Implement strmem() and use it in join(1)

We want our delimiters to also contain 0 characters and have them
handled gracefully.
To accomplish this, I wrote a function strmem(), which looks for a
certain, arbitrarily long memory subset in a given string.
memmem() is a GNU extension and forces you to call strlen every time.

Makefile

@@ -75,6 +75,7 @@ LIBUTILSRC =\
 	libutil/strcasestr.c\
 	libutil/strlcat.c\
 	libutil/strlcpy.c\
+	libutil/strmem.c\
 	libutil/strsep.c\
 	libutil/strtonum.c\
 	libutil/unescape.c

join.c

@@ -225,7 +225,7 @@ makeline(char *s, size_t len)
 		beg = sp;
 
 		if (sep) {
-			if (!(end = utfutf(sp, sep)))
+			if (!(end = strmem(sp, sep, seplen)))
 				eol = 1;
 
 			if (!eol) {

libutil/strmem.c

@@ -0,0 +1,23 @@
+/* See LICENSE file for copyright and license details. */
+#include <stddef.h>
+#include <string.h>
+
+char *
+strmem(char *haystack, char *needle, size_t needlelen)
+{
+	size_t i;
+
+	for (i = 0; i < needlelen; i++) {
+		if (haystack[i] == '\0') {
+			return NULL;
+		}
+	}
+
+	for (; haystack[i]; i++) {
+		if (!(memcmp(haystack + i - needlelen, needle, needlelen))) {
+			return (haystack + i - needlelen);
+		}
+	}
+
+	return NULL;
+}

util.h

@@ -58,6 +58,8 @@ size_t estrlcpy(char *, const char *, size_t);
 #undef strsep
 char *strsep(char **, const char *);
 
+char *strmem(char *, char *, size_t);
+
 /* regex */
 int enregcomp(int, regex_t *, const char *, int);
 int eregcomp(regex_t *, const char *, int);