Count occurrences of a substring: Difference between revisions
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m (→{{header|PHP}}: Avoiding "?>" at the end of your source file makes extra whitespace to appear less likely.) |
(→{{header|Prolog}}: version using DCG) |
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X = 2. |
X = 2. |
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</lang> |
</lang> |
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=== version using DCG === |
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{{works with|SWI-Prolog|7.6.4}} |
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<lang prolog> |
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:- system:set_prolog_flag(double_quotes,chars) . |
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occurrences(TARGETz0,SUBSTRINGz0,COUNT) |
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:- |
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prolog:phrase(occurrences(SUBSTRINGz0,0,COUNT),TARGETz0) |
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. |
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occurrences("",_,_) |
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--> |
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! , |
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{ false } |
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. |
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occurrences(SUBSTRINGz0,COUNT0,COUNT) |
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--> |
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SUBSTRINGz0 , |
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! , |
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{ COUNT1 is COUNT0 + 1 } , |
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occurrences(SUBSTRINGz0,COUNT1,COUNT) |
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. |
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occurrences(SUBSTRINGz0,COUNT0,COUNT) |
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--> |
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[_] , |
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! , |
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occurrences(SUBSTRINGz0,COUNT0,COUNT) |
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. |
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occurrences(_SUBSTRINGz0_,COUNT,COUNT) |
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--> |
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! |
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. |
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</lang> |
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{{out}} |
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<pre> |
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/* |
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?- occurrences("the three truths","th",COUNT) . |
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COUNT = 3. |
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?- occurrences("the three truths","",COUNT) . |
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false . |
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?- occurrences("ababababab","abab",COUNT) . |
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COUNT = 2. |
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?- occurrences("ababababab","",COUNT) . |
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false . |
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?- |
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*/ |
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</pre> |
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=={{header|PureBasic}}== |
=={{header|PureBasic}}== |
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