http://d.puremagic.com/issues/show_bug.cgi?id=6942

           Summary: lazy parameters can break purity
           Product: D
           Version: D2
          Platform: All
        OS/Version: All
            Status: NEW
          Keywords: accepts-invalid
          Severity: normal
          Priority: P2
         Component: DMD
        AssignedTo: nobody@puremagic.com
        ReportedBy: timon.gehr@gmx.ch


--- Comment #0 from timon.gehr@gmx.ch 2011-11-13 04:48:46 PST ---
int foo(lazy int x) pure{
    return x()+x();
}

void main(){
    auto a=foo((writeln("impure"),1));
}

This compiles and prints
impure
impure

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--- Comment #1 from timon.gehr@gmx.ch 2011-11-13 04:54:18 PST ---
note that

int foo(int delegate() x) pure{
    return x()+x();
}

void main(){
    auto a=foo({return writeln("impure"),1;});
}

fails with

Error: pure function 'foo' cannot call impure delegate 'x'

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Kenji Hara <k.hara.pg@gmail.com> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|NEW                         |RESOLVED
         Resolution|                            |INVALID


--- Comment #2 from Kenji Hara <k.hara.pg@gmail.com> 2011-12-02 18:56:15 PST ---
The pureness of lazy parameter belongs to the *caller side*, not callee side. It is a design.

One use case is std.exception.enforce. It receives a condition as a lazy parameter, but whole enforce function can become pure with the design.

Delegate parameter is similar to lazy parameter, but it is different in this point.

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