blocks|key|1139367|text|600851475143不能表示为32位整数(类型为int)。它可以表示为64位整数(类型为long)。Java中的长文字以“L”结尾：600851475143L|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1139368|entityMap^0|0|C|Q|3|1A|4|1W|D|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]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

<code>600851475143</code> cannot be represented as a 32-bit integer (type <code>int</code>). It can be represented as a 64-bit integer (type <code>long</code>). long literals in Java end with an "L": <code>600851475143L</code>

blocks|key|1139437|text|追加后缀L：23423429L。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1139438|默认情况下，java将所有数字字面值解释为32位整数值。如果您想明确指定这是大于32位整数的值，则应该对长值使用后缀L。|1139439|entityMap^0|4|1|6|9|0|1M|1|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

Append suffix <code>L</code>: <code>23423429L</code>.

By default, java interpret all numeral literals as 32-bit integer values. If you want to explicitely specify that this is something bigger then 32-bit integer you should use suffix <code>L</code> for long values.

blocks|key|1139316|text|您需要使用一个长文字：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1139317|obj.function(600851475143l);++//+note+the+"l"+at+the+end|code-block|syntax|javascript|1139318|但我预计该函数会耗尽内存(或时间)……|1139319|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You need to use a long literal: 

<pre><code>obj.function(600851475143l); // note the "l" at the end
</code></pre>

But I would expect that function to run out of memory (or time) ...

blocks|key|1139561|text|缺省情况下，java编译器试图将600851475143解释为int类型的常量值。这会导致错误，因为600851475143不能用int表示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1139562|要告诉编译器您希望将数字解释为长整型，您必须在后面添加l或L。你的号码应该看起来像这样的600851475143L。|offset|length|style|CODE|1139563|由于某些字体很难区分"1“和小写"L”，所以您应该始终使用大写"l“。|1139564|entityMap^0|0|R|1|T|1|18|D|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]|$D|P|E|Q|F|G]|$D|R|E|S|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|K|$]]

The java compiler tries to interpret 600851475143 as a constant value of type int by default. This causes an error since 600851475143 can not be represented with an int.

To tell the compiler that you want the number interpretet as a long you have to add either <code>l</code> or <code>L</code> after it. Your number should then look like this <code>600851475143L</code>.

Since some Fonts make it hard to distinguish "1" and lower case "l" from each other you should always use the upper case "L".

blocks|key|924945|text|您需要40位来表示整数文字600851475143。然而，在Java中，最大整数值是2%5E31-1+(即整数是32位的，参见http://download.oracle.com/javase/1.4.2/docs/api/java/lang/Integer.html)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|924946|这与function无关。尝试使用长整型文字(如其他答案中所建议的)。|style|CODE|924947|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/1.4.2/docs/api/java/lang/Integer.html^0|1P|1Z|0|0|2|8|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@$A|U|B|V|F|G]]|9|@]|C|$]]|$1|H|3|-4|5|6|7|W|8|@]|9|@]|C|$]]]|I|$J|$5|K|L|M|C|$N|O]]]]

You need 40 bits to represent the integer literal 600851475143. In Java, the maximum integer value is 2^31-1 however (i.e. integers are 32 bit, see <a href="http://download.oracle.com/javase/1.4.2/docs/api/java/lang/Integer.html" rel="noreferrer">http://download.oracle.com/javase/1.4.2/docs/api/java/lang/Integer.html</a>).

This has nothing to do with <code>function</code>. Try using a long integer literal instead (as suggested in the other answers).

blocks|key|1139619|text|在编译时，数字"600851475143“是以32位整数表示的，尝试在数字末尾使用长文字来解决这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1139620|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

At compile time the number "600851475143" is represented in 32-bit integer, try long literal instead at the end of your number to get over from this problem.

blocks|key|925125|text|除了所有其他答案之外，您可以做的是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|925126|long+l+=+Long.parseLong("600851475143");|code-block|syntax|javascript|925127|例如：|925128|obj.function(Long.parseLong("600851475143"));|925129|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Apart from all the other answers, what you can do is : 

<pre><code>long l = Long.parseLong("600851475143");
</code></pre>

for example : 

<pre><code>obj.function(Long.parseLong("600851475143"));
</code></pre>

blocks|key|925190|text|或者，您可以将输入数字声明为long，然后让它执行代码探戈:D+...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|925191|public+static+void+main(String[]+args)+{

++++Scanner+in+=+new+Scanner(System.in);
++++System.out.println("Enter+a+number");
++++long+n+=+in.nextLong();

++++for+(long+i+=+2;+i+<=+n;+i%2B%2B)+{
++++++++while+(n+%25+i+==+0)+{
++++++++++++System.out.print(",+"+%2B+i);
++++++++++++n+/=+i;
++++++++}
++++}
}|code-block|syntax|javascript|925192|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|$]]

Or, you can declare input number as long, and then let it do the code tango :D ...

<pre><code>public static void main(String[] args) {

 Scanner in = new Scanner(System.in);
 System.out.println("Enter a number");
 long n = in.nextLong();

 for (long i = 2; i &lt;= n; i++) {
 while (n % i == 0) {
 System.out.print(", " + i);
 n /= i;
 }
 }
}
</code></pre>

<pre><code>public class Three {
 public static void main(String[] args) {
 Three obj = new Three();
 obj.function(600851475143);
 }

 private Long function(long i) {
 Stack&lt;Long&gt; stack = new Stack&lt;Long&gt;();

 for (long j = 2; j &lt;= i; j++) {
 if (i % j == 0) {
 stack.push(j);
 }
 }
 return stack.pop();
 }
}
</code></pre>

When the code above is run, it produces an error on the line <code>obj.function(600851475143);</code>. Why?

"Integer number too large" error message for 600851475143

编译器

Java

public class Three { public static void main(String[] args) { Three obj = new Three(); obj.function(600851475143); } private Long function(long i) { Stack<Long> stack =

问600851475143的"Integer number too large“错误消息
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回答 8

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问600851475143的"Integer number too large“错误消息EN

回答 8

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问600851475143的"Integer number too large“错误消息
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