tuple_modtuple_modTupleModTupleModtuple_mod (算子)
名称
tuple_modtuple_modTupleModTupleModtuple_mod — 计算两个元组整数除法的余数。
签名
Herror tuple_mod(const Hlong T1, const Hlong T2, Hlong* Mod)
Herror T_tuple_mod(const Htuple T1, const Htuple T2, Htuple* Mod)
void TupleMod(const HTuple& T1, const HTuple& T2, HTuple* Mod)
HTuple HTuple::TupleMod(const HTuple& T2) const
def tuple_mod(t1: MaybeSequence[int], t2: MaybeSequence[int]) -> Sequence[int]
def tuple_mod_s(t1: MaybeSequence[int], t2: MaybeSequence[int]) -> int
描述
tuple_modtuple_modTupleModTupleModTupleModtuple_mod returns the remainder of the integer
division of the input tuples T1T1T1T1t1t1/T2T2T2T2t2t2 in the
output tuple ModModModModmodmod. If both
tuples have the same length the division is performed for the
corresponding elements of both tuples. Otherwise, either
T1T1T1T1t1t1 or T2T2T2T2t2t2 must have length 1. In this case, the
division is performed for each element of the longer tuple with the
single element of the other tuple. The result is always an integer
number. The division of strings is not allowed.
例外：空输入元组
If either or both of the input tuples are empty, the operator returns an
empty tuple.
HDevelop 内联操作
HDevelop provides an in-line operation for tuple_modtuple_modTupleModTupleModTupleModtuple_mod,
which can be used in an expression in the following syntax:
Mod := T1 % T2
执行信息
- 多线程类型：独立（即使使用独占算子也能并行运行）。
- 多线程作用域：全局（可从任何线程调用）。
- 未采用并行化处理。
参数
T1T1T1T1t1t1 (输入控制) number(-array) → HTupleMaybeSequence[int]HTupleHtuple (integer) (int / long) (Hlong) (Hlong)
输入元组 1。
T2T2T2T2t2t2 (输入控制) number(-array) → HTupleMaybeSequence[int]HTupleHtuple (integer) (int / long) (Hlong) (Hlong)
输入元组 2。
限制：
T2 != 0
ModModModModmodmod (输出控制) number(-array) → HTupleSequence[int]HTupleHtuple (integer) (int / long) (Hlong) (Hlong)
Remainder of the division of the input tuples.
另见
tuple_fmodtuple_fmodTupleFmodTupleFmodTupleFmodtuple_fmod,
tuple_divtuple_divTupleDivTupleDivTupleDivtuple_div
模块
基础