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Home/ Questions/Q 8790157
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T22:34:47+00:00 2026-06-13T22:34:47+00:00

I’m trying to understand flex/bison, but the documentation is a bit difficult for me,

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I’m trying to understand flex/bison, but the documentation is a bit difficult for me, and I’ve probably grossly misunderstood something. Here’s a test case: http://namakajiri.net/misc/bison_charlit_test/

File “a” contains the single character ‘a’. “foo.y” has a trivial grammar like this:

%%

file: 'a' ;

The generated parser can’t parse file “a”; it gives a syntax error.

The grammar “bar.y” is almost the same, only I changed the character literal for a named token:

%token TOK_A;

%%

file: TOK_A;

and then in bar.lex:

a       { return TOK_A; }

This one works just fine.

What am I doing wrong in trying to use character literals directly as bison terminals, like in the docs?

I’d like my grammar to look like “statement: selector ‘{‘ property ‘:’ value ‘;’ ‘}'” and not “statement: selector LBRACE property COLON value SEMIC RBRACE”…

I’m running bison 2.5 and flex 2.5.35 in debian wheezy.

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  1. Editorial Team
    Editorial Team
    2026-06-13T22:34:48+00:00Added an answer on June 13, 2026 at 10:34 pm

    Rewrite

    The problem is a runtime problem, not a compile time problem.

    The trouble is that you have two radically different lexical analyzers.

    The bar.lex analyzer recognizes an a in the input and returns it as a TOK_A and ignores everything else.

    The foo.lex analyzer echoes every single character, but that’s all.

    foo.lex — as written

    %{
    #include "foo.tab.h"
    %}
    
    %%
    

    foo.lex — equivalent

    %{
    #include "foo.tab.h"
    %}
    
    %%
    . { ECHO; }
    

    foo.lex — required

    %{
    #include "foo.tab.h"
    %}
    
    %%
    . { return *yytext; }
    

    Working code

    Here’s some working code with diagnostic printing in place.

    foo-lex.l

    %%
    . { printf("Flex: %d\n", *yytext); return *yytext; }
    

    foo.y

    %{
    #include <stdio.h>
    void yyerror(char *s);
    %}
    
    %%
    
    file: 'a' { printf("Bison: got file!\n") }
        ;
    
    %%
    
    int main(void)
    {
        yyparse();
    }
    
    void yyerror(char *s)
    {
        fprintf(stderr, "%s\n", s);
    }
    

    Compilation and execution

    $ flex foo-lex.l
    $ bison foo.y
    $ gcc -o foo foo.tab.c lex.yy.c -lfl
    $ echo a | ./foo
    Flex: 97
    Bison: got file!
    
    $
    

    Point of detail: how did that blank line get into the output? Answer: the lexical analyzer put it there. The pattern . does not match a newline, so the newline was treated as if there was a rule:

    \n    { ECHO; }
    

    This is why the input was accepted. If you change the foo-lex.l file to:

    %%
    .       { printf("Flex-1: %d\n", *yytext); return *yytext; }
    \n      { printf("Flex-2: %d\n", *yytext); return *yytext; }
    

    and then recompile and run again, the output is:

    $ echo a | ./foo
    Flex-1: 97
    Bison: got file!
    Flex-2: 10
    syntax error
    $
    

    with no blank lines. This is because the grammar doesn’t allow a newline to appear in a valid ‘file’.

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