http://d.puremagic.com/issues/show_bug.cgi?id=11044

           Summary: Escaping references to lazy argument are allowed
           Product: D
           Version: unspecified
          Platform: All
        OS/Version: All
            Status: NEW
          Keywords: accepts-invalid
          Severity: normal
          Priority: P2
         Component: DMD
        AssignedTo: nobody@puremagic.com
        ReportedBy: maxim@maxim-fomin.ru


--- Comment #0 from Maxim Fomin <maxim@maxim-fomin.ru> 2013-09-15 13:17:04 PDT ---
Currently lazy parameter implies creating delegate which references local variable. Allowing to escape lazy argument without allocating local variable at heap leads to leaked pointers to some stack zone which is a memory error.

extern(C) int printf(const char*,...) @safe;

auto foo(lazy double i) @safe
{
   return { return i; } ;
}

auto bar() @safe
{
   double i = 4.0;
   return foo(i);
}

void baz() @safe
{
   double[2] i = 3.14; // this value overwrites 4.0
                       // replace with char[16] val = 'f';
                       // to print deterministic garbage
}

void main() @safe
{
   auto x = bar();
   baz();
   printf("%f\n", x());
}

This prints 3.14 which is wrong. Potentially compiler can recognize that lazy argument escapes and allocate i = 4.0 on heap, but due to separate compilation model, it cannot do this in general case, so the code should be rejected.

This is a more complex variation of known 'ref-ref' bug which consists in returning reference to local object which has gone out of scope:

ref int foo(ref int i) { return i; }
ref int bar() { int val; return foo(i); }

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