What is an "illegal primary" in awk?

Question

Awk gives me the following error:

awk: illegal primary in regular expression (?<=\>)(.*?)(?=\<) at <=\>)(.*?)(?=\<)
source line number 10 source file transpile.awk
context is
    match($0, >>>  /(?<=\>)(.*?)(?=\<)/) <<<

But what is an "illegal primary"?

Lookarounds are not supported by `awk`. Hence the syntax error. — revo, May 01 '18 at 20:20
In POSIX regex, `\>` and `\<` are word boundaries. What did you want to achieve with that pattern of yours? — Wiktor Stribiżew, May 01 '18 at 20:29
@WiktorStribiżew The input lines are text inside HTML tags. I want to match the text without the tags. — 303, May 01 '18 at 20:57
@WiktorStribiżew Thank you for pointing to a question about matching HTML contents, but my question is not a duplicate. I actually asked what an illegal primary is, because I want to learn about awk and how to read its error messages. Please remove the mark. — 303, May 01 '18 at 21:08

score 6 · Accepted Answer · answered May 02 '18 at 05:32

A "primary", in awk parlance, is the basic unit of a regex.

A regex consists of an alternative of (1 or more) branches. Each branch consists of a concatenation of (0 or more) primaries.

A primary is either a normal character (e.g. a), or an escaped special character (e.g. \*), or a character class ([...]), or a dot (.), or an anchor (^ or $), or a parenthesized subexpression ((...)). Most of these can have a quantifier (?, +, *), too.

The problem with your regex is that (?<=\>) parses as ( first, which starts a subgroup. The next item then needs to be a primary. ? is not a valid primary, hence you get an error.

Awk does not support look-ahead or look-behind.

Sufiyan Ghori · Answer 2 · 2018-05-02T05:40:32.320

If you have a look at the awk source code, you can see that illegal primary is a default FATAL error when none of the cases for the regex tokens matched. It is not a Syntax Error.

Here is the specific code from b.c file of awk (stripped),

    case NCCL:
        np = op2(NCCL, NIL, (Node *) cclenter((char *) rlxstr));
        rtok = relex();
        return (unary(np));
    case '^':
        rtok = relex();
        return (unary(op2(CHAR, NIL, itonp(HAT))));
    case '$':
        rtok = relex();
        return (unary(op2(CHAR, NIL, NIL)));
    case '(':
        rtok = relex();
        if (rtok == ')') {  /* special pleading for () */
            rtok = relex();
            return unary(op2(CCL, NIL, (Node *) tostring("")));
        }
        np = regexp();
        if (rtok == ')') {
            rtok = relex();
            return (unary(np));
        }
        else
            FATAL("syntax error in regular expression %s at %s", lastre, prestr);
    default:
        FATAL("illegal primary in regular expression %s at %s", lastre, prestr);

This code is also available on github here,

https://github.com/onetrueawk/awk/blob/master/b.c

in your specific case the reason is because Lookarounds, i.e, Lookahead: (?=...), (?!...) and Lookbehind: (?<=...), (?<!...) are not supported by awk.

Hope this helps

What is an "illegal primary" in awk?

2 Answers2

Linked