我有一个字符串;
String value = "(5+5) + ((5+8 + (85*4))+524)";
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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
...
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任何的想法?一切都很受欢迎
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抽象语法树(简称AST).
事实证明,CUP手册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, DIVIDE, 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; :}
;
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