@rem
@rem Copyright 2015 the original author or authors.
@rem
@rem Licensed under the Apache License, Version 2.0 (the "License");
@rem you may not use this file except in compliance with the License.
@rem You may obtain a copy of the License at
@rem
@rem      https://www.apache.org/licenses/LICENSE-2.0
@rem
@rem Unless required by applicable law or agreed to in writing, software
@rem distributed under the License is distributed on an "AS IS" BASIS,

@REM
@REM    http://www.apache.org/licenses/LICENSE-2.0
@REM
@REM Unless required by applicable law or agreed to in writing,
@REM software distributed under the License is distributed on an
@REM "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
@REM KIND, either express or implied.  See the License for the
@REM specific language governing permissions and limitations
@REM under the License.

// constant instead of dividing by the constant.
// rem = remainder(arg0, cst)
// for i in 0 to len(arg0):
//    rem[i] = (arg0[i] - rem[i]) * 1 / cst
//    if (rem[i] != 0 && sign(cst) != sign(rem[i]))
//       rem[i] += -1.0
// return round_nearest_afz(rem)
// As a dag this looks like the following:
//                       round
//                         |

@rem
@rem Copyright 2015 the original author or authors.
@rem
@rem Licensed under the Apache License, Version 2.0 (the "License");
@rem you may not use this file except in compliance with the License.
@rem You may obtain a copy of the License at
@rem
@rem      https://www.apache.org/licenses/LICENSE-2.0
@rem
@rem Unless required by applicable law or agreed to in writing, software
@rem distributed under the License is distributed on an "AS IS" BASIS,

			}
			if ret64 != int64(tc.ret) {
				t.Errorf("%d / %d got ret %d rem %d want ret %d rem %d", tc.num, tc.div, ret64, rem64, tc.ret, tc.rem)
			}

			var rem int32
			ret := Timediv(tc.num, tc.div, &rem)
			if ret != tc.ret || rem != tc.rem {
				t.Errorf("timediv %d / %d got ret %d rem %d want ret %d rem %d", tc.num, tc.div, ret, rem, tc.ret, tc.rem)
			}
		})
	}

/*

Multi-precision division. Here be dragons.

Given u and v, where u is n+m digits, and v is n digits (with no leading zeros),
the goal is to return quo, rem such that u = quo*v + rem, where 0 ≤ rem < v.
That is, quo = ⌊u/v⌋ where ⌊x⌋ denotes the floor (truncation to integer) of x,
and rem = u - quo·v.


Long Division

Division in a computer proceeds the same as long division in elementary school,

        assertAnnotationConsumed("""@Deprecated(message="Use rem(other) instead",replaceWith=ReplaceWith("rem(other)"))""")
        assertAnnotationConsumed("""@Deprecated ( message = "Use rem(other) instead" , replaceWith = ReplaceWith ( "rem(other)" ) )""")

        assertAnnotationConsumed("""@Throws(IOException::class)""")

				offset = 32
				rem -= 16
			}
		}

		// Generate all the remaining load and store pairs, starting with
		// as many 8 byte moves as possible, then 4, 2, 1.
		for rem > 0 {
			op, size := ppc64.AMOVB, int64(1)
			switch {
			case rem >= 8:
				op, size = ppc64.AMOVD, 8
			case rem >= 4:
				op, size = ppc64.AMOVWZ, 4
			case rem >= 2:
				op, size = ppc64.AMOVH, 2
			}

// quo = (hi, lo)/y, rem = (hi, lo)%y with the dividend bits' upper
// half in parameter hi and the lower half in parameter lo.
// Div32 panics for y == 0 (division by zero) or y <= hi (quotient overflow).
func Div32(hi, lo, y uint32) (quo, rem uint32) {
	if y != 0 && y <= hi {
		panic(overflowError)
	}
	z := uint64(hi)<<32 | uint64(lo)
	quo, rem = uint32(z/uint64(y)), uint32(z%uint64(y))
	return
}

     * that floor(floor(x)/i) == floor(x/i) for any real x and integer i != 0. The proof is not
     * quite trivial.
     */
    long quotient = ((dividend >>> 1) / divisor) << 1;
    long rem = dividend - quotient * divisor;
    return rem - (compare(rem, divisor) >= 0 ? divisor : 0);
  }

  /**
   * Returns the unsigned {@code long} value represented by the given decimal string.
   *