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Another breaking unit test for ebnf

db4
Chris Double 2008-06-20 12:35:33 +12:00
parent 91768f171c
commit 6bd761e460
1 changed files with 11 additions and 4 deletions
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15
extra/peg/ebnf/ebnf-tests.factor
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@ -2,7 +2,7 @@
! See http://factorcode.org/license.txt for BSD license. ! See http://factorcode.org/license.txt for BSD license.
! !
USING: kernel tools.test peg peg.ebnf words math math.parser USING: kernel tools.test peg peg.ebnf words math math.parser
sequences accessors peg.parsers ; sequences accessors peg.parsers parser namespaces ;
IN: peg.ebnf.tests IN: peg.ebnf.tests
{ T{ ebnf-non-terminal f "abc" } } [ { T{ ebnf-non-terminal f "abc" } } [
@ -443,10 +443,17 @@ foo=<foreign any-char> 'd'
"ad" parser4 ast>> "ad" parser4 ast>>
] unit-test ] unit-test
{ V{ "a" "\n" } } [ { t } [
"a\n" [EBNF foo='a' '\n' => [[ drop '\n' ]] EBNF] call ast>> "USING: kernel peg.ebnf ; [EBNF foo='a' '\n' => [[ drop '\n' ]] EBNF]" eval drop t
] unit-test ] unit-test
[
"USING: peg.ebnf ; [EBNF foo='a' foo='b' EBNF]" eval
] must-fail
{ t } [ { t } [
[EBNF foo='a' foo='b' EBNF] drop t #! Rule lookup occurs in a namespace. This causes an incorrect duplicate rule
#! if a var in a namespace is set. This unit test is to remind me to fix this.
[ "fail" "foo" set "foo='a'" 'ebnf' parse ast>> transform drop t ] with-scope
] unit-test ] unit-test