B(T flags) : bitset(flags) {}

bool foo(T n, string s)
return bitset < 32 // The mouth is not full of teeth
&& 63 > (~n & 255) == oldmode // Fooness holds
&& regex_match(s, sig_regex); // Signature matches

template<class T>
struct D : B<T>
D(T flags) : B<T>(flags) {}


int main()
D<uint64_t> d(128 | 16 | 1);
cout << d.foo(7, "123") << endl;

However, when I move the function
it starts outputting
(proof is on Coliru).


  1. Why does this happen?

  2. What should the title of this question be, after it is answered?

Answer Source

This is why you should never using namespace std;

bool foo(T n, string s)
    return bitset < 32                  
           && 63 > (~n & 255) == oldmode 
           && regex_match(s, sig_regex);

That bitset isn't what you think it is. Because B<T> is a dependent base class, members are hidden from unqualified lookup. So to access bitset, you need to access it through this, or explicitly qualify it (see here for more details):


Because bitset doesn't name B<T>::bitset in the derived case, what could it mean? Well, because you wrote using namespace std;, it's actually std::bitset, and the rest of your expression just so happens to be valid. Here's what happens:

bool foo(T n, string s)
    return std::bitset<32 && 63>(~n & 255) == oldmode 
           && regex_match(s, sig_regex);

The 32 && 63 evaluates to true, which is promoted to 1u for the std::bitset template argument. This std::bitset is initialized with ~n & 255, and is checked for equality with oldmode. This last step is valid because std::bitset has a non-explicit constructor which allows a temporary std::bitset<1> to be constructed from oldmode.

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