Integer number too large error message for 600851475143

Question

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              When the code above is run  it produces an error on the line obj function 600851475143    Why

User · Answer

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 http://download.oracle.com/javase/1.4.2/docs/api/java/lang/Integer.html).

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

User · Answer

Append suffix L  23423429L   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 L for long values

User · Answer

You need to use a long literal    obj function 600851475143l       note the  l  at the end   But I would expect that function to run out of memory  or time

User · Answer

600851475143 cannot be represented as a 32-bit integer  type int   It can be represented as a 64-bit integer  type long   long literals in Java end with an  L   600851475143L

User · Answer

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

User · Answer

Apart from all the other answers  what you can do is     long l   Long parseLong  600851475143      for example     obj function Long parseLong  600851475143

User · Answer

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 l or L after it  Your number should then look like this 600851475143L   Since some Fonts make it hard to distinguish  1  and lower case  l  from each other you should always use the upper case  L

User · Answer

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

[java] "Integer number too large" error message for 600851475143

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