SWI-prolog semweb library processing of URI
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Being new to prolog I am reading existing code (as well as trying to write some code). Having some prior background in semweb I started to play with it and see something that is confusing me. Example assertion:
?- rdf_assert(ex:bob, rdf:type, foaf:'Person').
I also did find the following in the documentation:
Remember: Internally, all resources are atoms. The transformations
above are realised at compile-time using rules for goal_expansion/2
provided by the rdf_db library
Am I correct in assuming that somehow the library is treating the three URIs as atoms? I thought that the compiler would treat this as module_name:predicate, but that does not seem to be the case. If that is true, could you please provide a simple example on how this could be done in prolog?
Thanks
prolog swi-prolog semweb
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up vote
1
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Being new to prolog I am reading existing code (as well as trying to write some code). Having some prior background in semweb I started to play with it and see something that is confusing me. Example assertion:
?- rdf_assert(ex:bob, rdf:type, foaf:'Person').
I also did find the following in the documentation:
Remember: Internally, all resources are atoms. The transformations
above are realised at compile-time using rules for goal_expansion/2
provided by the rdf_db library
Am I correct in assuming that somehow the library is treating the three URIs as atoms? I thought that the compiler would treat this as module_name:predicate, but that does not seem to be the case. If that is true, could you please provide a simple example on how this could be done in prolog?
Thanks
prolog swi-prolog semweb
New contributor
stan_plogic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Being new to prolog I am reading existing code (as well as trying to write some code). Having some prior background in semweb I started to play with it and see something that is confusing me. Example assertion:
?- rdf_assert(ex:bob, rdf:type, foaf:'Person').
I also did find the following in the documentation:
Remember: Internally, all resources are atoms. The transformations
above are realised at compile-time using rules for goal_expansion/2
provided by the rdf_db library
Am I correct in assuming that somehow the library is treating the three URIs as atoms? I thought that the compiler would treat this as module_name:predicate, but that does not seem to be the case. If that is true, could you please provide a simple example on how this could be done in prolog?
Thanks
prolog swi-prolog semweb
New contributor
stan_plogic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Being new to prolog I am reading existing code (as well as trying to write some code). Having some prior background in semweb I started to play with it and see something that is confusing me. Example assertion:
?- rdf_assert(ex:bob, rdf:type, foaf:'Person').
I also did find the following in the documentation:
Remember: Internally, all resources are atoms. The transformations
above are realised at compile-time using rules for goal_expansion/2
provided by the rdf_db library
Am I correct in assuming that somehow the library is treating the three URIs as atoms? I thought that the compiler would treat this as module_name:predicate, but that does not seem to be the case. If that is true, could you please provide a simple example on how this could be done in prolog?
Thanks
prolog swi-prolog semweb
prolog swi-prolog semweb
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stan_plogic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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edited Nov 17 at 19:57
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asked Nov 17 at 19:52
stan_plogic
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1 Answer
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Prolog is not a functional language. This implies 2+3 does not evaluate to 5 and is just a term that gets meaning from the predicate that processes it. Likewise, ex:bob is just a term that has no direct relations to modules or
predicates. Only predicates such call/1 will interpret this as "call bob in the module ex".
Next to that, (SWI-)Prolog (most Prolog's, but not all) have term expansion that allows you to rewrite the term that is read before it is handed to the compiler. That is used to rewrite the argument of rdf/3: each appearance of prefix:local is expanded to a full atom. You can check that by using listing/1 on predicates that call rdf/3 using the prefix notation.
See also rdf_meta
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Prolog is not a functional language. This implies 2+3 does not evaluate to 5 and is just a term that gets meaning from the predicate that processes it. Likewise, ex:bob is just a term that has no direct relations to modules or
predicates. Only predicates such call/1 will interpret this as "call bob in the module ex".
Next to that, (SWI-)Prolog (most Prolog's, but not all) have term expansion that allows you to rewrite the term that is read before it is handed to the compiler. That is used to rewrite the argument of rdf/3: each appearance of prefix:local is expanded to a full atom. You can check that by using listing/1 on predicates that call rdf/3 using the prefix notation.
See also rdf_meta
add a comment |
up vote
1
down vote
Prolog is not a functional language. This implies 2+3 does not evaluate to 5 and is just a term that gets meaning from the predicate that processes it. Likewise, ex:bob is just a term that has no direct relations to modules or
predicates. Only predicates such call/1 will interpret this as "call bob in the module ex".
Next to that, (SWI-)Prolog (most Prolog's, but not all) have term expansion that allows you to rewrite the term that is read before it is handed to the compiler. That is used to rewrite the argument of rdf/3: each appearance of prefix:local is expanded to a full atom. You can check that by using listing/1 on predicates that call rdf/3 using the prefix notation.
See also rdf_meta
add a comment |
up vote
1
down vote
up vote
1
down vote
Prolog is not a functional language. This implies 2+3 does not evaluate to 5 and is just a term that gets meaning from the predicate that processes it. Likewise, ex:bob is just a term that has no direct relations to modules or
predicates. Only predicates such call/1 will interpret this as "call bob in the module ex".
Next to that, (SWI-)Prolog (most Prolog's, but not all) have term expansion that allows you to rewrite the term that is read before it is handed to the compiler. That is used to rewrite the argument of rdf/3: each appearance of prefix:local is expanded to a full atom. You can check that by using listing/1 on predicates that call rdf/3 using the prefix notation.
See also rdf_meta
Prolog is not a functional language. This implies 2+3 does not evaluate to 5 and is just a term that gets meaning from the predicate that processes it. Likewise, ex:bob is just a term that has no direct relations to modules or
predicates. Only predicates such call/1 will interpret this as "call bob in the module ex".
Next to that, (SWI-)Prolog (most Prolog's, but not all) have term expansion that allows you to rewrite the term that is read before it is handed to the compiler. That is used to rewrite the argument of rdf/3: each appearance of prefix:local is expanded to a full atom. You can check that by using listing/1 on predicates that call rdf/3 using the prefix notation.
See also rdf_meta
answered 9 hours ago
Jan Wielemaker
1,106612
1,106612
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