blocks|key|1366207|text|因为在整数除法中，如果答案不是完全整数，小数点后的数字将被删除(整数除法产生整数值)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1366208|请注意，您不必强制转换这两个整数，您可以仅强制转换一个，第二个将被隐式转换。|1366209|为什么在造型后它会起作用呢？|1366210|因为cast比/具有更高的优先级。所以它首先进行转换，然后再进行除法运算。如果不是这样的话，你会得到0.0。|offset|length|style|CODE|1366211|entityMap^0|0|0|0|7|1|1E|3|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Q|8|@$H|R|I|S|J|K]|$H|T|I|U|J|K]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|M|$]]

Because in integer division, if the answer is not a perfect integer, the digits after the decimal point will be removed (integer division yields integer value).

Note that you don't have to cast the both integers, you can cast only one, the second will be implicitly converted.

Why after the cast it works?

Because cast has higher precedence than <code>/</code>. So first it cast, then it divides. If this were not the case, you would get <code>0.0</code>.

blocks|key|223560|text|与往常一样，赋值运算符右侧的任何计算都是在将其结果赋给左侧指定的变量之前执行的。当执行计算时，它是两个int之间的除法，这是一个整数除法(我相信您知道)，它产生一个整数结果(即没有任何余数的商)。只有在执行此计算之后，才会将整数除法的结果-0-转换为双精度。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|223561|正如您所观察到的，将一个或两个int更改为double将使计算成为双除法，而不是整数除法。|223562|entityMap^0|1F|3|0|F|3|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|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

As always, any calculation on the right side of the assignment operator is executed before assigning its result to the variable specified on the left side. When the calculation is performed, it is a division between two <code>int</code>s, which is an integer division, which (as I'm sure you know), produces an integer result (i.e. a quotient without any remainder). Only after this calculation is performed is the result of the integer division - which is 0 - cast to a double. 

As you observed, changing one or both of the <code>int</code>s to a double makes the calculation a double division, rather than an integer division.

blocks|key|223567|text|A是int，b是int，除以a/b会得到一个int值，因此0.35转换为int就变成了0。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|223568|double+d+=+(double)a/b;|code-block|syntax|javascript|223569|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|$]]

a is int, and b is int, dividing a/b results an int value so 0.35 converted to int becomes 0.So you have to cast it like

<pre><code>double d = (double)a/b;
</code></pre>

blocks|key|1079843|text|因为你现在的做法是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1079844|你对a和b做整数除法，结果是整数0。之后，将结果赋值给double。|1079845|您只需将a或b转换为double，Java就会识别这一点，并为您提供double。|1079846|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Because the way you are doing it now is like this:

You do an integerdivision on a and b. The result of that is the integer 0. After that you assign the result to a double.

You could just cast either a or b to a double and Java will recognize that and give you a double as a result.

I'm having a weird problem.

This code simply divides an int by another int, stores result in a double variable and prints it:

<pre><code>int a = 200;
int b = 557;

double divisionResult = a / b;

System.out.println("Result: " + divisionResult);
</code></pre>

After executing this code, the output is:

<pre class="lang-none prettyprint-override"><code>Result: 0
</code></pre>

This is weird, because <code>200/557</code> is <code>0.3590664272890485</code>

I noticed that if i cast <code>a</code> and <code>b</code> to <code>double</code> in the division line

<pre><code>double divisionResult = (double) a / (double) b;
</code></pre>

It works perfectly.

Why do i have to cast my variables to double to get the real division result?

Unexpected result while dividing int by int and storing result into double

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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我有个奇怪的问题。这段代码简单地将一个int除以另一个int，将结果存储在一个双精度变量中并打印出来：int a = 200;int b = 557;double divisionResult = a / b;System.out.println("Result: " + divisionResult);执行此代码后，...

问将int除以int并将结果存储为双精度时出现意外结果
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问将int除以int并将结果存储为双精度时出现意外结果EN