blocks|key|330227|text|你将失去商的小数部分。因此，是的，数字越大，相对精度就越高，比如与5000/3000相比。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|330228|但是，5+/+3将返回整数，而不是双精度。要强制将其除以双精度，请将被除数类型转换为static_cast<double>(5)+/+3。|330229|entityMap^0|X|9|0|3|5|16|Q|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

You will lose the fractional portion of the quotient. So yes, with greater numbers you will have more relative precision, such as compared with <code>5000/3000</code>.

However, <code>5 / 3</code> will return an integer, not a double. To force it to divide as double, typecast the dividend as <code>static_cast&lt;double&gt;(5) / 3</code>.

blocks|key|330245|text|5/3将始终生成1(整数)，如果您执行5.0/3或5/3.0，则结果将是双精度。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|330246|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

5/3 will always produce 1 (an integer), if you do 5.0/3 or 5/3.0 the result will be a double.

blocks|key|331442|text|因为在C和C%2B%2B中，正如其他人所说，/是整数除法，所以它将返回一个整数。特别是，它将返回双重答案的发言权...(C和C%2B%2B总是截断)所以，基本上5/3就是你想要的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|331443|它可能会在负片中变得有点奇怪，因为-5/3+=>+-2可能是你想要的，也可能不是你想要的。|331444|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Since in C and C++, as others have said, / is integer division, it will return an int. in particular, it will return the floor of the double answer... (C and C++ always truncate) So, basically 5/3 is exactly what you want.

It may get a little weird in negatives as -5/3 => -2 which may or may not be what you want...

blocks|key|1623322|text|整数除法得到整数结果，因此5/3是1，5%253是2(余数运算符)。然而，这并不一定适用于负数。在最初的C%2B%2B标准中，-5+/3可以是-1+(舍入到零)或-2+(下限)，但建议使用-1。在最新的C%2B%2B0B草案中(几乎可以肯定，它非常接近最终标准)，它是-1，因此查找具有负数的下限更加复杂。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1623323|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Integer division gives integer results, so 5 / 3 is 1 and 5 % 3 is 2 (the remainder operator). However, this doesn't necessarily hold with negative numbers. In the original C++ standard, -5 / 3 could be either -1 (rounding towards zero) or -2 (the floor), but -1 was recommended. In the latest C++0B draft (which is almost certainly very close to the final standard), it is -1, so finding the floor with negative numbers is more involved.

blocks|key|3031758|text|据我所知，没有用于此目的的预定义函数。如果由于某种原因浮点计算是不可能的(例如，int64_t的精度高于double可以正确表示的精度)，则可能有必要使用这样的函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3031759|我们可以这样定义这个函数：|3031760|#include+<cmath>

inline+long
floordiv+(long+num,+long+den)
{
++if+(0+<+(num%5Eden))
++++return+num/den;
++else
++++{
++++++ldiv_t+res+=+ldiv(num,den);
++++++return+(res.rem)?+res.quot-1+
++++++++++++++++++++++:+res.quot;
++++}
}|code-block|syntax|javascript|3031761|其思想是使用正常的整数除法，但调整为负结果以匹配double+floor(double)函数的行为。要点是始终向下一个较低的整数截断，而不考虑零点的位置。如果意图是创建均匀大小的间隔，这可能非常重要。|3031762|计时测量表明，与内置的/操作符相比，这个函数在这里只产生了很小的开销，但是基于浮点的floor函数当然要快得多。|3031763|entityMap^0|14|7|1G|6|0|0|0|O|K|0|B|1|16|5|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$]]

As far as I know, there is no predefined function for this purpose.
It might be necessary to use such a function, if for some reason floating-point calculations are out of question (e.g. <code>int64_t</code> has a higher precision than <code>double</code> can represent without error)

We could define this function as follows:

<pre><code>#include &lt;cmath&gt;

inline long
floordiv (long num, long den)
{
 if (0 &lt; (num^den))
 return num/den;
 else
 {
 ldiv_t res = ldiv(num,den);
 return (res.rem)? res.quot-1 
 : res.quot;
 }
}
</code></pre>

The idea is to use the normal integer divison, but adjust for negative results to match the behaviour of the <code>double floor(double)</code> function. The point is to truncate always towards the next lower integer, irrespective of the position of the zero point. This can be very important if the intention is to create even sized intervals.

Timing measurements show that this function here only creates a small overhead compared with the built-in <code>/</code> operator, but of course the floating point based <code>floor</code> function is significantly faster....

I want to implement greatest integer function. [The "greatest integer function" is a <a href="http://www.google.com/search?q=greatest%20integer%20function" rel="noreferrer">quite</a> <a href="http://books.google.com/books?q=greatest+integer+function" rel="noreferrer">standard name</a> for what is also known as the floor function.]

<pre><code>int x = 5/3;
</code></pre>

My question is with greater numbers could there be a loss of precision as 5/3 would produce a double?

EDIT: Greatest integer function is integer less than or equal to X.
Example:

<pre><code>4.5 = 4
4 = 4
3.2 = 3
3 = 3
</code></pre>

What I want to know is 5/3 going to produce a double? Because if so I will have loss of precision when converting to int.

Hope this makes sense.

C++ integer floor function

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想实现最大整数函数。[“最大整数函数”是也称为floor函数的 。]int x = 5/3;我的问题是，对于更大的数字，是否会有精度损失，因为5/3会产生一个双精度？编辑:最大整数函数是小于或等于X的整数。示例：4.5 = 44 = 43.2 = 33 = 3我想知道的是5/3会产生一个双倍吗？因为如果是这样的话，当...

问C++整数下限函数
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