March 20, 2007 [Issue 1073] New: CTFE: mixin expression with a function argument doesn't work | ||||
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http://d.puremagic.com/issues/show_bug.cgi?id=1073 Summary: CTFE: mixin expression with a function argument doesn't work Product: D Version: 1.009 Platform: PC OS/Version: Windows Status: NEW Severity: normal Priority: P2 Component: DMD AssignedTo: bugzilla@digitalmars.com ReportedBy: torhu@yahoo.com This code checks if TEST is defined, and defines it if not. If using the line that's commented out instead of the one above it, it works as expected. --- const int TEST = 1; char[] def(char[] s) { if (!is(typeof(mixin(s)))) // doesn't detect that TEST is defined //if (!is(typeof(mixin("TEST")))) // this works fine return "const int " ~ s ~ " = 3;\n"; else return ""; } mixin(def("TEST")); void main() { } --- test.d: variable test.TEST conflicts with test.TEST at test.d(1) I tried to recreate this problem without CTFE, just with a static if, but it doesn't happen then. -- |
March 31, 2007 [Issue 1073] CTFE: mixin expression with a function argument doesn't work | ||||
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Posted in reply to d-bugmail | http://d.puremagic.com/issues/show_bug.cgi?id=1073 bugzilla@digitalmars.com changed: What |Removed |Added ---------------------------------------------------------------------------- Status|NEW |RESOLVED Resolution| |INVALID ------- Comment #1 from bugzilla@digitalmars.com 2007-03-31 17:29 ------- What this code is trying to do is to dynamically change the parse, rather than the evaluated result, at compile time. This isn't going to work (should use templates for that). I'll put some clarifications in the spec about this. -- |
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