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Home/ Questions/Q 677743
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T01:04:22+00:00 2026-05-14T01:04:22+00:00

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

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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 an 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

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T01:04:23+00:00Added an answer on May 14, 2026 at 1:04 am

    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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