这个功能是什么以及它可以用来做什么？ [英] What does the function remquo do and what can it be used for?

问题描述

通过C规范阅读我发现这个函数：

  double remquo（double x，double y，int * quo） ; 
 float remquof（float x，float y，int * quo）; 
 long double remquol（long double x，long double y，
 int * quo）; 
  

remquo 函数计算与 余数 函数相同的余数。在
中，指向quo的对象存储一个值，其符号是 x / y 的符号，其
幅度是对于 x / y 的积分商数的模一致模2 ^ n
，其中
n 是一个实现定义的大于或等于3的整数。
$ b

remquo 函数返回 x REM y 即可。如果 y 为零，则存储在 quo 指向的对象
中的值code> 是未指定的，并且是否发生域错误或函数
返回零是实现定义的。


我知道它返回的是什么，它返回 fmod（x，y），但我不明白整个 quo 部分。它在语义上等于这个吗？

  * quo =（int）x / y; 
 * quo％= n; / * n实现定义* / 
  

我的最后一个问题是，这个函数有什么用处？ / p>


解决方案

编辑：正如Jeffrey Scofield所说他的答案，返回的商并不是 x / y ，但是低3位（加号）



它相当于（最大类型差异）：



  quo = x / y; 
 rem = x％y; 
  

其中 rem 是返回值，并且 quo 作为输出参数返回。

与上述语法相比，它的优势在于它只执行一次除法操作。


Reading through the C specs I found this function:

double remquo(double x, double y, int *quo);
float remquof(float x, float y, int *quo);
long double remquol(long double x, long double y,
    int *quo);

The remquo functions compute the same remainder as the remainder functions. In the object pointed to by quo they store a value whose sign is the sign of x/y and whose magnitude is congruent modulo 2^n to the magnitude of the integral quotient of x/y, where n is an implementation-defined integer greater than or equal to 3.

The remquo functions return x REM y. If y is zero, the value stored in the object pointed to by quo is unspecified and whether a domain error occurs or the functions return zero is implementation defined.

I understand what it returns, it returns fmod(x, y), but I don't understand the whole quo part. Is it semantically equal to this?

*quo = (int) x/y;
*quo %= n; /* n implementation defined */

And my last question, for what could this function be useful?

解决方案

EDIT: As Jeffrey Scofield said in his answer, the returned quotient really isn't x/y, but the low 3 bits (plus sign) of the quotient.

It's equivalent to (up to type differences):

quo = x/y;
rem = x%y;

Where rem is the return value, and quo is returned as an output parameter.
Its advantage over the above syntax is that it does just one division operation.

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