[hackers] [sbase][PATCH 2/2] expr: don't evaluate matched substr as a number

[Randy Palamar]

POSIX specifies that if the pattern contains a subexpression then
the first matched subexpression should be returned if it exists.

This fixes things like the following:

./expr 00003 : '\(.*\)'
Before: 3
After: 00003
---
 expr.c | 13 ++-----------
 1 file changed, 2 insertions(+), 11 deletions(-)

diff --git a/expr.c b/expr.c
index ae32b9f..3afb94b 100644
--- a/expr.c
+++ b/expr.c
@@ -59,11 +59,9 @@ match(struct val *vstr, struct val *vregx, struct val *ret)
 {
        regex_t re;
        regmatch_t matches[2];
-       long long d;
        size_t anchlen;
        char *s, *p, *anchreg;
        char *str = vstr->str, *regx = vregx->str;
-       const char *errstr;
 
        /* anchored regex */
        anchlen = strlen(regx) + 1 + 1;
@@ -83,15 +81,8 @@ match(struct val *vstr, struct val *vregx, struct val *ret)
                s = str + matches[1].rm_so;
                p = str + matches[1].rm_eo;
                *p = '\0';
-
-               d = strtonum(s, LLONG_MIN, LLONG_MAX, &errstr);
-               if (!errstr) {
-                       ret->num = d;
-                       return;
-               } else {
-                       ret->str = enstrdup(3, s);
-                       return;
-               }
+               ret->str = enstrdup(3, s);
+               return;
        } else {
                regfree(&re);
                str += matches[0].rm_so;
-- 
2.41.0