From: Minsoo Choo <minsoochoo0122@proton.me>
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(-)
@@ -86,39 +86,6 @@ div (int num,
r.quot = num / denom;
r.rem = num % denom;
- /*
- * The ANSI standard says that |r.quot| <= |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 >= 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 >= 0 && r.rem < 0) {
- ++r.quot;
- r.rem -= denom;
- }
- else if (num < 0 && r.rem > 0) {
- --r.quot;
- r.rem += denom;
- }
+
return (r);
}
@@ -37,11 +37,6 @@ imaxdiv(intmax_t numer, intmax_t denom)
retval.quot = numer / denom;
retval.rem = numer % denom;
-#if !defined(__STDC_VERSION__) || (__STDC_VERSION__ < 199901L)
- if (numer >= 0 && retval.rem < 0) {
- retval.quot++;
- retval.rem -= denom;
- }
-#endif
+
return (retval);
}
@@ -89,13 +89,6 @@ ldiv (long num,
r.quot = num / denom;
r.rem = num % denom;
- if (num >= 0 && r.rem < 0) {
- ++r.quot;
- r.rem -= denom;
- }
- else if (num < 0 && r.rem > 0) {
- --r.quot;
- r.rem += denom;
- }
+
return (r);
}
@@ -72,29 +72,6 @@ No supporting OS subroutines are required.
#include <stdlib.h>
-/*
- * The ANSI standard says that |r.quot| <= |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 >= 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)
retval.quot = numer / denom;
retval.rem = numer % denom;
- if (numer >= 0 && retval.rem < 0) {
- retval.quot++;
- retval.rem -= denom;
- }
+
return (retval);
}
-