blocks|key|562543|text|devtop+%2B=+Math.pow(x[i]+-+mean,+2);会将操作Math.pow(x[i]+-+mean,+2)的结果添加到devtop变量中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|562544|一个更简单的例子：|562545|int+devtop+=+2;
devtop+%2B=+3;+//+devtop+now+equals+5|code-block|syntax|javascript|562546|entityMap^0|0|Z|13|O|1X|6|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]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

<code>devtop += Math.pow(x[i] - mean, 2);</code> will add the result of the operation <code>Math.pow(x[i] - mean, 2)</code> to the <code>devtop</code> variable.

A more simple example:

<pre><code>int devtop = 2;
devtop += 3; // devtop now equals 5
</code></pre>

blocks|key|562591|text|devtop+%2B=+Math.pow(x[i]+-+mean,+2);将Math.pow(x[i]+-+mean,+2)添加到devtop。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|562592|entityMap^0|0|Z|10|O|1R|6|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>devtop += Math.pow(x[i] - mean, 2);</code> adds <code>Math.pow(x[i] - mean, 2)</code> to <code>devtop</code>.

blocks|key|779119|text|这是assignment+operators中的一个。它将x的值加上0.1，然后将(x+%2B+0.1)的结果存储回x中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|779120|所以：|779121|double+x+=+1.3;
x+%2B=+0.1;++++//+sets+'x'+to+1.4|code-block|syntax|javascript|779122|它在功能上与以下内容相同，但比：|779123|double+x+=+1.3;
x+=+x+%2B+0.1;|779124|注意:在进行浮点数学运算时，使用don't+always+work+the+way+you+think+they+will。|779125|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/tutorial/java/nutsandbolts/operators.html|1|http://support.microsoft.com/kb/42980^0|T|1|1K|1|2|K|0|0|0|0|0|0|G|19|1|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|F|3|G|5|6|7|1A|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1C|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1D|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1E|8|@]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|S|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]|10|$5|V|W|X|E|$Y|11]]]]

It's one of the <a href="http://download.oracle.com/javase/tutorial/java/nutsandbolts/operators.html" rel="noreferrer">assignment operators</a>. It takes the value of <code>x</code>, adds 0.1 to it, and then stores the result of (x + 0.1) back into <code>x</code>.

So:

<pre><code>double x = 1.3;
x += 0.1; // sets 'x' to 1.4
</code></pre>

It's functionally identical to, but shorter than:

<pre><code>double x = 1.3;
x = x + 0.1;
</code></pre>

NOTE: When doing floating-point math, things <a href="http://support.microsoft.com/kb/42980" rel="noreferrer">don't always work the way you think they will</a>.

blocks|key|779218|text|在c#中，像2或-2(没有小数部分)这样的数字的默认类型是int，而与c#不同的是，这不是一个对象，我们不能像java那样做像2.tostring这样的事情，而像2.5(有小数部分)这样的数字的默认类型是double；所以如果你写：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|779219|short+s+=+2;
s+=+s+%2B+4;|code-block|syntax|javascript|779220|如果你像下面这样做，你会得到一个编译错误，int也不能转换成短整型：|779221|float+f+=+4.6;
f+=+f+%2B+4.3;|779222|在两行中设置双精度'4.6‘为浮点型变量会出现两个编译错误，第一行的错误是合乎逻辑的，因为浮点型和双精度使用不同的数字存储系统，使用一个而不是另一个会导致数据丢失；上面提到的两个示例可以这样更改：|779223|s+%2B=+4
f+%2B=+4.3|779224|它们在代码后面都有一个隐式的类型转换，没有编译错误；另一个值得考虑的问题是'+byte‘数据类型范围内的数字被缓存在java中，因此数字-128到127在java中是byte类型的，所以这段代码没有任何编译错误：|779225|byte+b+=+127|779226|但这个确实有一个错误：|779227|byte+b+=+128|779228|因为128在java中是一个整数；对于长数字，我们建议在数字后面使用一个L来处理整数溢出问题，如下所示：|779229|long+l+=+2134324235234235L|779230|在java中，我们没有像c%2B%2B那样的操作符重载，但是%2B=只对String重载，而不是对StringBuilder或StringBuffer重载，我们可以使用它来代替String+'concat‘方法，但我们知道String是不可变的，这将产生另一个对象，而不会像以前那样改变同一个对象：|779231|String+str+=+"Hello";
str+%2B=+"World";|779232|这很好；|779233|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1A|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1C|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1G|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1K|8|@]|9|@]|A|$]]|$1|12|3|13|5|D|7|1L|8|@]|9|@]|A|$E|F]]|$1|14|3|15|5|6|7|1M|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|17|$]]

In java the default type of numbers like 2 or -2(without a fractional component) is int and unlike c# that's not an object and we can't do sth like 2.tostring as in c# and the default type of numbers like 2.5(with a fractional component) is double;
So if you write:

<pre><code>short s = 2;
s = s + 4;
</code></pre>

you will get a compilation error that int cannot be cast into short also if you do sth like below:

<pre><code>float f = 4.6;
f = f + 4.3;
</code></pre>

you will get two compilation errors for setting double '4.6' to a float variable at both lines and the error of first line is logical because float and double use different system of storing numbers and using one instead of another can cause data loss;
two examples mentioned can be changed like this:

<pre><code>s += 4
f += 4.3
</code></pre>

which both have an implicit cast behind code and have no compile errors;
Another point worthy of consideration is numbers in the range of 'byte' data type are cached in java and thus numbers -128 to 127 are of type byte in java and so this code doesn't have any compile errors:

<pre><code>byte b = 127
</code></pre>

but this one has an error indeed:

<pre><code>byte b = 128
</code></pre>

because 128 is an int in java;
about long numbers we are recommended to use an L after the number for the matter of integer overflow like this:

<pre><code>long l = 2134324235234235L
</code></pre>

in java we don't have operator overloading like c++ but += is overloaded only for String and not for the let's say StringBuilder or StringBuffer and we can use it instead of String 'concat' method but as we know String is immutable and that will make another object and will not change the same object as before : 

<pre><code>String str = "Hello";
str += "World";
</code></pre>

It's fine;

blocks|key|562770|text|您可以随时查看字节码，以了解java操作符是如何工作的。下面是你编译的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|562771|int+x+=+0;
x+%2B=+0.1;|code-block|syntax|javascript|562772|字节码可以通过jdk命令javap+-c+[*.class]访问：(有关字节码的更多解释，请参考Java+bytecode+instruction+listings+)|offset|length|style|CODE|562773|0:+iconst_0+//++load+the+int+value+0+onto+the+stack
1:+istore_1+//++store+int+value+into+variable+1+(x)
2:+iload_1+//+load+an+int+value+from+local+variable+1+(x)
3:+i2d+//+convert+an+int+into+a+double+(cast+x+to+double)
4:+ldc2_w++++++++#2++++++++++++++++++//+double+0.1d+->+push+a+constant+value+(0.1)+from+a+constant+pool+onto+the+stack
7:+dadd+//++add+two+doubles+(pops+two+doubles+from+stack,+adds+them,+and+pushes+the+answer+onto+stack)
8:+d2i+//+convert+a+double+to+an+int+(pops+a+value+from+stack,+casts+it+to+int+and+pushes+it+onto+stack)
9:+istore_1+//+store+int+value+into+variable+1+(x)|562774|现在很明显，java编译器将x提升为double，然后用0.1将其添加。|562775|最后，它给出了integer的答案。|562776|当你写的时候，我发现了一个有趣的事实：|562777|byte+b+=+10;
b+%2B=+0.1;|562778|编译器将b转换为double，然后将其与0.1相加，将结果double转换为integer，最后将其转换为byte，这是因为没有直接将double转换为byte的指令。|562779|如果您有疑问，可以检查字节码:)|562780|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Java_bytecode_instruction_listings^0|0|0|C|I|1C|Y|0|0|0|E|1|S|3|0|7|7|0|0|0|4|1|K|3|T|6|12|7|1H|4|1V|6|24|4|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1A|8|@$I|1B|J|1C|K|L]]|9|@$I|1D|J|1E|1|1F]]|A|$]]|$1|M|3|N|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1H|8|@$I|1I|J|1J|K|L]|$I|1K|J|1L|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1M|8|@$I|1N|J|1O|K|L]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1P|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1Q|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1R|8|@$I|1S|J|1T|K|L]|$I|1U|J|1V|K|L]|$I|1W|J|1X|K|L]|$I|1Y|J|1Z|K|L]|$I|20|J|21|K|L]|$I|22|J|23|K|L]|$I|24|J|25|K|L]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|26|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|27|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]]]

You can take a look at the bytecode whenever you want to understand how java operators work. Here if you compile:

<pre><code>int x = 0;
x += 0.1;
</code></pre>

the bytecode will be accessible with jdk command <code>javap -c [*.class]</code>:(you can refer to <a href="https://en.wikipedia.org/wiki/Java_bytecode_instruction_listings" rel="nofollow noreferrer">Java bytecode instruction listings</a> for more explanation about bytecode)

<pre><code>0: iconst_0 // load the int value 0 onto the stack
1: istore_1 // store int value into variable 1 (x)
2: iload_1 // load an int value from local variable 1 (x)
3: i2d // convert an int into a double (cast x to double)
4: ldc2_w #2 // double 0.1d -&gt; push a constant value (0.1) from a constant pool onto the stack
7: dadd // add two doubles (pops two doubles from stack, adds them, and pushes the answer onto stack)
8: d2i // convert a double to an int (pops a value from stack, casts it to int and pushes it onto stack)
9: istore_1 // store int value into variable 1 (x)
</code></pre>

Now it is clear that java compiler promotes <code>x</code> to double and then adds it with <code>0.1</code>. 
Finally it casts the answer to <code>integer</code> . 
There is one interesting fact I found out that when you write:

<pre><code>byte b = 10;
b += 0.1;
</code></pre>

compiler casts <code>b</code> to double, adds it with <code>0.1</code>, casts the result which is <code>double</code> to <code>integer</code>, and finally casts it to <code>byte</code> and that is because there is no instruction to cast <code>double</code> to <code>byte</code> directly. 
You can check the bytecode if you doubt :)

blocks|key|779284|text|它将变量的值加以%2B=之后的值。例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|779285|float+x+=+0;
x+%2B=+0.1;
//x+is+now+0.1
x+%2B=+0.1;
//x+is+now+0.2|code-block|syntax|javascript|779286|它只是一个更短的版本：|779287|x+=+x%2B0.1;|779288|entityMap^0|8|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

It increases the value of the variable by the the value after <code>+=</code>.
For example:

<pre><code>float x = 0;
x += 0.1;
//x is now 0.1
x += 0.1;
//x is now 0.2
</code></pre>

It's just a shorter version of:

<pre><code>x = x+0.1;
</code></pre>

Can you please help me understand what the following code means:

<pre><code>x += 0.1;
</code></pre>

What does the "+=" operator do in Java?

能否请您帮助我理解以下代码的含义：x += 0.1;

问"+=“操作符在Java语言中是做什么的？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问"+=“操作符在Java语言中是做什么的？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问"+=“操作符在Java语言中是做什么的？
EN