blocks|key|958092|text|assert(isnumeric(input)+&&+round(input)+==+input,+'That+number+is+not+an+integer.')|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|958093|您也可以很容易地添加其他检查(如正性检查)。|958094|使用isinteger编辑。感谢@SolarStatistics，我没有注意到他们添加了这个功能。再次编辑回原始答案，因为isinteger不合适(请参阅下面的注释)。|958095|entityMap^0|0|2B|0|0|2|9|G|G|1P|9|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|K|$]]

<code>assert(isnumeric(input) &amp;&amp; round(input) == input, 'That number is not an integer.')</code>

You could add other checks, (like for positivity) easily as well.

Edited using <code>isinteger</code>. Thanks <code>@SolarStatistics</code>, I hadn't noticed they added this functionality.
Edited back to original answer again as <code>isinteger</code> isn't appropriate (see comments below).

blocks|key|742253|text|您可以使用mod函数，该函数返回除法后的余数。所有整数都可以被1整除。因此，对非整数的一个很好的测试是|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|742254|integerTest=~mod(value,1);|code-block|syntax|javascript|742255|如果value不是整数，则返回0；如果是，则返回1。然后，您可以将其用作拒绝非整数用户输入的条件。|742256|entityMap|0|LINK|mutability|MUTABLE|url|http://www.mathworks.com/help/techdoc/ref/mod.html^0|5|3|V|1|5|3|0|0|0|2|5|F|1|O|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

You can use the <a href="http://www.mathworks.com/help/techdoc/ref/mod.html" rel="noreferrer"><code>mod</code></a> function, which returns the remainder after division. All integers are divisible by <code>1</code>. So a good test for non-integer would be 

<pre><code>integerTest=~mod(value,1);
</code></pre>

This returns <code>0</code> if <code>value</code> is not an integer and <code>1</code> if it is. You can then use this as a conditional to reject non-integer user inputs.

blocks|key|958231|text|下面是另一个变体(您可以在ISIND函数：edit+isind.m中看到它的用法)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|958232|integerTest+=+(+x+==+floor(x)+);|code-block|syntax|javascript|958233|在我的机器上，它比其他建议的解决方案更快：|958234|%25#+create+a+vector+of+doubles,+containing+integers+and+non-integers
x+=+(1:100000)';+++++++++++++++++++++++%25'
idx+=+(+rand(size(x))+<+0.5+);
x(idx)+=+x(idx)+%2B+rand(sum(idx),1);

%25#+test+for+integers
tic,+q1+=+~mod(x,+1);+toc
tic,+q2+=+x==double(uint64(x));+toc
tic,+q3+=+x==floor(x);+toc

%25#+compare+results
assert(+isequal(q1,q2,q3)+)|958235|计时：|958236|Elapsed+time+is+0.012253+seconds.
Elapsed+time+is+0.014201+seconds.
Elapsed+time+is+0.005665+seconds.|958237|entityMap|0|LINK|mutability|MUTABLE|url|http://www.mathworks.com/help/toolbox/images/ref/isind.html^0|L|C|D|5|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|19|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

Here is another variation (you can see it being used in <a href="http://www.mathworks.com/help/toolbox/images/ref/isind.html" rel="noreferrer">ISIND</a> function: <code>edit isind.m</code>):

<pre><code>integerTest = ( x == floor(x) );
</code></pre>

On my machine, it is faster than the other proposed solutions:

<pre><code>%# create a vector of doubles, containing integers and non-integers
x = (1:100000)'; %'
idx = ( rand(size(x)) &lt; 0.5 );
x(idx) = x(idx) + rand(sum(idx),1);

%# test for integers
tic, q1 = ~mod(x, 1); toc
tic, q2 = x==double(uint64(x)); toc
tic, q3 = x==floor(x); toc

%# compare results
assert( isequal(q1,q2,q3) )
</code></pre>

Timings:

<pre><code>Elapsed time is 0.012253 seconds.
Elapsed time is 0.014201 seconds.
Elapsed time is 0.005665 seconds.
</code></pre>

blocks|key|742383|text|通过double命令，您无法获得正确答案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|742384|>>+double(uint64(21/22))
ans+=
+++++1
>>+double(uint64(22/22))
ans+=
+++++1|code-block|syntax|javascript|742385|还有floor，round，...对这样的情况有问题：|742386|floor(22/22)==21.99999999999999999999999999999999999/22|742387|但是mod似乎可以区分22/22和21.99999999999999999999999999999999999/22|742388|>>+mod(22,22)
ans+=
+++++0
>>+(21.99999999999999999999999999999999999/22)
ans+=
+++++1|742389|entityMap^0|2|6|0|0|2|5|8|5|0|0|2|3|B|5|H|15|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|T|$]]

By <code>double</code> command, you cannot get the correct answer:

<pre><code>&gt;&gt; double(uint64(21/22))
ans =
 1
&gt;&gt; double(uint64(22/22))
ans =
 1
</code></pre>

also <code>floor</code>,<code>round</code>,... have problem with such cases:

<pre><code>floor(22/22)==21.99999999999999999999999999999999999/22
</code></pre>

but <code>mod</code> seems can distinguish <code>22/22</code> and <code>21.99999999999999999999999999999999999/22</code>:

<pre><code>&gt;&gt; mod(22,22)
ans =
 0
&gt;&gt; (21.99999999999999999999999999999999999/22)
ans =
 1
</code></pre>

I'm writing a program that will calculate factorials of integers. However, the part I'm stuck on is if someone enters a non-integer such as <code>1.3</code>, I'd like to be able to test the input and display <code>"The number you have entered is not an integer"</code>

How do I test for integers in MATLAB?

我正在写一个程序，它将计算整数的阶乘。但是，我遇到的问题是，如果有人输入一个非整数，比如1.3，我希望能够测试输入并显示"The number you have entered is not an integer"

问如何在MATLAB中测试整数？
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