How to interpret a g ++ warning

Question

I have a very strange g ++ warning when trying to compile the following code:

#include <map>
#include <set>

class A {
public:

    int x;
    int y;

    A(): x(0), y(0) {}
    A(int xx, int yy): x(xx), y(yy) {}

    bool operator< (const A &a) const {
        return (x < a.x || (!(a.x < x) && y < a.y));
    }
};

struct B {
    std::set<A> data;
};

int
main()
{
    std::map<int, B> m;

    B b;

    b.data.insert(A(1, 1));
    b.data.insert(A(1, 2));
    b.data.insert(A(2, 1));

    m[1] = b;

    return 0;
}

Output:

$ g++ -Wall -W -O3 t.cpp -o /tmp/t
t.cpp: In function ‘int main()’:
t.cpp:14: warning: dereferencing pointer ‘__x.52’ does break strict-aliasing rules
t.cpp:14: warning: dereferencing pointer ‘__x.52’ does break strict-aliasing rules
/usr/lib/gcc/i686-redhat-linux/4.4.2/../../../../include/c++/4.4.2/bits/stl_tree.h:525: note: initialized from here

This has nothing to do with me at all. How do I interpret this? I do not see what is wrong with the submitted code.

Forget specifying compiler details:

$ gcc --version
gcc (GCC) 4.4.2 20091027 (Red Hat 4.4.2-7)

Answer 1

gcc 4.4 has an error in which std :: map breaks incorrectly warns about strict anti-aliasing rules.

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=39390

Your code is valid C ++. Strict anti-aliasing simply allows a subset of optimizations that are enabled by default when used -O3.

-fno-strict-aliasing gcc.

, , .

Answer 2

:

return (x < a.x || (!(a.x < x) && y < a.y));

return (x < a.x || (a.x == x && y < a.y));

, g++ 3.4.2, .

Answer 3

What version of g ++? g ++ 4.3.2 compiles this without complaint.