C++ 如何一步得到商和余数？
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How can I get the quotient and the remainder in a single step?
提问by dtech
Possible Duplicate:
Divide and Get Remainder at the same time?
可能的重复：同时进行
除法和获取余数？
Is it possible to get both the quotient and the remainder of integer division in a single step, i.e., without performing integer division twice?
是否可以在一个步骤中同时获得整数除法的商和余数，即不执行两次整数除法？
回答by John Dibling
divwill do this. See referenceand example:
div会这样做。请参阅参考和示例：
/* div example */
#include <stdio.h>
#include <stdlib.h>
int main ()
{
div_t divresult;
divresult = div (38,5);
printf ("38 div 5 => %d, remainder %d.\n", divresult.quot, divresult.rem);
return 0;
}
Output:
输出：
38 div 5 => 7, remainder 3.
EDIT:
编辑：
The C Specification says:
C 规范说：
7.20 General utilities
7.20 一般公用事业
The types declared are size_t and wchar_t (both described in 7.17),
div_t
which is a structure type that is the type of the value returned by the div function,
ldiv_t
which is a structure type that is the type of the value returned by the ldiv function, and
lldiv_t
which is a structure type that is the type of the value returned by the lldiv function.
... but it doesn't say what the definition of div_tis.
...但它没有说明的定义div_t是什么。
