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'whatever' has no declared type

开发者 https://www.devze.com 2022-12-24 12:22 出处:网络
i am developing parser using bison...in my grammar i am getting this error Here is a code extern NodePtrCreateNode(NodeType, ...);

i am developing parser using bison...in my grammar i am getting this error

Here is a code

extern NodePtr  CreateNode(NodeType, ...);
extern NodePtr  ReplaceNode(NodeType, NodePtr); 
extern NodePtr  MergeSubTrees(NodeType, ...); 


            ...................


NodePtr   rootNodePtr = NULL; /* pointer to the root of the parse tree */
NodePtr   nodePtr = NULL; /* pointer to a开发者_如何转开发n error node */


                         ...........................

NodePtr   mainMethodDecNodePtr = NULL;

                   ................

/* YYSTYPE */

%union {
 NodePtr nodePtr;
}

i am getting this error whenever i use like $$.nodePtr or $1.nodePtr ... I am getting Parser.y:1302.32-33: $1 of `Expressi on' has no declared type


That means that the first item (terminal or non-terminal) on the RHS of the Expression rule on line 1302 of parser.y has no type declared for it. If its a terminal, you need to add %token declarations for it, and if its a non-terminal, you need to add a %type declaration for it. When you do that (probably either $type <nodePtr> or %token <nodePtr>), you will access the value as just $1 (no .nodePtr suffix)

edit

sounds like line 1302 should be $$ = $1;. The %type <nodePtr> XXX should go in the first section, where XXX is the non-terminal for this rule. When you use %union in a .y file, the tags declared in the union should ONLY be used in %type and %token declarations -- they should not appear in any action in the .y file

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