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Array reference and implicit constructor: bug in gcc or not? [duplicate]

I'm seeing some odd behaviour when returning a string literal from a function that should perform an implicit conversion with g++ (version 4.7.3). Can anyone explain why the following code:

#include <stdio.h>

class Test
{
public:
  template <unsigned int N>
  Test(const char (&foo)[N])
  {
    printf("Template const char array constructor\n");
  }

  Test(char* foo)
  {
    printf("char* constructor\n");
  }
};

Test fn()
{
  return "foo";
}

int main()
{
  Test t("bar");
  Test u = fn();

  return 0;
}

produces the result:

Template const char array constructor
char* constructor

on g++? The surprising thing being that the char* constructor is chosen in preference to the const char array constructor when generating the return value from fn(). Admittedly there is a warning, "deprecated conversion from string constant to 'char*'"

Even more surprisingly if you remove the char* constructor then the code doesn't compile with g++.

It works as expected with clang (Template constructor used both times), which makes me think this is a compiler bug, but maybe it's just a weird corner of the C++ spec - could anyone confirm?

like image 788
mikebell Avatar asked Feb 26 '14 22:02

mikebell


2 Answers

It appears that this is a bug affecting several versions of gcc which has been reported over and over again, most recently about a month ago against the most recent version, 4.8.2. See http://gcc.gnu.org/bugzilla/show_bug.cgi?id=24666

like image 121
Brian Bi Avatar answered Sep 28 '22 01:09

Brian Bi


If you want a reusable C++03 workaround (i.e. one in which you won't have to care about what the return type is, as long it is constructible from a char array), you will have to use some kind of a char array wrapper.

template <size_t N>
struct char_array_ref
{
    typedef const char (&ref_type)[N];
    ref_type ref;

    template <typename T>
    operator T() const
    {
        return T(ref);
    }
};

template <size_t N>
char_array_ref<N> stupid_gxx_use_array_reference(const char (&chars)[N])
{
    return char_array_ref<N> { chars };
}


Test fn()
{
  return stupid_gxx_use_array_reference("foo");
}

Should be easy to regex propagate this across your codebase, too.

Obviously, in your code you can change stupid_gxx_use_array_reference into something less verbose.

like image 25
gwiazdorrr Avatar answered Sep 28 '22 01:09

gwiazdorrr