blocks|key|588468|text|157/32为int/int，这会导致int。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|588469|尝试使用双字面量-+157/32d，它是int/double，这将产生一个double。|588470|entityMap^0|7|7|J|3|0|A|7|K|A|11|6|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|I|$]]

157/32 is <code>int/int</code>, which results in an <code>int</code>.

Try using the double literal - <code>157/32d</code>, which is <code>int/double</code>, which results in a <code>double</code>.

blocks|key|804349|text|157/32是一个整数除法，因为除非另外使用后缀指定，否则所有的数字文字都是整数(d表示双精度，l表示长整型)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|804350|除法向下舍入(到4)，然后转换为双精度(4.0)，然后向上舍入(到4.0)|804351|如果你使用一个变量，你可以避免这种情况|804352|double+a1=157;
double+a2=32;
int+total+=+(int)+Math.ceil(a1/a2);|code-block|syntax|javascript|804353|entityMap^0|0|6|15|1|1C|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|Y|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|Z|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

<code>157/32</code> is an integer division because all numerical literals are integers unless otherwise specified with a suffix (<code>d</code> for double <code>l</code> for long)

the division is rounded down (to 4) before it is converted to a double (4.0) which is then rounded up (to 4.0)

if you use a variables you can avoid that

<pre><code>double a1=157;
double a2=32;
int total = (int) Math.ceil(a1/a2);
</code></pre>

blocks|key|588563|text|int+total+=+(int)+Math.ceil((double)157/32);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|588564|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int total = (int) Math.ceil((double)157/32);
</code></pre>

blocks|key|804480|text|在Java语言中，添加一个.0将使它成为一个双精度...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|804481|int+total+=+(int)+Math.ceil(157.0+/+32.0);|code-block|syntax|javascript|804482|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In Java adding a .0 will make it a double...

<pre><code>int total = (int) Math.ceil(157.0 / 32.0);
</code></pre>

blocks|key|804696|text|当除以两个整数时，例如，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|804697|int+c+=+(int)+a+/+(int)+b;|offset|length|style|CODE|804698|结果是一个int，它的值是a除以b，四舍五入为零。因为结果已经四舍五入了，所以ceil()不做任何事情。请注意，这种舍入与floor()不同，后者舍入为负无穷大。因此，3/2等于1+(+floor(1.5)等于1.0，但(-3)/2等于-1，但floor(-1.5)等于-2.0)。|804699|这一点很重要，因为如果a/b始终与floor(a+/+(double)+b)相同，那么您可以将a/b的ceil()实现为-(+(-a)+/+b)。|804700|获取ceil(a/b)的建议|804701|int+n+=+(a+%2B+b+-+1)+/+b;，相当于a+/+b+%2B+(b+-+1)+/+b或(a+-+1)+/+b+%2B+1|804702|之所以有效，是因为ceil(a/b)始终比floor(a/b)大1，除非a/b是整数。因此，除非a/b是一个整数，否则您希望将其提升到(或超过)下一个整数。添加1+-+1+/+b可以做到这一点。对于整数，它不会将它们推高到下一个整数。对于其他任何事情，它都会。|804703|啊呀。希望这是有意义的。我相信有一种更优雅的数学方式来解释它。|804704|entityMap^0|0|0|Q|0|5|3|D|1|G|1|13|6|1P|7|2C|3|2H|1|2L|A|2X|3|32|6|3A|2|3E|B|3R|4|0|B|3|H|L|1B|3|1F|6|1O|C|0|2|9|0|0|O|S|J|1C|F|0|9|9|L|A|10|3|1C|3|28|9|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@$D|X|E|Y|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Z|8|@$D|10|E|11|F|G]|$D|12|E|13|F|G]|$D|14|E|15|F|G]|$D|16|E|17|F|G]|$D|18|E|19|F|G]|$D|1A|E|1B|F|G]|$D|1C|E|1D|F|G]|$D|1E|E|1F|F|G]|$D|1G|E|1H|F|G]|$D|1I|E|1J|F|G]|$D|1K|E|1L|F|G]|$D|1M|E|1N|F|G]|$D|1O|E|1P|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|1Q|8|@$D|1R|E|1S|F|G]|$D|1T|E|1U|F|G]|$D|1V|E|1W|F|G]|$D|1X|E|1Y|F|G]|$D|1Z|E|20|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|21|8|@$D|22|E|23|F|G]]|9|@]|A|$]]|$1|N|3|O|5|6|7|24|8|@$D|25|E|26|F|G]|$D|27|E|28|F|G]|$D|29|E|2A|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|2B|8|@$D|2C|E|2D|F|G]|$D|2E|E|2F|F|G]|$D|2G|E|2H|F|G]|$D|2I|E|2J|F|G]|$D|2K|E|2L|F|G]]|9|@]|A|$]]|$1|R|3|S|5|6|7|2M|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|2N|8|@]|9|@]|A|$]]]|U|$]]

When dividing two integers, e.g.,

<code>int c = (int) a / (int) b;</code>

the result is an <code>int</code>, the value of which is <code>a</code> divided by <code>b</code>, rounded toward zero. Because the result is already rounded, <code>ceil()</code> doesn't do anything. Note that this rounding is not the same as <code>floor()</code>, which rounds towards negative infinity. So, <code>3/2</code> equals <code>1</code> (and <code>floor(1.5)</code> equals <code>1.0</code>, but <code>(-3)/2</code> equals <code>-1</code> (but <code>floor(-1.5)</code> equals <code>-2.0</code>).

This is significant because if <code>a/b</code> were always the same as <code>floor(a / (double) b)</code>, then you could just implement <code>ceil()</code> of <code>a/b</code> as <code>-( (-a) / b)</code>.

The suggestion of getting <code>ceil(a/b)</code> from

<code>int n = (a + b - 1) / b;</code>, which is equivalent to <code>a / b + (b - 1) / b</code>, or <code>(a - 1) / b + 1</code>

works because <code>ceil(a/b)</code> is always one greater than <code>floor(a/b)</code>, except when <code>a/b</code> is a whole number. So, you want to bump it to (or past) the next whole number, unless <code>a/b</code> is a whole number. Adding <code>1 - 1 / b</code> will do this. For whole numbers, it won't quite push them up to the next whole number. For everything else, it will.

Yikes. Hopefully that makes sense. I'm sure there's a more mathematically elegant way to explain it.

blocks|key|804420|text|此外，要将数字从整数转换为实数，您可以添加一个点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|804421|int+total+=+(int)+Math.ceil(157/32.);|code-block|syntax|javascript|804422|和(157/32)也会是真实的。;)|804423|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Also to convert a number from integer to real number you can add a dot:

<pre><code>int total = (int) Math.ceil(157/32.);
</code></pre>

And the result of (157/32.) will be real too. ;)

blocks|key|804435|text|int+total+=+(int)+Math.ceil(+(double)157/+(double)+32);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|804436|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int total = (int) Math.ceil( (double)157/ (double) 32);
</code></pre>

blocks|key|804752|text|对于您的问题，请检查以下解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|804753|int+total+=+(int)+Math.ceil(157/32);|code-block|syntax|javascript|804754|在这里你应该将分子乘以1.0，然后它会给出你的答案。|804755|int+total+=+(int)+Math.ceil(157*1.0/32);|804756|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Check the solution below for your question:

<pre><code>int total = (int) Math.ceil(157/32);
</code></pre>

Here you should multiply Numerator with 1.0, then it will give your answer.

<pre><code>int total = (int) Math.ceil(157*1.0/32);
</code></pre>

blocks|key|588857|text|使用double类型转换为|type|unstyled|depth|inlineStyleRanges|entityRanges|data|588858|Math.ceil((double)value)或类似的|offset|length|style|CODE|588859|Math.ceil((double)value1/(double)value2);|code-block|syntax|javascript|588860|entityMap^0|0|0|O|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Use double to cast like 

<code>Math.ceil((double)value)</code> or like

<pre><code>Math.ceil((double)value1/(double)value2);
</code></pre>

blocks|key|804811|text|我知道这是一个古老的问题，但在我看来，我们有一个更好的方法，那就是使用BigDecimal来避免精度损失。顺便说一句，使用这个解决方案，我们可以使用几种舍入和缩放策略。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|804812|final+var+dividend+=+BigDecimal.valueOf(157);
final+var+divisor+=+BigDecimal.valueOf(32);
final+var+result+=+dividend.divide(divisor,+RoundingMode.CEILING).intValue();|code-block|syntax|javascript|804813|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/math/BigDecimal.html^0|Z|A|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

I know this is an old question but in my opinion, we have a better approach which is using <a href="https://docs.oracle.com/javase/8/docs/api/java/math/BigDecimal.html" rel="nofollow noreferrer">BigDecimal</a> to avoid precision loss. By the way, using this solution we have the possibility to use several rounding and scale strategies.
<pre><code>final var dividend = BigDecimal.valueOf(157);
final var divisor = BigDecimal.valueOf(32);
final var result = dividend.divide(divisor, RoundingMode.CEILING).intValue();
</code></pre>

blocks|key|804559|text|int+total+=+(157-1)/32+%2B+1|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|804560|或更一般的|unstyled|804561|(a-1)/b+%2B1+|804562|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>int total = (157-1)/32 + 1
</code></pre>

or more general

<pre><code>(a-1)/b +1 
</code></pre>

<pre><code>int total = (int) Math.ceil(157/32);
</code></pre>

Why does it still return 4? <code>157/32 = 4.90625</code>, I need to round up, I've looked around and this seems to be the right method.

I tried <code>total</code> as <code>double</code> type, but get 4.0.

What am I doing wrong?

Java rounding up to an int using Math.ceil

Java

int total = (int) Math.ceil(157/32);为什么它仍然返回4？157/32 = 4.90625，我需要四舍五入，我已经看过了，这似乎是正确的方法。我尝试将total作为double类型，但得到了4.0。我做错了什么？

问使用Math.ceil将Java舍入为整数
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