python/sepolgen: allow any policy statement in if(n)def
"ifdef/ifndef" statements can be used to conditionally define an interface, but this syntax is not recognised by sepolgen-ifgen. Fix sepolgen-ifgen to allow any policy statement inside an "ifdef/ifndef" statement. Fixes: $ cat <<EOF > i.if ifndef(`apache_manage_pid_files',` interface(`apache_manage_pid_files',` manage_files_pattern($1, httpd_var_run_t, httpd_var_run_t) ') ') #sepolgen-ifgen --interface=i.if i.if: Syntax error on line 2 interface [type=INTERFACE] i.if: Syntax error on line 4 ' [type=SQUOTE] Signed-off-by: Vit Mojzis <vmojzis@redhat.com> [OM: s/fidef/ifdef/] Signed-off-by: Ondrej Mosnacek <omosnace@redhat.com>
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@ -433,9 +433,9 @@ def p_ifelse(p):
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def p_ifdef(p):
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'''ifdef : IFDEF OPAREN TICK IDENTIFIER SQUOTE COMMA TICK interface_stmts SQUOTE CPAREN optional_semi
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| IFNDEF OPAREN TICK IDENTIFIER SQUOTE COMMA TICK interface_stmts SQUOTE CPAREN optional_semi
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| IFDEF OPAREN TICK IDENTIFIER SQUOTE COMMA TICK interface_stmts SQUOTE COMMA TICK interface_stmts SQUOTE CPAREN optional_semi
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'''ifdef : IFDEF OPAREN TICK IDENTIFIER SQUOTE COMMA TICK statements SQUOTE CPAREN optional_semi
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| IFNDEF OPAREN TICK IDENTIFIER SQUOTE COMMA TICK statements SQUOTE CPAREN optional_semi
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| IFDEF OPAREN TICK IDENTIFIER SQUOTE COMMA TICK statements SQUOTE COMMA TICK statements SQUOTE CPAREN optional_semi
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'''
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x = refpolicy.IfDef(p[4])
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if p[1] == 'ifdef':
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