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From: "ian at airs dot com" <gcc-bugzilla@gcc.gnu.org>
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Subject: [Bug tree-optimization/47579] New: STL size() == 0 does unnecessary shift
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=47579

           Summary: STL size() == 0 does unnecessary shift
           Product: gcc
           Version: 4.6.0
            Status: UNCONFIRMED
          Keywords: missed-optimization
          Severity: normal
          Priority: P3
         Component: tree-optimization
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: ian@airs.com


Consider this C++ code:

#include <vector>
extern void b1(), b2();
void foo(const std::vector<int>& v) { if (v.size() == 0) b1(); else b2(); }

When I compile it with current mainline with -O2 on x86_64, I get this:

        movq    8(%rdi), %rax
        subq    (%rdi), %rax
        sarq    $2, %rax
        testq   %rax, %rax
        ...

That sarq instruction is useless.  We know that the two values being subtracted
are both aligned pointers, so we should know that the two least significant
bits of the result are zero.  And that should be enough to let us know that we
don't need to shift before comparing for equality with zero.