[PATCH v2 6/8] libc: Remove support for pre-C99 C standards

From: Minsoo Choo <[email protected]>

Reviewed by:	jhb
Differential Revision:	https://reviews.freebsd.org/D43254
---
 newlib/libc/stdlib/div.c     | 35 +----------------------------------
 newlib/libc/stdlib/imaxdiv.c |  7 +------
 newlib/libc/stdlib/ldiv.c    |  9 +--------
 newlib/libc/stdlib/lldiv.c   | 29 +----------------------------
 4 files changed, 4 insertions(+), 76 deletions(-)

diff --git a/newlib/libc/stdlib/div.c b/newlib/libc/stdlib/div.c
index 4296f1e80..467f25aa3 100644
--- a/newlib/libc/stdlib/div.c
+++ b/newlib/libc/stdlib/div.c
@@ -86,39 +86,6 @@ div (int num,
=20
 	r.quot =3D num / denom;
 	r.rem =3D num % denom;
-	/*
-	 * The ANSI standard says that |r.quot| <=3D |n/d|, where
-	 * n/d is to be computed in infinite precision.  In other
-	 * words, we should always truncate the quotient towards
-	 * 0, never -infinity or +infinity.
-	 *
-	 * Machine division and remainer may work either way when
-	 * one or both of n or d is negative.  If only one is
-	 * negative and r.quot has been truncated towards -inf,
-	 * r.rem will have the same sign as denom and the opposite
-	 * sign of num; if both are negative and r.quot has been
-	 * truncated towards -inf, r.rem will be positive (will
-	 * have the opposite sign of num).  These are considered
-	 * `wrong'.
-	 *
-	 * If both are num and denom are positive, r will always
-	 * be positive.
-	 *
-	 * This all boils down to:
-	 *	if num >=3D 0, but r.rem < 0, we got the wrong answer.
-	 * In that case, to get the right answer, add 1 to r.quot and
-	 * subtract denom from r.rem.
-	 *      if num < 0, but r.rem > 0, we also have the wrong answer.
-	 * In this case, to get the right answer, subtract 1 from r.quot and
-	 * add denom to r.rem.
-	 */
-	if (num >=3D 0 && r.rem < 0) {
-		++r.quot;
-		r.rem -=3D denom;
-	}
-	else if (num < 0 && r.rem > 0) {
-		--r.quot;
-		r.rem +=3D denom;
-	}
+
 	return (r);
 }
diff --git a/newlib/libc/stdlib/imaxdiv.c b/newlib/libc/stdlib/imaxdiv.c
index 70808bcc1..1bb73a025 100644
--- a/newlib/libc/stdlib/imaxdiv.c
+++ b/newlib/libc/stdlib/imaxdiv.c
@@ -37,11 +37,6 @@ imaxdiv(intmax_t numer, intmax_t denom)
=20
 	retval.quot =3D numer / denom;
 	retval.rem =3D numer % denom;
-#if !defined(__STDC_VERSION__) || (__STDC_VERSION__ < 199901L)
-	if (numer >=3D 0 && retval.rem < 0) {
-		retval.quot++;
-		retval.rem -=3D denom;
-	}
-#endif
+
 	return (retval);
 }
diff --git a/newlib/libc/stdlib/ldiv.c b/newlib/libc/stdlib/ldiv.c
index c02711a78..4981e3b21 100644
--- a/newlib/libc/stdlib/ldiv.c
+++ b/newlib/libc/stdlib/ldiv.c
@@ -89,13 +89,6 @@ ldiv (long num,
=20
 	r.quot =3D num / denom;
 	r.rem =3D num % denom;
-	if (num >=3D 0 && r.rem < 0) {
-		++r.quot;
-		r.rem -=3D denom;
-	}
-	else if (num < 0 && r.rem > 0) {
-		--r.quot;
-		r.rem +=3D denom;
-	}
+
 	return (r);
 }
diff --git a/newlib/libc/stdlib/lldiv.c b/newlib/libc/stdlib/lldiv.c
index 65ce67ad1..f8b49e91e 100644
--- a/newlib/libc/stdlib/lldiv.c
+++ b/newlib/libc/stdlib/lldiv.c
@@ -72,29 +72,6 @@ No supporting OS subroutines are required.
=20
 #include <stdlib.h>
=20
-/*
- * The ANSI standard says that |r.quot| <=3D |n/d|, where
- * n/d is to be computed in infinite precision.  In other
- * words, we should always truncate the quotient towards
- * 0, never -infinity.
- *
- * Machine division and remainer may work either way when
- * one or both of n or d is negative.  If only one is
- * negative and r.quot has been truncated towards -inf,
- * r.rem will have the same sign as denom and the opposite
- * sign of num; if both are negative and r.quot has been
- * truncated towards -inf, r.rem will be positive (will
- * have the opposite sign of num).  These are considered
- * `wrong'.
- *
- * If both are num and denom are positive, r will always
- * be positive.
- *
- * This all boils down to:
- *      if num >=3D 0, but r.rem < 0, we got the wrong answer.
- * In that case, to get the right answer, add 1 to r.quot and
- * subtract denom from r.rem.
- */
 lldiv_t
 lldiv (long long numer, long long denom)
 {
@@ -102,10 +79,6 @@ lldiv (long long numer, long long denom)
=20
 	retval.quot =3D numer / denom;
 	retval.rem =3D numer % denom;
-	if (numer >=3D 0 && retval.rem < 0) {
-		retval.quot++;
-		retval.rem -=3D denom;
-	}
+
 	return (retval);
 }
-
--=20
2.51.0