blocks|key|3771736|text|这里有两个独立的问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3771737|3771738|You应该使用equals()+方法(或它的近亲)来比较字符串-否则，您将比较两个字符串是否是相同的引用(实例)，而不是equals()的等效序列应该避免使用比较双精度，因为双精度相等更加细微差别。查看this+paper以了解更多原因。|ordered-list-item|offset|length|style|BOLD|CODE|3771739|3771740|请参阅this+question，了解为什么在java中使用==进行浮点比较不是一个好主意。|3771741|除此之外，我还要提到，您的实现假设stuName和stuGpa都是长度相同的数组。这很可能不是这样的。在开始迭代数组之前，它可能是值得断言的东西。|3771742|entityMap|0|LINK|mutability|MUTABLE|url|http://www.validlab.com/goldberg/paper.pdf|1|https://stackoverflow.com/questions/1088216/whats-wrong-with-using-to-compare-floats-in-java^0|0|0|0|F|G|2B|7|8|1O|8|2T|A|0|0|0|U|2|3|D|1|0|H|7|P|6|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|11|8|@$F|12|G|13|H|I]|$F|14|G|15|H|I]|$F|16|G|17|H|J]|$F|18|G|19|H|J]]|9|@$F|1A|G|1B|1|1C]]|A|$]]|$1|K|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1E|8|@$F|1F|G|1G|H|J]]|9|@$F|1H|G|1I|1|1J]]|A|$]]|$1|N|3|O|5|6|7|1K|8|@$F|1L|G|1M|H|J]|$F|1N|G|1O|H|J]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W]]|X|$5|S|T|U|A|$V|Y]]]]

You have two separate problems here:

<ol>
<li>You should be comparing strings using the <code>equals()</code> method (or one of it's kin) - otherwise you are comparing whether two strings are the same reference (instance) rather than equivalent sequences of characters.</li>
<li>You should avoid comparing doubles using <code>==</code> as equality for doubles is more nuanced. Check out <a href="http://www.validlab.com/goldberg/paper.pdf" rel="nofollow noreferrer">this paper</a> for more information about why.</li>
</ol>

See <a href="https://stackoverflow.com/questions/1088216/whats-wrong-with-using-to-compare-floats-in-java">this question</a> about why using <code>==</code> for floating point comparison is a bad idea in java.

Aside from that, I would also mention that your implementation makes the assumption that both <code>stuName</code> and <code>stuGpa</code> are arrays of the same length. This could easily not be the case ... and is probably something worth asserting before you begin iterating over the arrays.

blocks|key|3771609|text|在Java中，字符串必须与.equals进行比较，而不是与==进行比较。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3771610|if(stuName[i].equals+(nameKey)+&&+stuGpa[i]+==+gpaKey)|code-block|syntax|javascript|3771611|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|$]]

Strings must be compared with .equals in Java, not ==.

<pre><code>if(stuName[i].equals (nameKey) &amp;&amp; stuGpa[i] == gpaKey)
</code></pre>

blocks|key|1330127|text|你可能想要|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1330128|if+(stuName[i].equals(nameKey)+&&+stuGpa[i].equals(gpaKey))|code-block|syntax|javascript|1330129|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|$]]

You probably want

<pre><code>if (stuName[i].equals(nameKey) &amp;&amp; stuGpa[i].equals(gpaKey))
</code></pre>

blocks|key|3771630|text|if(stuName[i]+==+nameKey不太可能是正确的，您正在比较对象标识而不是字符串内容。试试if(stuName[i].equals(nameKey)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3771631|entityMap^0|0|O|1H|T|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

<code>if(stuName[i] == nameKey</code> is unlikely to be right, you are comparing object identities not string content. Try <code>if(stuName[i].equals(nameKey)</code>

blocks|key|5239936|text|您正在比较两个字符串。字符串是不可变的。请使用"equalsIgnoreCase()“或"equals()”比较字符串|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5239937|请参阅此处的示例http://www.java-samples.com/showtutorial.php?tutorialid=224|offset|length|5239938|entityMap|0|LINK|mutability|MUTABLE|url|http://www.java-samples.com/showtutorial.php?tutorialid=224^0|0|8|1N|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]]]]

You are comparing two Strings.
Strings are immutable.
Please use "equalsIgnoreCase()" or "equals()" to compare Strings

See examples here
<a href="http://www.java-samples.com/showtutorial.php?tutorialid=224" rel="nofollow">http://www.java-samples.com/showtutorial.php?tutorialid=224</a>

blocks|key|3771676|text|一个重要的问题是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3771677|stuName[i]+==+nameKey|code-block|syntax|javascript|3771678|只是比较对象在内存中是否为相同的字符串对象。|3771679|实际上，您希望使用nameKey.equals(stuName[i])来比较实际的字符串值。而且您可能希望使用.equalsIgnoreCase来区分大小写。|offset|length|style|CODE|3771680|entityMap^0|0|0|0|9|Q|1J|H|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]|$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

An essential problem is that

<pre><code>stuName[i] == nameKey
</code></pre>

Is only comparing whether the objects are the same String Object in memory.

You actually want to use <code>nameKey.equals(stuName[i])</code>, to compare the actual string values.
And you might want to use <code>.equalsIgnoreCase</code> for case insensitivity.

blocks|key|5239962|text|下面的if语句是正确的。stuNamei是一个字符串，所以请与.equals进行比较。stuGpai是双精度的，所以使用==。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5239963|if(stuName[i].equals(nameKey_+&&+stuGpa[i]+==+gpaKey)|code-block|syntax|javascript|5239964|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|$]]

The following is correct for the if statement. stuName[i] is a string so compare with .equals. stuGpa[i] is a double so use ==. 

<pre><code>if(stuName[i].equals(nameKey_ &amp;&amp; stuGpa[i] == gpaKey)
</code></pre>

blocks|key|3771792|text|您的问题不是条件if语句，而是条件运算符==。==指的是对象的指针值，其中.equals方法返回由对象计算的值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3771793|就像大家之前说过的那样，在下一行中将您的==切换为.equals：|3771794|public+static+int+seqSearch(int+numRecords,+String[]+stuName,
++double[]+stuGpa,+String+nameKey,+double+gpaKey)

++++for(int+i+=+0;+i+<+numRecords;+i%2B%2B)
+++++++if(stuName[i].equals(nameKey)+&&+stuGpa[i]+==+gpaKey)
++++++++++return+i;
++++return+-1;|code-block|syntax|javascript|3771795|entityMap^0|K|2|N|2|11|7|0|K|2|P|7|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|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

Your problem is not the conditional if statement, but the conditional operator <code>==</code>. <code>==</code> refers to the pointer value of the object where as the <code>.equals</code> method returns something computed by the object.

Like everyone has said before, switch your <code>==</code> to <code>.equals</code> in this next line:

<pre><code>public static int seqSearch(int numRecords, String[] stuName,
 double[] stuGpa, String nameKey, double gpaKey)

 for(int i = 0; i &lt; numRecords; i++)
 if(stuName[i].equals(nameKey) &amp;&amp; stuGpa[i] == gpaKey)
 return i;
 return -1;
</code></pre>

blocks|key|1330279|text|要实际回答有关if语句控制的问题...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1330280|我相信你所做的对多个返回语句来说是很好的，但是...|1330281|我个人更喜欢我的方法只有一个入口点和一个出口点。对于我来说，拥有多个出口点总是让我感觉很脏。|1330282|因此，我将考虑以下代码：|1330283|public+static+int+seqSearch(int+numRecords,+String[]+stuName,+double[]+stuGpa,+String+nameKey,+double+gpaKey)

++++int+value+=+-1;

++++for(int+i+=+0;+i+<+numRecords;+i%2B%2B)+{++//+Don't+forget+your+braces,+they+aren't+required,+but+wait+until+you+add+a+newline+and+forget+to+add+them...
+++++++if(some.boolean().equals(comparison.here()))+{
++++++++++value+=+i;
++++++++++break;
+++++++}
++++}
++++return+value;
}|code-block|syntax|javascript|1330284|祝你好运。|1330285|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|T|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|U|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

To actually answer the question about the control of the if statement...

I believe what you're doing is fine with the the multiple return statements, BUT...

I personally prefer one entry point and only one exit point for my methods. It always feels dirty to me having multiple exit points.

So, I would consider the following code instead:

<pre><code>public static int seqSearch(int numRecords, String[] stuName, double[] stuGpa, String nameKey, double gpaKey)

 int value = -1;

 for(int i = 0; i &lt; numRecords; i++) { // Don't forget your braces, they aren't required, but wait until you add a newline and forget to add them...
 if(some.boolean().equals(comparison.here())) {
 value = i;
 break;
 }
 }
 return value;
}
</code></pre>

Best of Luck.

<pre><code>public static int seqSearch(int numRecords, String[] stuName,
 double[] stuGpa, String nameKey, double gpaKey)

 for(int i = 0; i &lt; numRecords; i++)
 if(stuName[i] == nameKey &amp;&amp; stuGpa[i] == gpaKey)
 return i;
 return -1;
</code></pre>

So, how would I used an if statement to control this? I'm doing sequential search to find if the name is found in the array and if the gpa is in the array, then it should return the position it was found in (i). But, all it does do is return -1 and print out that none were found.

Using an IF statement to control the return statement?

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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public static int seqSearch(int numRecords, String[] stuName, double[] stuGpa, String nameKey, double gpaKey) for(int i = 0; i < numRecords; i++) if(stuName[i] == nameKey && stuG

问使用IF语句来控制return语句？
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问使用IF语句来控制return语句？
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