如何建立与野牛/ Yacc的和递归规则数组 [英] How to build an Array with Bison/Yacc and a Recursive Rule

问题描述

通过野牛，我想通了，如何让一切都变成一个长字符串如下：

With Bison, I figured out how to get everything into one long string as follows:

arg_list:
    WORD arg_list { strcat( $1, "IFS" ); $$ = strcat($1, $2); }  |
    WORD
    ;

和

WORD arg_list { printf("%s, %s\n", $1, $2); }

但问题是，我便要再次分裂了$ 2中的第二条规则解析它。有没有一种方法来填充一个数组，而不是仅仅使用串联？我要对这个错误的方式？

But the problem is that I will then have to split up $2 in the second rule again to parse it. Is there a way to populate an array instead of just using concatenation? Am I going about this the wrong way?

如果我需要建立像一个链表，可能是有意义的，只是不知道什么是绑定到arg_list的正确方法，然后清理内存。

If I need to build something like a linked list that could make sense, just not sure what would be the proper way to bind to arg_list, and then clean up the memory.

推荐答案

如果你有一个push_front操作数组类型，这是平凡：

If you have an array type with a push_front operation, this is trivially:

arg_list:
    WORD arg_list { $$ = $2.push_front($1); }
    WORD { $$ = new Array<string>($1); }

不说，它需要更多的工作。可以使用的载体，并添加在末端的字符串（这将是在相反的顺序）。或者你可以使用一个链表（这是比较容易，如果你使用直C）：

without that, it requires more work. You can use a vector and add the strings on the end (which will be in the reversed order). Or you can use a linked list (which is easier if you're using straight C):

arg_list:
    WORD arg_list { $$ = malloc(sizeof(struct list_elem));
                    $$->next = $2;
                    $$->val = $1; }
    WORD          { $$ = malloc(sizeof(struct list_elem));
                    $$->next = 0;
                    $$->val = $1; }

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