Question

我想通过bison与节点树的公式制作AST：

以下是制作节点的函数和节点＆＃34;：

的结构

typedef enum
{null, opera, var, val} NodeType;

 typedef struct Node {
 float val;
 char * name; // val_name
 char * opr; // operator
 NodeType node_type;
 struct Node * left;
 struct Node * right;
 } Node;

 Node create_op_node(char * op){
 Node node;
 node.opr = op;
 node.node_type = opera;
 return node;
 }

 Node create_var_node(char * op){
 Node node;
 node.name = op;
 node.node_type = var;
 return node;
 }

 Node create_val_node(float value){
 Node node;
 node.val = value;
 node.node_type = val;
 return node;
 }

贯穿树的功能：

void run_through_tree (Node *node){
    switch (node->node_type){
        case var:
            printf("varable name:%s\n",node->name);
            break;
        case val:
            printf("num value:%f\n",node->val);
            break;
        case opera:
            printf("\noperator:%s\n",node->opr);
            printf("LFS\n");
            run_through_tree(node->left);
            printf("RHS\n");
            run_through_tree(node->right);
    }

}

和语法规则以及创建和连接的方式如下：

S：S E T_NEWLINE {main_node = $ 2; run_through_tree（main_node）;} //遍历节点树 | {} ;

E    :  E T_ADD E {Node * op_node = (Node *) malloc(sizeof(Node));*op_node = create_op_node("+"); connect_node(op_node,$1,$3);$$ = op_node;}
     |  E T_SUB E {Node * op_node = (Node *) malloc(sizeof(Node));*op_node = create_op_node("-"); connect_node(op_node,$1,$3);$$ = op_node;}
     |  E T_MUL E {Node * op_node = (Node *) malloc(sizeof(Node));*op_node = create_op_node("*"); connect_node(op_node,$1,$3);$$ = op_node;}
     |  E T_DIV E {Node * op_node = (Node *) malloc(sizeof(Node));*op_node = create_op_node("/"); connect_node(op_node,$1,$3);$$ = op_node;} 
     | T_SUB E %prec NEG {
                        Node * op_node = (Node *) malloc(sizeof(Node));
                        Node * zero_node = (Node *) malloc(sizeof(Node));
                        * zero_node = create_val_node(0.0);
                        * op_node = create_op_node("-");
                        connect_node(op_node,zero_node,$2);
                        $$ = op_node;}
     | T_NUM {Node * val_node = (Node *) malloc(sizeof(Node));* val_node = create_val_node($1); $$ = val_node;}
     | T_VAR {Node * var_node = (Node *) malloc(sizeof(Node));* var_node = create_var_node($1); $$ = var_node;}
     | T_LPATH E T_RPATH {$$ = $2;}
     ;

解析没有变量的公式的结果是预期的： 2 *（3 + 2）

operator:*
LFS
num value:2.000000
RHS

operator:+
LFS
num value:3.000000
RHS
num value:2.000000

但是，使用变量解析公式是不正确的：

2 * a - a + 7 操作：+ LFS

operator:-
LFS

operator:*
LFS
num value:2.000000
RHS
varable name:a - a + 7

RHS
varable name:a + 7

RHS
num value:7.000000

应该是：

2 * a - a + 7 操作：+ LFS

operator:-
LFS

operator:*
LFS
num value:2.000000
RHS
varable name:a

RHS
varable name:a

RHS
num value:7.000000

我怎么能解决这个问题？

使用Bison时将带有标识符的公式转换为抽象语法树时出错

0 个答案: