diff --git a/lib/libc/string/strlen.c b/lib/libc/string/strlen.c
index 0bdc81d7bb9a..a862ffc245ca 100644
--- a/lib/libc/string/strlen.c
+++ b/lib/libc/string/strlen.c
@@ -35,6 +35,10 @@ __FBSDID("$FreeBSD$");
 /*
  * Portable strlen() for 32-bit and 64-bit systems.
  *
+ * Rationale: it is generally much more efficient to do word length
+ * operations and avoid branches on modern computer systems, as
+ * compared to byte-length operations with a lot of branches.
+ *
  * The expression:
  *
  *	((x - 0x01....01) & ~x & 0x80....80)
@@ -42,13 +46,18 @@ __FBSDID("$FreeBSD$");
  * would evaluate to a non-zero value iff any of the bytes in the
  * original word is zero.
  *
+ * On multi-issue processors, we can divide the above expression into:
+ *	a)  (x - 0x01....01)
+ *	b) (~x & 0x80....80)
+ *	c) a & b
+ *
+ * Where, a) and b) can be partially computed in parallel.
+ *
  * The algorithm above is found on "Hacker's Delight" by
  * Henry S. Warren, Jr.
- *
- * Note: this leaves performance on the table and each architecture
- * would be best served with a tailor made routine instead.
  */
 
+/* Magic numbers for the algorithm */
 #if LONG_BIT == 32
 static const unsigned long mask01 = 0x01010101;
 static const unsigned long mask80 = 0x80808080;
@@ -61,45 +70,62 @@ static const unsigned long mask80 = 0x8080808080808080;
 
 #define	LONGPTR_MASK (sizeof(long) - 1)
 
-#if BYTE_ORDER == LITTLE_ENDIAN
-#define	FINDZERO __builtin_ctzl
-#else
-#define	FINDZERO __builtin_clzl
-#endif
+/*
+ * Helper macro to return string length if we caught the zero
+ * byte.
+ */
+#define testbyte(x)				\
+	do {					\
+		if (p[x] == '\0')		\
+		    return (p - str + x);	\
+	} while (0)
 
 size_t
 strlen(const char *str)
 {
+	const char *p;
 	const unsigned long *lp;
-	unsigned long mask;
 	long va, vb;
-	long val;
 
-	lp = (unsigned long *) (uintptr_t) str;
-	if ((uintptr_t)lp & LONGPTR_MASK) {
-		lp = (__typeof(lp)) ((uintptr_t)lp & ~LONGPTR_MASK);
-#if BYTE_ORDER == LITTLE_ENDIAN
-		mask = ~(~0UL << (((uintptr_t)str & LONGPTR_MASK) << 3));
-#else
-		mask = ~(~0UL >> (((uintptr_t)str & LONGPTR_MASK) << 3));
-#endif
-		val = *lp | mask;
-		va = (val - mask01);
-		vb = ((~val) & mask80);
-		if (va & vb) {
-			return ((const char *)lp - str + (FINDZERO(va & vb) >> 3));
-		}
-		lp++;
-	}
+	/*
+	 * Before trying the hard (unaligned byte-by-byte access) way
+	 * to figure out whether there is a nul character, try to see
+	 * if there is a nul character is within this accessible word
+	 * first.
+	 *
+	 * p and (p & ~LONGPTR_MASK) must be equally accessible since
+	 * they always fall in the same memory page, as long as page
+	 * boundaries is integral multiple of word size.
+	 */
+	lp = (const unsigned long *)((uintptr_t)str & ~LONGPTR_MASK);
+	va = (*lp - mask01);
+	vb = ((~*lp) & mask80);
+	lp++;
+	if (va & vb)
+		/* Check if we have \0 in the first part */
+		for (p = str; p < (const char *)lp; p++)
+			if (*p == '\0')
+				return (p - str);
 
+	/* Scan the rest of the string using word sized operation */
 	for (; ; lp++) {
 		va = (*lp - mask01);
 		vb = ((~*lp) & mask80);
 		if (va & vb) {
-			return ((const char *)lp - str + (FINDZERO(va & vb) >> 3));
+			p = (const char *)(lp);
+			testbyte(0);
+			testbyte(1);
+			testbyte(2);
+			testbyte(3);
+#if (LONG_BIT >= 64)
+			testbyte(4);
+			testbyte(5);
+			testbyte(6);
+			testbyte(7);
+#endif
 		}
 	}
 
-	__builtin_unreachable();
+	/* NOTREACHED */
 	return (0);
 }
diff --git a/sys/libkern/strlen.c b/sys/libkern/strlen.c
index 8fa5f3927ea9..a8c7964f69a3 100644
--- a/sys/libkern/strlen.c
+++ b/sys/libkern/strlen.c
@@ -34,6 +34,10 @@ __FBSDID("$FreeBSD$");
 /*
  * Portable strlen() for 32-bit and 64-bit systems.
  *
+ * Rationale: it is generally much more efficient to do word length
+ * operations and avoid branches on modern computer systems, as
+ * compared to byte-length operations with a lot of branches.
+ *
  * The expression:
  *
  *	((x - 0x01....01) & ~x & 0x80....80)
@@ -41,10 +45,18 @@ __FBSDID("$FreeBSD$");
  * would evaluate to a non-zero value iff any of the bytes in the
  * original word is zero.
  *
+ * On multi-issue processors, we can divide the above expression into:
+ *	a)  (x - 0x01....01)
+ *	b) (~x & 0x80....80)
+ *	c) a & b
+ *
+ * Where, a) and b) can be partially computed in parallel.
+ *
  * The algorithm above is found on "Hacker's Delight" by
  * Henry S. Warren, Jr.
  */
 
+/* Magic numbers for the algorithm */
 #if LONG_BIT == 32
 static const unsigned long mask01 = 0x01010101;
 static const unsigned long mask80 = 0x80808080;
@@ -57,45 +69,62 @@ static const unsigned long mask80 = 0x8080808080808080;
 
 #define	LONGPTR_MASK (sizeof(long) - 1)
 
-#if BYTE_ORDER == LITTLE_ENDIAN
-#define	FINDZERO __builtin_ctzl
-#else
-#define	FINDZERO __builtin_clzl
-#endif
+/*
+ * Helper macro to return string length if we caught the zero
+ * byte.
+ */
+#define testbyte(x)				\
+	do {					\
+		if (p[x] == '\0')		\
+		    return (p - str + x);	\
+	} while (0)
 
 size_t
 (strlen)(const char *str)
 {
+	const char *p;
 	const unsigned long *lp;
-	unsigned long mask;
 	long va, vb;
-	long val;
 
-	lp = (unsigned long *) (uintptr_t) str;
-	if ((uintptr_t)lp & LONGPTR_MASK) {
-		lp = (__typeof(lp)) ((uintptr_t)lp & ~LONGPTR_MASK);
-#if BYTE_ORDER == LITTLE_ENDIAN
-		mask = ~(~0UL << (((uintptr_t)str & LONGPTR_MASK) << 3));
-#else
-		mask = ~(~0UL >> (((uintptr_t)str & LONGPTR_MASK) << 3));
-#endif
-		val = *lp | mask;
-		va = (val - mask01);
-		vb = ((~val) & mask80);
-		if (va & vb) {
-			return ((const char *)lp - str + (FINDZERO(va & vb) >> 3));
-		}
-		lp++;
-	}
+	/*
+	 * Before trying the hard (unaligned byte-by-byte access) way
+	 * to figure out whether there is a nul character, try to see
+	 * if there is a nul character is within this accessible word
+	 * first.
+	 *
+	 * p and (p & ~LONGPTR_MASK) must be equally accessible since
+	 * they always fall in the same memory page, as long as page
+	 * boundaries is integral multiple of word size.
+	 */
+	lp = (const unsigned long *)((uintptr_t)str & ~LONGPTR_MASK);
+	va = (*lp - mask01);
+	vb = ((~*lp) & mask80);
+	lp++;
+	if (va & vb)
+		/* Check if we have \0 in the first part */
+		for (p = str; p < (const char *)lp; p++)
+			if (*p == '\0')
+				return (p - str);
 
+	/* Scan the rest of the string using word sized operation */
 	for (; ; lp++) {
 		va = (*lp - mask01);
 		vb = ((~*lp) & mask80);
 		if (va & vb) {
-			return ((const char *)lp - str + (FINDZERO(va & vb) >> 3));
+			p = (const char *)(lp);
+			testbyte(0);
+			testbyte(1);
+			testbyte(2);
+			testbyte(3);
+#if (LONG_BIT >= 64)
+			testbyte(4);
+			testbyte(5);
+			testbyte(6);
+			testbyte(7);
+#endif
 		}
 	}
 
-	__builtin_unreachable();
+	/* NOTREACHED */
 	return (0);
 }