blocks|key|2036021|text|依赖于它的优先顺序|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2036022|当两个操作符共享一个操作数时，优先级较高的操作符优先。例如，1%2B2*3被视为1%2B+(2+*+3)，而1*2%2B3被视为(1+*+2)+%2B3，因为乘法比加法+(%2B)具有更高的优先级。|blockquote|style|BOLD|2036023|entityMap|0|LINK|mutability|MUTABLE|url|http://introcs.cs.princeton.edu/java/11precedence/^0|3|6|0|0|1W|8|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@$A|V|B|W|G|H]]|9|@]|C|$]]|$1|I|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Depends on <a href="http://introcs.cs.princeton.edu/java/11precedence/" rel="nofollow">It's precedence order</a>

<blockquote>
 When two operators share an operand the operator with the higher precedence goes first. For example, 1 + 2 * 3 is treated as 1 + (2 * 3), whereas 1 * 2 + 3 is treated as (1 * 2) + 3 since multiplication has a higher precedence than addition (+).
</blockquote>

blocks|key|2035973|text|如果您想在System.out.println中执行任何数学操作，请用大括号包装它，因为Java在第一位看到字符串。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2035974|试试System.out.println+("number"+%2B+(6+%2B+4+*+5))。|2035975|第二个示例使用：System.out.print(s+%2B+(a+%2B+b));|2035976|在本例中，您有a和b之和。|2035977|但在System.out.print(b+%2B+a+%2B+s);，b和a则是第一名。编译器第一次执行a%2Bb，在添加字符串之后，不需要大括号|2035978|entityMap^0|5|I|0|2|17|0|8|U|0|7|1|9|1|0|2|S|V|1|X|1|1C|3|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|O|$]]

If you want to do any Math into <code>System.out.println</code>, wrap it with braces, because Java sees String at 1st place.

Try <code>System.out.println ("number" + (6 + 4 * 5))</code>.

For 2nd example use: <code>System.out.print(s + (a + b));</code>

in this case you have sum of <code>a</code> and <code>b</code>. 

but in <code>System.out.print(b + a + s);</code> <code>b</code> and <code>a</code> stay at the 1st place. Compiler does <code>a+b</code> 1st and after add String, you don't need braces

blocks|key|55697|text|"*“具有比"%2B”更高的运算符优先级，这意味着表达式"4+*+5“是在字符串连接发生之前计算的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|55698|请参阅http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html|offset|length|55699|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html^0|0|3|1Z|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

"*" has a higher operator precedence than "+", this means the expression "4 * 5" is calculated before the String concatenation happens.

See <a href="http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html" rel="nofollow">http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html</a>

blocks|key|2036032|text|它与操作符优先级有关：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2036033|第一，4*5=+20|2036034|第二，将"number“与6连接起来，然后将其与20连接起来。|2036035|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|$]]

It has something to do with the operator precedence:

First, 4 * 5 = 20

Second, "number" is concatenated with 6, which is further concatenated with 20.

blocks|key|55706|text|在第一个例子中，由于使用了乘法运算符，在连接到其他字符串之前，4被乘以5。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|55707|对于第二个示例，您从字符串之前的2个整数开始，该字符串将在连接到字符串之前先计算。|55708|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|$]]

On your first example, since you used a multiplicative operator, 4 is being multiplied to 5 before concatenated to other string. 
For the second example, you started with 2 integer before the String which will be calculated first before concatenated to a String.

blocks|key|55720|text|为什么System.out.println+("number“%2B6%2B4*+5)会导致number620|type|blockquote|depth|inlineStyleRanges|entityRanges|data|55721|因为，*的优先级高于%2B，所以结果是number620。|unstyled|offset|length|style|CODE|55722|字符串s=+"crunch"；int+a=+3，b=+1；System.out.print(s+%2Ba%2B+b)；System.out.println(b+%2Ba%2B+s)；|55723|在这里，%2B用作操作符重载，而不是作为二进制操作。所以，'%2B‘做连接操作，而不是求和操作。因此，结果是crunch314crunch。|55724|entityMap^0|0|H|9|0|0|1E|F|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@$E|Q|F|R|G|H]]|9|@]|A|$]]|$1|I|3|J|5|6|7|S|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|T|8|@$E|U|F|V|G|H]]|9|@]|A|$]]|$1|M|3|-4|5|D|7|W|8|@]|9|@]|A|$]]]|N|$]]

<blockquote>
 Why does System.out.println ("number" + 6 + 4 * 5) result in number620
</blockquote>

Because, * has higher precedence than +, so the result is <code>number620</code>.

<blockquote>
 String s = "crunch"; int a = 3, b = 1; System.out.print(s + a + b);
 System.out.println(b + a + s); result in crunch314crunch?
</blockquote>

Here, + is used as operator overloading not as binary operation. So, '+' do concat operation, not sum operation. So, the result is <code>crunch314crunch</code>.

blocks|key|959842|text|有两件事：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|959843|运算符优先|unordered-list-item|959844|字符串级联与加法|959845|%2B有以下规则：|offset|length|style|CODE|959846|int+%2B+int+=>+int
int+%2B+string+=>+String+
String+%2B+int+=>+String+
String+%2B+String+=>+String|code-block|syntax|javascript|959847|也就是说，一旦涉及到一个String，%2B就意味着连接。|959848|优先级相同的运算符从左到右计算。因此|959849|String+%2B+int+%2B+int+=>+String+%2B+int+=>+String|959850|但|959851|int+%2B+int+%2B+String+=>+int+%2B+String+=>+String|959852|第一种情况只使用串联，而第二种情况在第一步中使用加法。|959853|在第一个示例中，*具有比%2B更高的优先级，因此首先执行乘法。|959854|String+%2B+int+*+int+=>+String+%2B+int+=>+String|959855|entityMap^0|0|0|0|0|1|0|0|C|6|J|1|0|0|0|0|0|0|8|1|C|1|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1A|8|@]|9|@]|A|$]]|$1|E|3|F|5|D|7|1B|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|1C|8|@$I|1D|J|1E|K|L]]|9|@]|A|$]]|$1|M|3|N|5|O|7|1F|8|@]|9|@]|A|$P|Q]]|$1|R|3|S|5|6|7|1G|8|@$I|1H|J|1I|K|L]|$I|1J|J|1K|K|L]]|9|@]|A|$]]|$1|T|3|U|5|6|7|1L|8|@]|9|@]|A|$]]|$1|V|3|W|5|O|7|1M|8|@]|9|@]|A|$P|Q]]|$1|X|3|Y|5|6|7|1N|8|@]|9|@]|A|$]]|$1|Z|3|10|5|O|7|1O|8|@]|9|@]|A|$P|Q]]|$1|11|3|12|5|6|7|1P|8|@]|9|@]|A|$]]|$1|13|3|14|5|6|7|1Q|8|@$I|1R|J|1S|K|L]|$I|1T|J|1U|K|L]]|9|@]|A|$]]|$1|15|3|16|5|O|7|1V|8|@]|9|@]|A|$P|Q]]|$1|17|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]]|18|$]]

It's about two things:

<ul>
<li>operator precedence</li>
<li>string concatenation vs addition</li>
</ul>

The <code>+</code> has the following rules:

<pre><code>int + int =&gt; int
int + string =&gt; String 
String + int =&gt; String 
String + String =&gt; String
</code></pre>

That is, as soon as a <code>String</code> is involved, <code>+</code> means concatenation.

Operators with the same precedence are evaluated left to right. Therefore

<pre><code>String + int + int =&gt; String + int =&gt; String
</code></pre>

but

<pre><code>int + int + String =&gt; int + String =&gt; String
</code></pre>

The first case uses concatenation only, whereas the second uses addition in the first step.

<hr>

In your first example, <code>*</code> has higher precedence than <code>+</code>, so the multiplication is performed first.

<pre><code>String + int * int =&gt; String + int =&gt; String
</code></pre>

blocks|key|55953|text|这都是关于算子优先和它们的结合性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|55954|您的第一个示例："number“%2B6%2B4*+5|offset|length|style|BOLD|55955|Acc.首先计算运算符优先级*，使其成为"number“%2B6%2B+20|CODE|55956|现在，%2B的结合性是左->右(L->R)，所以%2B成为一个级联运算符，因为它与String一起使用，所以表达式变成"number6“%2B+20，然后是"number620"。|55957|(实际上，int在级联之前被转换为String+)|55958|同样，第二个例子是：|55959|只有%2B运算符并从L->R开始执行|55960|“紧缩”%2B3%2B+1+=+"crunch3“%2B1=+"crunch31”|55961|1+%2B3%2B“紧缩”=4%2B“紧缩”=“4+crunch”|55962|entityMap^0|0|8|F|0|E|1|K|E|0|3|1|11|6|1J|D|20|B|0|5|3|H|6|0|0|2|1|0|0|9|0|0|9|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|12|8|@$D|13|E|14|F|J]|$D|15|E|16|F|G]]|9|@]|A|$]]|$1|K|3|L|5|6|7|17|8|@$D|18|E|19|F|J]|$D|1A|E|1B|F|J]|$D|1C|E|1D|F|G]|$D|1E|E|1F|F|G]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1G|8|@$D|1H|E|1I|F|J]|$D|1J|E|1K|F|J]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1L|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1M|8|@$D|1N|E|1O|F|J]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1P|8|@$D|1Q|E|1R|F|G]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1S|8|@$D|1T|E|1U|F|G]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1V|8|@]|9|@]|A|$]]]|X|$]]

<h2>It's all about operator precedence and their associativity.</h2>

Your first example: "number" + 6 + 4 * 5

Acc. to operator precedence <code>*</code> is calculated first, so it becomes "number" + 6 + 20

Now, Associativity for <code>+</code> is Left -> Right (L->R), so + becomes a concatenation operator cause it is used with <code>String</code>, so the expression becomes "number6" + 20, and then "number620"

(Actually the <code>int</code> are converted to <code>String</code> before concatenation)

Similarly, your 2nd example:

Only <code>+</code> operator and start execution from L->R

"crunch" + 3 + 1 = "crunch3" + 1 = "crunch31"

1 + 3 + "crunch" = 4 + "crunch" = "4crunch"

blocks|key|55767|text|根据你的问题和答案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|55768|解释是|55769|1.
++A)System.out.println+("number"+%2B+6+%2B+4+*+5);
++B)System.out.println+("number6"+%2B+4+*+5);
++C)System.out.println+("number6"+%2B+20);
++D)System.out.println+("number620");|code-block|syntax|javascript|55770|它输出的输出就像|55771|number620|55772|第二个是|55773|2.
++A)System.out.print("crunch"+%2B+3+%2B+1);
++++System.out.print(1+%2B+3+%2B+"crunch");
++B)System.out.print("crunch3"+%2B+1);
++++System.out.print(4+%2B+"crunch");
++C)System.out.print("crunch31");
++++System.out.print("4crunch");|55774|它以一行的形式输出输出，为什么使用print()语句？|offset|length|style|CODE|55775|crunch314crunch|55776|entityMap^0|0|0|0|0|0|0|0|H|7|0|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]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|17|8|@$S|18|T|19|U|V]]|9|@]|A|$]]|$1|W|3|X|5|F|7|1A|8|@]|9|@]|A|$G|H]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

According to your question and answer

explanation is

<pre><code>1.
 A)System.out.println ("number" + 6 + 4 * 5);
 B)System.out.println ("number6" + 4 * 5);
 C)System.out.println ("number6" + 20);
 D)System.out.println ("number620");
</code></pre>

And it prints output like

<pre><code>number620
</code></pre>

And Second one is

<pre><code>2.
 A)System.out.print("crunch" + 3 + 1);
 System.out.print(1 + 3 + "crunch");
 B)System.out.print("crunch3" + 1);
 System.out.print(4 + "crunch");
 C)System.out.print("crunch31");
 System.out.print("4crunch");
</code></pre>

And it prints output with in a line, why because you have used <code>print()</code> statement

<pre><code>crunch314crunch
</code></pre>

I'm just starting out in AP Comp sci in high school and I stumbled across a question regarding the + operator in strings

Why does 
<code>System.out.println ("number" + 6 + 4 * 5)</code>
result in number620 

whereas 

<pre><code>String s = "crunch";
int a = 3, b = 1;
System.out.print(s + a + b);
System.out.print(b + a + s);
</code></pre>

result in <code>crunch314crunch</code>?

Thanks

+ operator and strings

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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我刚从高中的AP Comp sci开始，我偶然发现了一个关于字符串中+操作符的问题为什么System.out.println ("number" + 6 + 4 * 5)会导致number620鉴于String s = "crunch";int a = 3, b = 1;System.out.print(s + a + b);System.out.print(b + a + s);crunch31

问+运算符和字符串
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回答 9

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