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Neil Booth wrote: >> *boggle* you mean this is broken even without varargs? Ouch! > >No, only in the varargs case. Ah, I see. Here is my simplified test case, probably like yours: #define half(x) ((x) / 2) #define apply(...) apply2 (__VA_ARGS__) #define apply2(f,x) f (x) apply (half, X) Expands to `half (X)' when it should expand to `((X) / 2)'. -- Jamie
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