序言中的语法

Question

I want to be able to translate arrays like [int(2), sep(<), int(1), sep(-), id(N)] into conditions (in this example [int(2) < int(1)-id(N)] ) using grammar. I have the grammar below:

cond(_, A > B) --> expr(_,A), [sep(>)], expr(_,B).
cond(_, A < B) --> expr(_,A), [sep(<)], expr(_,B).
expr(_, int(X)) --> [int(X)].
expr(_, id(X)) --> [id(X)].
expr(_, X + Y) --> expr(_,X), [sep(+)], expr(_,Y).
expr(_, X - Y) --> expr(_,X), [sep(-)], expr(_,Y).

What I don't understand is the fact that if I have the array: [int(2), sep(>), int(1), sep(+), id(N)] everything is fine, the condition is translated correctly. I make some changes then, replacing + by - or > by < (take account of the fact that in my grammar > is one line over < , same as + and - ) and this isn't translated, it falls into infinite loop:

Call:  (18)  ? [sep(-), id(N)]=[sep(+)|_G2290032]creep
Fail:  (18)  ? [sep(-), id(N)]=[sep(+)|_G2290032]creep

It is always comparing to line with sep(+) . What is wrong in this grammar?

--EDIT added tracing

phrase(cond(_, X), [int(2), sep(<), int(1), sep(-),id(N)]).
   Call:  (10)  ? cond(_G2313, _G2314, [int(2), sep(<), int(1), sep(-), id(_G2336)], [])creep
   Call:  (11)  ? expr(_L254, _G2482, [int(2), sep(<), int(1), sep(-), id(_G2336)], _L235)creep
   Exit:  (11)  ? expr(_L254, int(2), [int(2), sep(<), int(1), sep(-), id(_G2336)], [sep(<), int(1), sep(-), id(_G2336)])creep
   Call:  (11)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(>)|_G2488]creep
   Fail:  (11)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(>)|_G2488]creep
   Redo:  (11)  ? expr(_L254, _G2482, [int(2), sep(<), int(1), sep(-), id(_G2336)], _L235)creep
   Call:  (12)  ? expr(_L279, _G2485, [int(2), sep(<), int(1), sep(-), id(_G2336)], _L260)creep
   Exit:  (12)  ? expr(_L279, int(2), [int(2), sep(<), int(1), sep(-), id(_G2336)], [sep(<), int(1), sep(-), id(_G2336)])creep
   Call:  (12)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(+)|_G2491]creep
   Fail:  (12)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(+)|_G2491]creep
   Redo:  (12)  ? expr(_L279, _G2485, [int(2), sep(<), int(1), sep(-), id(_G2336)], _L260)creep
   Call:  (13)  ? expr(_L304, _G2488, [int(2), sep(<), int(1), sep(-), id(_G2336)], _L285)creep
   Exit:  (13)  ? expr(_L304, int(2), [int(2), sep(<), int(1), sep(-), id(_G2336)], [sep(<), int(1), sep(-), id(_G2336)])creep
   Call:  (13)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(+)|_G2494]creep
   Fail:  (13)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(+)|_G2494]creep
   Redo:  (13)  ? expr(_L304, _G2488, [int(2), sep(<), int(1), sep(-), id(_G2336)], _L285)creep
   Call:  (14)  ? expr(_L329, _G2491, [int(2), sep(<), int(1), sep(-), id(_G2336)], _L310)creep
   Exit:  (14)  ? expr(_L329, int(2), [int(2), sep(<), int(1), sep(-), id(_G2336)], [sep(<), int(1), sep(-), id(_G2336)])creep
   Call:  (14)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(+)|_G2497]creep
   Fail:  (14)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(+)|_G2497]creep
   Redo:  (14)  ? expr(_L329, _G2491, [int(2), sep(<), int(1), sep(-), id(_G2336)], _L310)creep
   Call:  (15)  ? expr(_L354, _G2494, [int(2), sep(<), int(1), sep(-), id(_G2336)], _L335)creep
   Exit:  (15)  ? expr(_L354, int(2), [int(2), sep(<), int(1), sep(-), id(_G2336)], [sep(<), int(1), sep(-), id(_G2336)])creep
   Call:  (15)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(+)|_G2500]creep
   Fail:  (15)  ? [sep(<), int(1), sep(-), id(_G2336)]=[sep(+)|_G2500]
   ...

Answer 1

As suggested by @TomasBy, left recursion in non terminal expr//2 leads to non termination. In SWI-Prolog there is library( tabling ), just import it and declare expr//2 as tabled.

:- use_module(library(tabling)).
:- table expr//2.