blocks|key|74119|text|第二行中t1的值为5|type|unstyled|depth|inlineStyleRanges|entityRanges|data|74120|t2+=+t1+%2B+(%2B%2Bt1)
t2+=+5+%2B+6;++++++//+t1+becomes+6+here|code-block|syntax|javascript|74121|The+evaluation+order+is+from+left+to+right.|offset|length|74122|因此，首先将t1求值为5，然后将%2B%2Bt1求值为6，因此结果为11|style|CODE|74123|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.7^0|0|0|4|13|0|0|B|1|G|4|N|1|U|2|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]]|$1|G|3|H|5|6|7|Y|8|@]|9|@$I|Z|J|10|1|11]]|A|$]]|$1|K|3|L|5|6|7|12|8|@$I|13|J|14|M|N]|$I|15|J|16|M|N]|$I|17|J|18|M|N]|$I|19|J|1A|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

The value of t1 in the second line is 5

<pre><code>t2 = t1 + (++t1)
t2 = 5 + 6; // t1 becomes 6 here
</code></pre>

The <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.7" rel="nofollow">evaluation order is from left to right.</a>

So first t1 is evaluated to <code>5</code> then <code>++t1</code> to <code>6</code> and hence the result as <code>11</code>

blocks|key|4580600|text|%2B的计算顺序是从左到右，因此|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4580601|t1+%2B+(%2B%2Bt1)+++++//+Left+side+is+evaluated+to+5,+right+side+evaluated+to+6...
5++%2B+(6)++++++++//+...and+as+a+side+effect+t1+becomes+6|code-block|syntax|javascript|4580602|结果为11。|4580603|entityMap^0|0|1|0|0|3|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

The <code>+</code> is evaluated from left to right, so

<pre><code>t1 + (++t1) // Left side is evaluated to 5, right side evaluated to 6...
5 + (6) // ...and as a side effect t1 becomes 6
</code></pre>

Results in <code>11</code>.

blocks|key|4580615|text|评估从左到右进行。所以实际上发生的事情如下所示|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4580616|t2+=+t1+%2B+(%2B%2Bt1);
t2+=+5+%2B+6;++++++
t2+=+11;|code-block|syntax|javascript|4580617|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|$]]

evaluation happen from left to right. So actually what happen is following

<pre><code>t2 = t1 + (++t1);
t2 = 5 + 6; 
t2 = 11;
</code></pre>

blocks|key|4580649|text|%2B%2Bx在您使用变量x之前执行，而x%2B%2B在您使用它之后增加x：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4580650|x=5;
y=x%2B%2B;
y+is+5;
and+x+is+6

x=5;
y=%2B%2Bx;
y+is+6;
and+x+is+6|code-block|syntax|javascript|4580651|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|$]]

++x is executed before you use the variable x, and x++ increase x after you used it:

<pre><code>x=5;
y=x++;
y is 5;
and x is 6

x=5;
y=++x;
y is 6;
and x is 6
</code></pre>

blocks|key|4580653|text|t2+=+t1+%2B+(%2B%2Bt1);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4580654|这相当于|unstyled|4580655|temp+=+t1+%2B+1
t2+=++t1+%2B+temp;
t1=+temp;|4580656|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>t2 = t1 + (++t1);
</code></pre>

This is equivalent to 

<pre><code>temp = t1 + 1
t2 = t1 + temp;
t1= temp;
</code></pre>

blocks|key|4580695|text|你几乎是正确的，但你微妙地误解了优先规则是如何工作的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4580696|比较这两种情况：|4580697|int+t1+=+5;
int+t2+=+t1+%2B+(%2B%2Bt1);
System.out.println+(t2);

t1+=+5;
t2+=+(%2B%2Bt1)+%2B+t1;
System.out.println+(t2);|code-block|syntax|javascript|4580698|结果是：|4580699|11
12|4580700|优先级确实要求先计算%2B%2B，然后再计算%2B，但直到它到达表达式的那一部分时才适用。|offset|length|style|CODE|4580701|表达式的格式为X+%2B+Y，其中X为t1，Y为(%2B%2Bt1)|4580702|首先计算左分支，即X。然后，计算右分支，即Y。只有在评估Y时，才会执行%2B%2B操作。|4580703|优先级规则只说明%2B%2B在Y表达式的“内部”，并没有说明操作的顺序。|4580704|entityMap^0|0|0|0|0|0|A|2|I|1|0|7|5|F|1|H|2|K|1|M|6|0|9|1|L|1|S|1|Z|2|0|8|2|B|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|15|8|@$O|16|P|17|Q|R]|$O|18|P|19|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1A|8|@$O|1B|P|1C|Q|R]|$O|1D|P|1E|Q|R]|$O|1F|P|1G|Q|R]|$O|1H|P|1I|Q|R]|$O|1J|P|1K|Q|R]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1L|8|@$O|1M|P|1N|Q|R]|$O|1O|P|1P|Q|R]|$O|1Q|P|1R|Q|R]|$O|1S|P|1T|Q|R]]|9|@]|A|$]]|$1|W|3|X|5|6|7|1U|8|@$O|1V|P|1W|Q|R]|$O|1X|P|1Y|Q|R]]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]]|Z|$]]

You are nearly correct but you are subtly misunderstanding how the precedence rules work.

Compare these two cases:

<pre><code>int t1 = 5;
int t2 = t1 + (++t1);
System.out.println (t2);

t1 = 5;
t2 = (++t1) + t1;
System.out.println (t2);
</code></pre>

The result is:

<pre><code>11
12
</code></pre>

The precedence does indeed say to evaluate the <code>++</code> before the <code>+</code>, but that doesn't apply until it reaches that part of the expression.

Your expression is of the form <code>X + Y</code>
Where <code>X</code> is <code>t1</code> and <code>Y</code> is <code>(++t1)</code>

The left branch, i.e. <code>X</code>, is evaluated first.
Afterwards the right branch, i.e. <code>Y</code>, is evaluated.
Only when it comes to evaluate <code>Y</code> the <code>++</code> operation is performed.

The precedence rules only say that the <code>++</code> is "inside" the <code>Y</code> expression, they don't say anything about the order of operations.

blocks|key|4580736|text|另一种想法是扩展%2B%2B表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4580737|%2B%2Bt1和the+t1+=+t1+%2B+1是一样的。|offset|length|style|CODE|4580738|1)+t1+=+5;
2)+t2+=+t1+%2B+(%2B%2Bt1);+
3)+t2+=+t1+%2B+(t1+=+t1+%2B+1),+
4)+t2+=+5+%2B+(t1+=+t1+%2B+1)
5)+t2+=+5+%2B+(t1+=+6)
6)+t2+=+5+%2B+6+=+11|code-block|syntax|javascript|4580739|如果将顺序颠倒为t2+=+(%2B%2Bt1)+%2B+t1;，则表达式将展开为：|4580740|1)+t2+=+(t1+=+t1+%2B+1)+%2B+t1+++++
2)+t2+=+(t1+=+5+%2B+1)+%2B+t1
3)+t2+=+(t1+=+6)+%2B+t1
4)+t2+=+6+%2B+6+=+12|4580741|entityMap^0|0|0|4|9|B|0|0|8|H|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|12|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

Another way of thinking of it is to expand the ++ expression:

<code>++t1</code> is the same as putting <code>t1 = t1 + 1</code>.

<pre><code>1) t1 = 5;
2) t2 = t1 + (++t1); 
3) t2 = t1 + (t1 = t1 + 1), 
4) t2 = 5 + (t1 = t1 + 1)
5) t2 = 5 + (t1 = 6)
6) t2 = 5 + 6 = 11
</code></pre>

If you were to reverse the order to <code>t2 = (++t1) + t1;</code> Then the expression would expand to:

<pre><code>1) t2 = (t1 = t1 + 1) + t1 
2) t2 = (t1 = 5 + 1) + t1
3) t2 = (t1 = 6) + t1
4) t2 = 6 + 6 = 12
</code></pre>

blocks|key|74281|text|enter+code+here
t1+=+5;
t2+=+t1+%2B+(%2B%2Bt1);
//+it+takes+5+for+t1+and+as+it+moves+further+to+(%2B%2Bt1),+it+increments+t1+to+6,+so+it+takes+6+for++(%2B%2Bt1)
t2+=+5+%2B+6;
//+(%2B%2Bt1)+this+increments+t1+by+1+then+return+new+value.+So+(%2B%2Bt1)=6
//+(t1%2B%2B)+this+returns+old+value+n+then+increments+t1+by+1.+So+(t1%2B%2B)=5|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|74282|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>enter code here
t1 = 5;
t2 = t1 + (++t1);
// it takes 5 for t1 and as it moves further to (++t1), it increments t1 to 6, so it takes 6 for (++t1)
t2 = 5 + 6;
// (++t1) this increments t1 by 1 then return new value. So (++t1)=6
// (t1++) this returns old value n then increments t1 by 1. So (t1++)=5
</code></pre>

I tried the following code in Java

<pre><code>t1 = 5;
t2 = t1 + (++t1);
System.out.println (t2);
</code></pre>

My view is since ++ has a higher precedence than +, the above becomes

<pre><code>t2 = t1 + (++t1);
t2 = t1 + 6; // t1 becomes 6 here
t2 = 6 + 6;
t2 = 12;
</code></pre>

However, I get the answer 11 for t2. Can someone explain?

In Java, which gets executed first, "+" or "++"?

Java

我在Java中尝试了以下代码t1 = 5;t2 = t1 + (++t1);System.out.println (t2);我的观点是，由于++具有比+更高的优先级，所以上面变成了t2 = t1 + (++t1);t2 = t1 + 6;      // t1 becomes 6 heret2 = 6 + 6;t2 =...

问在java语言中，"+“和"++”哪个先被执行？
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问在java语言中，"+“和"++”哪个先被执行？EN

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问在java语言中，"+“和"++”哪个先被执行？
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