我有一个字符串;
String value = "(5+5) + ((5+8 + (85*4))+524)";
如何从括号内的此字符串中拆分/提取逻辑值为;
(85*4) as one (5+8 + one) as two (two+524) as three ((5+5) + three) as four ...
任何的想法？一切都很受欢迎
解决方法
这不能使用一些切割器正则表达式来完成(正则表达式不能“计算括号”).您最好的选择是使用一些解析器生成器并将字符串解析为
abstract syntax tree(简称AST).
事实证明,CUP manual实际上有一个例子涵盖你的情况：
// CUP specification for a simple expression evaluator (w/ actions)
import java_cup.runtime.*;
/* Preliminaries to set up and use the scanner. */
init with {: scanner.init(); :};
scan with {: return scanner.next_token(); :};
/* Terminals (tokens returned by the scanner). */
terminal SEMI,PLUS,MINUS,TIMES,DIVIDE,MOD;
terminal UMINUS,LPAREN,RPAREN;
terminal Integer NUMBER;
/* Non-terminals */
non terminal expr_list,expr_part;
non terminal Integer expr;
/* Precedences */
precedence left PLUS,MINUS;
precedence left TIMES,MOD;
precedence left UMINUS;
/* The grammar */
expr_list ::= expr_list expr_part
|
expr_part;
expr_part ::= expr:e
{: System.out.println("= " + e); :}
SEMI
;
expr ::= expr:e1 PLUS expr:e2
{: RESULT = new Integer(e1.intValue() + e2.intValue()); :}
|
expr:e1 MINUS expr:e2
{: RESULT = new Integer(e1.intValue() - e2.intValue()); :}
|
expr:e1 TIMES expr:e2
{: RESULT = new Integer(e1.intValue() * e2.intValue()); :}
|
expr:e1 DIVIDE expr:e2
{: RESULT = new Integer(e1.intValue() / e2.intValue()); :}
|
expr:e1 MOD expr:e2
{: RESULT = new Integer(e1.intValue() % e2.intValue()); :}
|
NUMBER:n
{: RESULT = n; :}
|
MINUS expr:e
{: RESULT = new Integer(0 - e.intValue()); :}
%prec UMINUS
|
LPAREN expr:e RPAREN
{: RESULT = e; :}
;
