awk: FS regex matches only non-empty separators (gawk compat)
function old new delta awk_split 484 553 +69 Signed-off-by: Denys Vlasenko <vda.linux@googlemail.com>
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@ -1763,6 +1763,29 @@ static void fsrealloc(int size)
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nfields = size;
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}
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static int regexec1_nonempty(const regex_t *preg, const char *s, regmatch_t pmatch[])
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{
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int r = regexec(preg, s, 1, pmatch, 0);
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if (r == 0 && pmatch[0].rm_eo == 0) {
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/* For example, happens when FS can match
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* an empty string (awk -F ' *'). Logically,
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* this should split into one-char fields.
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* However, gawk 5.0.1 searches for first
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* _non-empty_ separator string match:
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*/
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size_t ofs = 0;
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do {
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ofs++;
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if (!s[ofs])
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return REG_NOMATCH;
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regexec(preg, s + ofs, 1, pmatch, 0);
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} while (pmatch[0].rm_eo == 0);
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pmatch[0].rm_so += ofs;
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pmatch[0].rm_eo += ofs;
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}
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return r;
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}
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static int awk_split(const char *s, node *spl, char **slist)
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{
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int n;
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@ -1788,17 +1811,11 @@ static int awk_split(const char *s, node *spl, char **slist)
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regmatch_t pmatch[2]; // TODO: why [2]? [1] is enough...
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l = strcspn(s, c+2); /* len till next NUL or \n */
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if (regexec(icase ? spl->r.ire : spl->l.re, s, 1, pmatch, 0) == 0
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if (regexec1_nonempty(icase ? spl->r.ire : spl->l.re, s, pmatch) == 0
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&& pmatch[0].rm_so <= l
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) {
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/* if (pmatch[0].rm_eo == 0) ... - impossible */
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l = pmatch[0].rm_so;
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if (pmatch[0].rm_eo == 0) {
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/* For example, happens when FS can match
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* an empthy string (awk -F ' *')
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*/
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l++;
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pmatch[0].rm_eo++;
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}
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n++; /* we saw yet another delimiter */
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} else {
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pmatch[0].rm_eo = l;
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@ -398,5 +398,12 @@ testing 'awk do not allow "str"++' \
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'' \
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'anything'
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#gawk compat: FS regex matches only non-empty separators:
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#with -*, the splitting is NOT f o o b a r, but foo bar:
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testing 'awk FS regex which can match empty string' \
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"awk -F '-*' '{print \$1 \"-\" \$2 \"=\" \$3 \"*\" \$4}'" \
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"foo-bar=*\n" \
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'' \
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'foo--bar'
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exit $FAILCOUNT
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