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((and (list? items) ;; thus we know our items are already calculated
(not itemdat)) ;; and not yet expanded into the list of things to be done
(debug:print-info 4 "OUTER COND: (and (list? items)(not itemdat))")
;; Must determine if the items list is valid. Discard the test if it is not.
(if (and (list? items)
(> (length items) 0)
(and (list? (car items))
(> (length (car items)) 0)))
(begin
(if (debug:debug-mode 1)(pp items))
(for-each
(lambda (my-itemdat)
(let* ((new-test-record (let ((newrec (make-tests:testqueue)))
(vector-copy! test-record newrec)
newrec))
(my-item-path (item-list->path my-itemdat)))
(if (tests:match test-patts hed my-item-path required: required-tests) ;; (patt-list-match my-item-path item-patts) ;; yes, we want to process this item, NOTE: Should not need this check here!
(let ((newtestname (runs:make-full-test-name hed my-item-path))) ;; test names are unique on testname/item-path
(tests:testqueue-set-items! new-test-record #f)
(tests:testqueue-set-itemdat! new-test-record my-itemdat)
(tests:testqueue-set-item_path! new-test-record my-item-path)
(hash-table-set! test-records newtestname new-test-record)
(set! tal (append tal (list newtestname))))))) ;; since these are itemized create new test names testname/itempath
items))
(debug:print-info 0 "Test " (tests:testqueue-get-testname test-record) " is itemized but has no items"))
;; At this point we have possibly added items to tal but all must be handed off to
;; INNER COND logic. I think loop without rotating the queue
;; (loop hed tal reg reruns))
;; (let ((newtal (append tal (list hed)))) ;; We should discard hed as it has been expanded into it's items? Yes, but only if this *is* an itemized test
;; (loop (car newtal)(cdr newtal) reg reruns)
(if (null? tal)
#f
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((and (list? items) ;; thus we know our items are already calculated
(not itemdat)) ;; and not yet expanded into the list of things to be done
(debug:print-info 4 "OUTER COND: (and (list? items)(not itemdat))")
;; Must determine if the items list is valid. Discard the test if it is not.
(if (and (list? items)
(> (length items) 0)
(and (list? (car items))
(> (length (car items)) 0))
(debug:debug-mode 1))
(pp items))
(for-each
(lambda (my-itemdat)
(let* ((new-test-record (let ((newrec (make-tests:testqueue)))
(vector-copy! test-record newrec)
newrec))
(my-item-path (item-list->path my-itemdat)))
(if (tests:match test-patts hed my-item-path required: required-tests) ;; (patt-list-match my-item-path item-patts) ;; yes, we want to process this item, NOTE: Should not need this check here!
(let ((newtestname (runs:make-full-test-name hed my-item-path))) ;; test names are unique on testname/item-path
(tests:testqueue-set-items! new-test-record #f)
(tests:testqueue-set-itemdat! new-test-record my-itemdat)
(tests:testqueue-set-item_path! new-test-record my-item-path)
(hash-table-set! test-records newtestname new-test-record)
(set! tal (append tal (list newtestname))))))) ;; since these are itemized create new test names testname/itempath
items)
;; (debug:print-info 0 "Test " (tests:testqueue-get-testname test-record) " is itemized but has no items")
;; At this point we have possibly added items to tal but all must be handed off to
;; INNER COND logic. I think loop without rotating the queue
;; (loop hed tal reg reruns))
;; (let ((newtal (append tal (list hed)))) ;; We should discard hed as it has been expanded into it's items? Yes, but only if this *is* an itemized test
;; (loop (car newtal)(cdr newtal) reg reruns)
(if (null? tal)
#f
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