正則表達式遞歸JSL是否支持遞歸正則表達式？
我試過了？R和 \ g <0>它們似乎都不起作用。
regex("test(1, (2, 3));", "\(([^()]|(?R))*\");Re：正則表達式遞歸幾分鐘後維基百科給出一組沒有左遞歸的規則......運行速度快。
text = "(1+2)*(3+4+5)*6*7*8*21*(0+(1+(2*3))+4*(5+6)*7*8+(((12+13+14)*15+16+17)*18+19+20))*22"; // plus, star, paren, integer. pattern does not handle spaces.
stack = {}; // the "working" stack... holds the RPN numbers
// the following add and mul routines pop two numbers and push the result...
addstack = Function( {},
{num1 = Num( Remove From( stack )[1] ), num2 = Num( Remove From( stack )[1] )},
Insert Into( stack, Char( num1 + num2 ) )
);
mulstack = Function( {},
{num1 = Num( Remove From( stack )[1] ), num2 = Num( Remove From( stack )[1] )},
Insert Into( stack, Char( num1 * num2 ) )
);
dummy = {}; // dummy variable with a subscript that will evaluate as needed
// BNF-like grammar of Expression, Term, Factor, Constant
// >? is the conditional assignment from LHS to RHS. The >>
// immediate assignment will not work because of back-up and retry
// which would push incorrect items to stack. The RHS of the >? is either
// the dummy[] list or working stack. The dummy stack reference uses the
// calculation of its subscript to manipulate the working stack.
// As the grammar/pattern runs, the expression is evaluated.
/*
E = (Expr( T ) + "+" + Expr( E )) >? dummy[addstack(); 1] | Expr( T );
T = (Expr( T ) + "*" + Expr( F )) >? dummy[mulstack(); 1] | Expr( F );
F = ("(" + Expr( E ) + ")") | Expr( C );
C = Pat Span( "0123456789" ) >? stack[N Items( stack ) + 1];
*/
// re-written without left recursion
// https://en.wikipedia.org/wiki/Left_recursion#Removing_left_recursion
//
E = Expr( T ) + Expr( EP );
EP = ("+" + Expr( T ) >? dummy[addstack(); 1] + Expr( EP )) | "";
T = Expr( F ) + Expr( TP );
TP = ("*" + Expr( F ) >? dummy[mulstack(); 1] + Expr( TP )) | "";
F = ("(" + Expr( E ) + ")") | Expr( C );
C = Pat Span( "0123456789" ) >? stack[N Items( stack ) + 1];
rc = Pat Match( text, Pat Pos( 0 ) + E + Pat R Pos( 0 ), NULL, FULLSCAN );
If( rc,
Show( rc, Num( stack[1] ), Eval( Parse( text ) ) ); // use JMP's parse to verify the answer
,
Show( rc )
);
如果你正在密切關注，你會想知道第一個版本是如何運行的，左遞歸。為什麼它不會永遠循環，或者至少在遞歸消耗可用內存之前？默認情況下，模式匹配器假定每個expr（...）必須至少匹配一個字符。在某些時候，遞歸足夠深，以至於沒有足夠的文本留下並且匹配備份。所有備份和重試都需要一個長字符串。在第二個示例中，沒有左遞歸，patmatch（）需要FULLSCAN選項來關閉該假設，因為expr（）現在可以匹配“”（零個字符）。
Re：正則表達式遞歸不，但模式匹配器喜歡遞歸。這是表達式解釋器的完整示例。我不是語法專家;我認為這是一個遞歸下降解析器 ，我認為它對中等長度的表達式進行了大量的後向跟踪，對於擁有超過10-20個運算符的表達式來說，它顯然很慢。
如果模式匹配，PatMatch返回1，如果失敗則返回0（不平衡的parens，無效的運算符，無效的數字）。
text = "3+4*((5+7)*(9+11*2+2)+13)"; // plus, star, paren, integer. pattern does not handle spaces.
stack = {}; // the "working" stack... holds the RPN numbers
// the following add and mul routines pop two numbers and push the result...
addstack = Function( {},
{num1 = Num( Remove From( stack )[1] ), num2 = Num( Remove From( stack )[1] )},
Insert Into( stack, Char( num1 + num2 ) )
);
mulstack = Function( {},
{num1 = Num( Remove From( stack )[1] ), num2 = Num( Remove From( stack )[1] )},
Insert Into( stack, Char( num1 * num2 ) )
);
dummy = {}; // dummy variable with a subscript that will evaluate as needed
// BNF-like grammar of Expression, Term, Factor, Constant
// >? is the conditional assignment from LHS to RHS. The >>
// immediate assignment will not work because of back-up and retry
// which would push incorrect items to stack. The RHS of the >? is either
// the dummy[] list or working stack. The dummy stack reference uses the
// calculation of its subscript to manipulate the working stack.
// As the grammar/pattern runs, the expression is evaluated.
E = (Expr( T ) + "+" + Expr( E )) >? dummy[addstack(); 1] | Expr( T );
T = (Expr( T ) + "*" + Expr( F )) >? dummy[mulstack(); 1] | Expr( F );
F = ("(" + Expr( E ) + ")") | Expr( C );
C = Pat Span( "0123456789" ) >? stack[N Items( stack ) + 1];
rc = Pat Match( text, Pat Pos( 0 ) + E + Pat R Pos( 0 ) );
If( rc,
Show( rc, Num( stack[1] ), Eval( Parse( text ) ) ); // use JMP's parse to verify the answer
,
Show( rc )
);
