> (equal? obj1 obj2) procedure > > The equal? procedure, when applied to pairs, vectors, strings and > bytevectors, recursively compares them, returning #t when the unfoldings of > its arguments into (possibly infinite) trees are equal as ordered trees, and > #f otherwise. It returns the same as eqv? when applied to booleans, symbols, > numbers, characters, ports, procedures, and the empty list. If two objects > are eqv?, they must be equal? as well. In all other cases, equal? may return > either #t or #f.
What, precisely, does it mean to say that the "trees are equal". I suggest: "returning #t when the unfoldings of its arguments into (possibly infinite) trees are equal as ordered trees, with corresponding leaves being compared by equal?, and #f otherwise" There may well be better ways to say this. Alternatively, I think one could say that the leafs are compared by eqv? (since structure has been unfolded). Regards, Alan _______________________________________________ Scheme-reports mailing list [email protected] http://lists.scheme-reports.org/cgi-bin/mailman/listinfo/scheme-reports
