blocks|key|427063|text|C%2B%2B标准(2003)在§5.6/4中规定，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|427064|427065|+...如果/或%25的第二个操作数为零，则行为为未定义的；...|blockquote|427066|427067|427068|也就是说，以下表达式调用未定义的行为(UB)：|427069|X+/+0;+//UB
X+%25+0;+//UB|code-block|syntax|javascript|427070|427071|还要注意的是，-5+%25+2并不等同于-(5+%25+2)+(正如Petar在他的答案的评论中所暗示的那样)。它是由实现定义的。规范上说(第5.6/4节)，|offset|length|style|CODE|427072|427073|+...如果两个操作数都是非负的，则余数是非负的；如果不是，则余数的符号为implementation-defined.|427074|427075|entityMap^0|0|0|0|0|0|0|0|0|7|6|I|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|13|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|14|8|@]|9|@]|A|$]]|$1|G|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|16|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|17|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|19|8|@$R|1A|S|1B|T|U]|$R|1C|S|1D|T|U]]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|W|3|X|5|E|7|1F|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|10|$]]

The C++ Standard(2003) says in §5.6/4,

<blockquote>
 [...] If the second operand of / or % is zero the behavior is undefined; [...]
</blockquote>

That is, following expressions invoke undefined-behavior(UB):

<pre><code>X / 0; //UB
X % 0; //UB
</code></pre>

<hr>

Note also that <code>-5 % 2</code> is NOT equal to <code>-(5 % 2)</code> (as Petar seems to suggest in his comment to his answer). It's implementation-defined. The spec says (§5.6/4), 

<blockquote>
 [...] If both operands are nonnegative then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined.
</blockquote>

blocks|key|2390510|text|这个答案不适合数学家。这个答案试图给出动机(以数学精度为代价)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2390511|Mathematicians:+See+here.|offset|length|style|BOLD|2390512|程序员：请记住，除以0是undefined。因此，依赖于除法的mod也是undefined。|CODE|2390513|2390514|这表示正X和D的除法；它由整数部分和小数部分组成：|2390515|(X+/+D)+=+++integer++++%2B++fraction
++++++++=+floor(X+/+D)+%2B+(X+%25+D)+/+D|code-block|syntax|javascript|2390516|重新排列，你会得到：|2390517|(X+%25+D)+=+D+*+(X+/+D)+-+D+*+floor(X+/+D)|2390518|用0代替D|2390519|(X+%25+0)+=+0+*+(X+/+0)+-+0+*+floor(X+/+0)|2390520|因为除以0是undefined|2390521|(X+%25+0)+=+0+*+undefined+-+0+*+floor(undefined)
++++++++=+undefined+-+undefined
++++++++=+undefined|2390522|entityMap|0|LINK|mutability|MUTABLE|url|https://math.stackexchange.com/questions/516251/why-is-n-mod-0-undefined^0|0|0|F|G|9|0|0|0|4|A|1|C|9|V|3|10|9|0|0|4|1|6|1|0|0|0|0|1|1|4|1|0|0|4|1|6|9|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1D|8|@$D|1E|E|1F|F|G]]|9|@$D|1G|E|1H|1|1I]]|A|$]]|$1|H|3|I|5|6|7|1J|8|@$D|1K|E|1L|F|G]|$D|1M|E|1N|F|J]|$D|1O|E|1P|F|J]|$D|1Q|E|1R|F|J]|$D|1S|E|1T|F|J]]|9|@]|A|$]]|$1|K|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1V|8|@$D|1W|E|1X|F|J]|$D|1Y|E|1Z|F|J]]|9|@]|A|$]]|$1|N|3|O|5|P|7|20|8|@]|9|@]|A|$Q|R]]|$1|S|3|T|5|6|7|21|8|@]|9|@]|A|$]]|$1|U|3|V|5|P|7|22|8|@]|9|@]|A|$Q|R]]|$1|W|3|X|5|6|7|23|8|@$D|24|E|25|F|J]|$D|26|E|27|F|J]]|9|@]|A|$]]|$1|Y|3|Z|5|P|7|28|8|@]|9|@]|A|$Q|R]]|$1|10|3|11|5|6|7|29|8|@$D|2A|E|2B|F|J]|$D|2C|E|2D|F|J]]|9|@]|A|$]]|$1|12|3|13|5|P|7|2E|8|@]|9|@]|A|$Q|R]]|$1|14|3|-4|5|6|7|2F|8|@]|9|@]|A|$]]]|15|$16|$5|17|18|19|A|$1A|1B]]]]

This answer is not for the mathematician. This answer attempts to give motivation (at the cost of mathematical precision).

Mathematicians: <a href="https://math.stackexchange.com/questions/516251/why-is-n-mod-0-undefined">See here.</a>

Programmers: Remember that division by <code>0</code> is <code>undefined</code>. Therefore, <code>mod</code>, which relies on division, is also <code>undefined</code>.

<hr>

This represents division for positive <code>X</code> and <code>D</code>; it's made up of the integral part and fractional part:

<pre><code>(X / D) = integer + fraction
 = floor(X / D) + (X % D) / D
</code></pre>

Rearranging, you get:

<pre><code>(X % D) = D * (X / D) - D * floor(X / D)
</code></pre>

Substituting <code>0</code> for <code>D</code>:

<pre><code>(X % 0) = 0 * (X / 0) - 0 * floor(X / 0)
</code></pre>

Since division by <code>0</code> is <code>undefined</code>:

<pre><code>(X % 0) = 0 * undefined - 0 * floor(undefined)
 = undefined - undefined
 = undefined
</code></pre>

blocks|key|427013|text|根据的定义，X+%25+D是a+number+0+<=+R+<+D，这样就存在Q，从而|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|427014|X+=+D*Q+%2B+R|code-block|syntax|javascript|427015|因此，如果为D+=+0，则不能存在这样的数字(因为0+<=+R+<+0)|427016|entityMap^0|2|A|6|5|L|A|11|1|0|0|6|5|P|A|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|D]|$9|U|A|V|B|D]|$9|W|A|X|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|Y|8|@]|E|@]|F|$J|K]]|$1|L|3|M|5|6|7|Z|8|@$9|10|A|11|B|D]|$9|12|A|13|B|D]]|E|@]|F|$]]|$1|N|3|-4|5|6|7|14|8|@]|E|@]|F|$]]]|O|$]]

<code>X % D</code> is by definition a number <code>0 &lt;= R &lt; D</code>, such that there exists <code>Q</code> so that

<pre><code>X = D*Q + R
</code></pre>

So if <code>D = 0</code>, no such number can exists (because <code>0 &lt;= R &lt; 0</code>)

blocks|key|4290543|text|我认为，因为要得到X+%25+0的余数，你需要首先计算产生无穷大的X+/+0，而试图计算无穷大的余数是不可能的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4290544|但是，符合您的想法的最佳解决方案是这样做|4290545|REMAIN+=+Y+?+X+%25+Y+:+X|code-block|syntax|javascript|4290546|entityMap^0|9|5|V|5|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

I think because to get the remainder of <code>X % 0</code> you need to first calculate <code>X / 0</code> which yields infinity, and trying to calculate the remainder of infinity is not really possible.

However, the best solution in line with your thinking would be to do something like this

<pre><code>REMAIN = Y ? X % Y : X
</code></pre>

blocks|key|4006360|text|另一种可能在概念上很容易理解这个问题的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4006361|暂时忽略参数符号的问题，可以很容易地将a+%25+b重写为a+-+((a+/+b)+*+b)。如果b为零，则表达式a+/+b未定义，因此在这种情况下，整个表达式也必须为零。|offset|length|style|CODE|4006362|最后，模数实际上是一种除法运算，所以如果a+/+b是未定义的，那么期望a+%25+b也是未定义的并不是不合理的。|4006363|entityMap^0|0|J|5|R|H|1B|1|1J|5|0|K|5|Z|5|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]|$D|P|E|Q|F|G]|$D|R|E|S|F|G]|$D|T|E|U|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|V|8|@$D|W|E|X|F|G]|$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|K|$]]

Another way that might be conceptually easy to understand the issue:

Ignoring for the moment the issue of argument sign, <code>a % b</code> could easily be re-written as <code>a - ((a / b) * b)</code>. The expression <code>a / b</code> is undefined if <code>b</code> is zero, so in that case the overall expression must be too.

In the end, modulus is effectively a divisive operation, so if <code>a / b</code> is undefined, it's not unreasonable to expect <code>a % b</code> to be as well.

blocks|key|4290373|text|X+%25+Y给出整数[+0,+Y+)范围内的结果。X+%25+0必须给出大于或等于零且小于零的结果。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4290374|entityMap^0|0|5|9|8|O|5|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

<code>X % Y</code> gives a result in the integer <code>[ 0, Y )</code> range. <code>X % 0</code> would have to give a result greater or equal to zero, and less than zero.

blocks|key|2390685|text|您可以避免(A%25B)的“被0除”的情况，因为它的类型float(+b)+=b=0.0的浮点标识mod(a，b)是未定义的，或者在任何2个实现之间以不同的方式定义，以避免逻辑错误(硬崩溃)而有利于算术错误...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2390686|通过计算mod([a*b],[b])==b*(a-floor(a))|offset|length|style|CODE|2390687|实例读取|2390688|计算mod([a],[b])|2390689|在x轴上，随着时间b的变化，a*b==your+==曲线的最大值(永远不会达到)，==一阶导数。|2390690|https://www.shadertoy.com/view/MslfW8|2390691|entityMap|0|LINK|mutability|MUTABLE|url^0|0|4|U|0|0|2|C|0|0|0|11|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|10|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|14|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|15|8|@]|9|@$D|16|E|17|1|18]]|A|$]]|$1|P|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|O]]]]

you can evade the "divivion by 0" case of (A%B) for its type float identity mod(a,b) for float(B)=b=0.0 , that is undefined, or defined differently between any 2 implementations, to avoid logic errors (hard crashes) in favor of arithmetic errors...

by computing <code>mod([a*b],[b])==b*(a-floor(a))</code> 
INSTREAD OF 
computing <code>mod([a],[b])</code>

where [a*b]==your x-axis, over time
[b] == the maximum of the seesaw curve (that will never be reached) == the first derivative of the seesaw function

<a href="https://www.shadertoy.com/view/MslfW8" rel="nofollow noreferrer">https://www.shadertoy.com/view/MslfW8</a>

blocks|key|2390718|text|我想是因为要得到X%250的余数，你需要首先计算产生无穷大的X/0，而试图计算无穷大的余数是不可能的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2390719|然而，符合您的想法的最佳解决方案是这样做，|2390720|ans+=+Y+?+X+%25+Y+:+X|code-block|syntax|javascript|2390721|此外，在C%2B%2B文档中写道，X%250或X/0会导致未定义的值。|2390722|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I suppose because to get the remainder of X % 0 you need to first calculate X / 0 which yields infinity, and trying to calculate the remainder of infinity is not really possible.
However, the best solution in line with your thinking would be to do something like this,
<pre><code>ans = Y ? X % Y : X
</code></pre>
Also, in C++ docs its written that X % 0 or X / 0 ,results in an undefined value.

blocks|key|4006452|text|计算机如何划分：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4006453|从被除数开始，减去除数，直到结果小于除数。减去的次数就是结果，剩下的就是余数。例如，要将10和3相除：|4006454|10+-+3+=+7
7+-+3+=+4
4+-+3+=+1|code-block|syntax|javascript|4006455|所以|4006456|10+/+3+=+3
10+%25+3+=+1|4006457|要除1和0，请执行以下操作：|4006458|1+/+0
1+-+0+=+1++
1+-+0+=+1++
1+-+0+=+1++
...|4006459|4006460|1+/+0+=+Infinity+(technically+even+infinity+is+too+small,+but+it's+easy+to+classify+it+as+that)
1+%25+0+=+NaN|4006461|如果没有什么可以阻止它，CPU将继续执行它，直到它重载并返回一个完全随机的结果。因此，在CPU级别有一条指令，如果除数为0，则返回NaN或Infinity+(取决于您的平台)。|offset|length|style|CODE|4006462|这将永远不会结束，因此其余部分是未定义的(对于计算机，这是NaN+)。|4006463|entityMap^0|0|0|0|0|0|0|0|0|0|1T|3|1X|8|0|T|3|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|14|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|15|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|16|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|17|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|19|8|@]|9|@]|A|$G|H]]|$1|Q|3|J|5|6|7|1A|8|@]|9|@]|A|$]]|$1|R|3|S|5|F|7|1B|8|@]|9|@]|A|$G|H]]|$1|T|3|U|5|6|7|1C|8|@$V|1D|W|1E|X|Y]|$V|1F|W|1G|X|Y]]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1H|8|@$V|1I|W|1J|X|Y]]|9|@]|A|$]]|$1|11|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]]|12|$]]

How computers divide:
Start with the dividend and subtract the divisor until the result is less then the divisor. The number of times you subtracted is the result and what you have left is the remainder. For example, to divide 10 and 3:
<pre><code>10 - 3 = 7
7 - 3 = 4
4 - 3 = 1
</code></pre>
So
<pre><code>10 / 3 = 3
10 % 3 = 1
</code></pre>
To divide 1 and 0:
<pre><code>1 / 0
1 - 0 = 1 
1 - 0 = 1 
1 - 0 = 1 
...
</code></pre>
So
<pre><code>1 / 0 = Infinity (technically even infinity is too small, but it's easy to classify it as that)
1 % 0 = NaN
</code></pre>
If there is nothing to stop it, the CPU will continue to execute this until it overloads and returns a totally random result. So there is an instruction at the CPU level that if the divisor is 0, return <code>NaN</code> or <code>Infinity</code> (depending on your platform).
This will never end so the remainder is undefined (which is <code>NaN</code> for computers).

Why is <code>X % 0</code> an invalid expression?

I always thought <code>X % 0</code> should equal X. Since you can't divide by zero, shouldn't the answer naturally be the remainder, X (everything left over)?

Can't Mod Zero?

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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社区新版编辑器体验调研

为什么X % 0是一个无效的表达式？我一直认为X % 0应该等于X。既然你不能被零除，答案自然不应该是余数，X(所有剩余的东西)？

问Mod Zero不能吗？
EN

回答 9

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问Mod Zero不能吗？EN

回答 9

Stack Overflow用户

Stack Overflow用户

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腾讯云开发者

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问Mod Zero不能吗？
EN