blocks|key|1313817|text|这有帮助吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1313818|a+=+5;
i=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;+=>
i=6+%2B+7+%2B+7;+(a=8)

a+=+5;
i=a%2B%2B+%2B+%2B%2Ba+%2B+%2B%2Ba;+=>
i=5+%2B+7+%2B+8;+(a=8)|code-block|syntax|javascript|1313819|要点是%2B%2Ba递增该值并立即返回它。|offset|length|style|CODE|1313820|a%2B%2B还会递增该值(在后台)，但返回未更改的变量值-它看起来像是稍后执行的值。|1313821|entityMap^0|0|0|3|3|0|0|3|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@$I|W|J|X|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

Does this help?

<pre><code>a = 5;
i=++a + ++a + a++; =&gt;
i=6 + 7 + 7; (a=8)

a = 5;
i=a++ + ++a + ++a; =&gt;
i=5 + 7 + 8; (a=8)
</code></pre>

The main point is that <code>++a</code> increments the value and immediately returns it. 

<code>a++</code> also increments the value (in the background) but returns unchanged value of the variable - what looks like it is executed later.

blocks|key|1314095|text|%2B%2Ba递增，然后使用变量。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1314096|a%2B%2B使用变量，然后递增该变量。|1314097|如果你有|1314098|a+=+1;|code-block|syntax|javascript|1314099|而你确实做到了|1314100|System.out.println(a%2B%2B);+//You+will+see+1

//Now+a+is+2

System.out.println(%2B%2Ba);+//You+will+see+3|1314101|codaddict+explains您的特定代码片段。|1314102|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/2371148/866022^0|0|3|0|0|3|0|0|0|0|0|0|I|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|18|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|19|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1A|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1B|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1C|8|@]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|U|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

<code>++a</code> increments and then uses the variable. 
<code>a++</code> uses and then increments the variable.

If you have

<pre><code>a = 1;
</code></pre>

and you do

<pre><code>System.out.println(a++); //You will see 1

//Now a is 2

System.out.println(++a); //You will see 3
</code></pre>

<a href="https://stackoverflow.com/a/2371148/866022">codaddict explains</a> your particular snippet.

blocks|key|1313699|text|在对a求值之前，%2B%2Ba会递增它。a%2B%2B对a求值，然后递增它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1313700|与您给定的表达式相关：|1313701|i+=+((%2B%2Ba)+%2B+(%2B%2Ba)+%2B+(a%2B%2B))+==+((6)+%2B+(7)+%2B+(7));+//+a+is+8+at+the+end
i+=+((a%2B%2B)+%2B+(%2B%2Ba)+%2B+(%2B%2Ba))+==+((5)+%2B+(7)+%2B+(8));+//+a+is+8+at+the+end|code-block|syntax|javascript|1313702|我上面使用的父类是由Java隐式使用的。如果你这样看这些术语，你可以很容易地看到，它们都是相同的，因为它们是可交换的。|1313703|entityMap^0|2|1|8|3|G|3|K|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]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

<code>++a</code> increments <code>a</code> before it is evaluated.
<code>a++</code> evaluates <code>a</code> and then increments it.

Related to your expression given:

<pre><code>i = ((++a) + (++a) + (a++)) == ((6) + (7) + (7)); // a is 8 at the end
i = ((a++) + (++a) + (++a)) == ((5) + (7) + (8)); // a is 8 at the end
</code></pre>

The parenteses I used above are implicitly used by Java. If you look at the terms this way you can easily see, that they are both the same as they are commutative.

blocks|key|1099238|text|在上面的例子中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1099239|int+a+=+5,i;

i=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;++++++++//Ans:+i+=+6+%2B+7+%2B+7+=+20+then+a+=+8+

i=a%2B%2B+%2B+%2B%2Ba+%2B+%2B%2Ba;++++++++//Ans:+i+=+8+%2B+10+%2B+11+=+29+then+a+=+11

a=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;++++++++//Ans:+a+=+12+%2B+13+%2B+13+=+38

System.out.println(a);++++//Ans:+a+=+38

System.out.println(i);++++//Ans:+i+=+29|code-block|syntax|javascript|1099240|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 the above example 

<pre><code>int a = 5,i;

i=++a + ++a + a++; //Ans: i = 6 + 7 + 7 = 20 then a = 8 

i=a++ + ++a + ++a; //Ans: i = 8 + 10 + 11 = 29 then a = 11

a=++a + ++a + a++; //Ans: a = 12 + 13 + 13 = 38

System.out.println(a); //Ans: a = 38

System.out.println(i); //Ans: i = 29
</code></pre>

blocks|key|1314502|text|但是，我相信如果您将所有语句组合在一起，并在Java+8.1中运行它，您将得到一个不同的答案，至少我的经验是这样说的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314503|代码的工作方式如下：|1314504|int+a=5,i;

i=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;++++++++++++/*a+=+5;
++++++++++++++++++++++++++++++++i=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;+=>
++++++++++++++++++++++++++++++++i=6+%2B+7+%2B+7;+(a=8);+i=20;*/

i=a%2B%2B+%2B+%2B%2Ba+%2B+%2B%2Ba;+++++++++++/*a+=+5;
++++++++++++++++++++++++++++++++i=a%2B%2B+%2B+%2B%2Ba+%2B+%2B%2Ba;+=>
++++++++++++++++++++++++++++++++i=8+%2B+10+%2B+11;+(a=11);+i=29;*/

a=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;++++++++++++/*a=5;
++++++++++++++++++++++++++++++++a=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;+=>
++++++++++++++++++++++++++++++++a=12+%2B+13+%2B+13;++a=38;*/

System.out.println(a);++++++++//output:+38
System.out.println(i);+++++++++//output:+29|code-block|syntax|javascript|1314505|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I believe however if you combine all of your statements and run it in Java 8.1 you will get a different answer, at least that's what my experience says.

The code will work like this:

<pre><code>int a=5,i;

i=++a + ++a + a++; /*a = 5;
 i=++a + ++a + a++; =&gt;
 i=6 + 7 + 7; (a=8); i=20;*/

i=a++ + ++a + ++a; /*a = 5;
 i=a++ + ++a + ++a; =&gt;
 i=8 + 10 + 11; (a=11); i=29;*/

a=++a + ++a + a++; /*a=5;
 a=++a + ++a + a++; =&gt;
 a=12 + 13 + 13; a=38;*/

System.out.println(a); //output: 38
System.out.println(i); //output: 29
</code></pre>

blocks|key|1314603|text|假设你的意思是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314604|int+a=5;+int+i;

i=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;

System.out.println(i);

a=5;

i=a%2B%2B+%2B+%2B%2Ba+%2B+%2B%2Ba;

System.out.println(i);

a=5;

a=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;

System.out.println(a);|code-block|syntax|javascript|1314605|其计算结果为：|1314606|i+=+(6,+a+is+now+6)+%2B+(7,+a+is+now+7)+%2B+(7,+a+is+now+8)|1314607|所以i是6%2B7%2B7=+20，所以打印了20。|1314608|i+=+(5,+a+is+now+6)+%2B+(7,+a+is+now+7)+%2B+(8,+a+is+now+8)|1314609|所以i是5%2B7%2B8=+20，所以20被再次打印出来。|1314610|a+=+(6,+a+is+now+6)+%2B+(7,+a+is+now+7)+%2B+(7,+a+is+now+8)|1314611|在评估完所有右侧之后(包括将a设置为8)，然后将a设置为6%2B7%2B7=+20，并最后打印20。|1314612|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Presuming that you meant

<pre><code>int a=5; int i;

i=++a + ++a + a++;

System.out.println(i);

a=5;

i=a++ + ++a + ++a;

System.out.println(i);

a=5;

a=++a + ++a + a++;

System.out.println(a);
</code></pre>

This evaluates to:

<pre><code>i = (6, a is now 6) + (7, a is now 7) + (7, a is now 8)
</code></pre>

so i is 6 + 7 + 7 = 20 and so 20 is printed.

<pre><code>i = (5, a is now 6) + (7, a is now 7) + (8, a is now 8)
</code></pre>

so i is 5 + 7 + 8 = 20 and so 20 is printed again.

<pre><code>a = (6, a is now 6) + (7, a is now 7) + (7, a is now 8)
</code></pre>

and after all of the right hand side is evaluated (including setting a to 8) THEN a is set to 6 + 7 + 7 = 20 and so 20 is printed a final time.

blocks|key|1313963|text|当a为5时，a%2B%2B为表达式提供5，之后递增a，而%2B%2Ba在将数字传递给表达式之前递增a+(这在本例中为表达式提供a+6)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1313964|所以你计算|1313965|i+=+6+%2B+7+%2B+7
i+=+5+%2B+7+%2B+8|code-block|syntax|javascript|1313966|entityMap^0|1|1|6|3|L|1|O|3|15|1|1J|1|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]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|N|$]]

when <code>a</code> is 5, then <code>a++</code> gives a 5 to the expression and increments <code>a</code> afterwards, while <code>++a</code> increments <code>a</code> before passing the number to the expression (which gives <code>a</code> 6 to the expression in this case).

So you calculate

<pre><code>i = 6 + 7 + 7
i = 5 + 7 + 8
</code></pre>

blocks|key|1314707|text|前增量意味着变量在表达式中求值之前被递增。递增后表示变量在计算后递增，以便在表达式中使用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314708|因此，仔细看一下，你会发现所有三个赋值在算术上是等价的。|1314709|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Pre-increment means that the variable is incremented BEFORE it's evaluated in the expression. Post-increment means that the variable is incremented AFTER it has been evaluated for use in the expression.

Therefore, look carefully and you'll see that all three assignments are arithmetically equivalent.

blocks|key|1314380|text|如果不在表达式中，则前增量和后增量是等效的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314381|int+j+=0;
int+r=0+++++++++
for(int+v+=+0;+v<10;+%2B%2Bv)+{+
++++++++++%2B%2Br;
++++++++++j%2B%2B;
++++++++++System.out.println(j%2B"+"%2Br);
++}++
+1+1++
+2+2++
+3+3+++++++
+4+4
+5+5
+6+6
+7+7
+8+8
+9+9
10+10|code-block|syntax|javascript|1314382|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|$]]

pre-increment and post increment are equivalent if not in an expression 

<pre><code>int j =0;
int r=0 
for(int v = 0; v&lt;10; ++v) { 
 ++r;
 j++;
 System.out.println(j+" "+r);
 } 
 1 1 
 2 2 
 3 3 
 4 4
 5 5
 6 6
 7 7
 8 8
 9 9
10 10
</code></pre>

blocks|key|1314338|text|a=5;+i=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1314339|是|unstyled|1314340|i+=+7+%2B+6+%2B+7|1314341|工作:+pre/post增量具有“从右到左”的关联性，并且pre优先于post，因此首先pre增量将被解析为(%2B%2Ba+%2B+%2B%2Ba)+=>+7+%2B+6。然后提供a=7来发布增量=>、7+%2B+6+%2B+7+=20和a+=8。|offset|length|style|CODE|1314342|a=5;+i=a%2B%2B+%2B+%2B%2Ba+%2B+%2B%2Ba;|1314343|1314344|i=7+%2B+7+%2B+6|1314345|工作:+pre/post增量具有“从右到左”的关联性，并且pre优先于post，因此首先将解决pre增量，因为为post增量=>+7+%2B+7+%2B+6+=20和a+=8提供了(%2B%2Ba+%2B+%2B%2Ba)+=>+7+%2B+6.then+a=7。|1314346|entityMap^0|0|0|0|1I|K|27|3|2I|D|2W|4|0|0|0|0|1T|D|27|4|2E|K|34|3|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Z|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|10|8|@$K|11|L|12|M|N]|$K|13|L|14|M|N]|$K|15|L|16|M|N]|$K|17|L|18|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|19|8|@]|9|@]|A|$B|C]]|$1|Q|3|E|5|F|7|1A|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1B|8|@]|9|@]|A|$B|C]]|$1|T|3|U|5|F|7|1C|8|@$K|1D|L|1E|M|N]|$K|1F|L|1G|M|N]|$K|1H|L|1I|M|N]|$K|1J|L|1K|M|N]]|9|@]|A|$]]|$1|V|3|-4|5|F|7|1L|8|@]|9|@]|A|$]]]|W|$]]

<pre><code>a=5; i=++a + ++a + a++;
</code></pre>

is

<pre><code>i = 7 + 6 + 7
</code></pre>

Working: pre/post increment has "right to left" Associativity , and pre has precedence over post , so first of all pre increment will be solve as <code>(++a + ++a) =&gt; 7 + 6</code> . then <code>a=7</code> is provided to post increment => <code>7 + 6 + 7 =20</code> and <code>a =8</code>.

<pre><code>a=5; i=a++ + ++a + ++a;
</code></pre>

is

<pre><code>i=7 + 7 + 6
</code></pre>

Working: pre/post increment has "right to left" Associativity , and pre has precedence over post , so first of all pre increment will be solve as <code>(++a + ++a) =&gt; 7 + 6</code>.then <code>a=7</code> is provided to post increment => <code>7 + 7 + 6 =20</code> and <code>a =8</code>.

blocks|key|1314628|text|我相信您正在以不同的方式执行所有这些语句|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314629|一起执行将产生=>+38、29|1314630|int+a=5,i;
i=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;
//this+means+i=+6%2B7%2B7=20+and+when+this+result+is+stored+in+i,
//then+last+*a*+will+be+incremented+ 
i=a%2B%2B+%2B+%2B%2Ba+%2B+%2B%2Ba;
//this+means+i=+5%2B7%2B8=20+(this+could+be+complicated,+
//but+its+working+like+this), 
a=%2B%2Ba+%2B+%2B%2Ba+%2B+a%2B%2B;
//as+a+is+6%2B7%2B7=20+(this+is+incremented+like+this)|code-block|syntax|javascript|1314631|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I believe you are executing all these statements differently 
 
executing together will result => 38 ,29 

<pre><code>int a=5,i;
i=++a + ++a + a++;
//this means i= 6+7+7=20 and when this result is stored in i,
//then last *a* will be incremented &lt;br&gt;
i=a++ + ++a + ++a;
//this means i= 5+7+8=20 (this could be complicated, 
//but its working like this),&lt;br&gt;
a=++a + ++a + a++;
//as a is 6+7+7=20 (this is incremented like this)
</code></pre>

Can you explain to me the output of this Java code?

<pre><code>int a=5,i;

i=++a + ++a + a++;
i=a++ + ++a + ++a;
a=++a + ++a + a++;

System.out.println(a);
System.out.println(i);
</code></pre>

The output is 20 in both cases

How do the post increment (i++) and pre increment (++i) operators work in Java?

Java

你能给我解释一下这段Java代码的输出吗？int a=5,i;i=++a + ++a + a++;i=a++ + ++a + ++a;a=++a + ++a + a++;System.out.println(a);System.out.println(i);两种情况下的输出都是20

问在Java语言中，后增量(i++)和前增量(++i)运算符是如何工作的？
EN

回答 11

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问在Java语言中，后增量(i++)和前增量(++i)运算符是如何工作的？
EN