how to convert a string into an int using java?
101861
The Integer class includes a method named parseInt() that serves just this purpose. An example of a simple program that performs this conversion is shown below:\n\n\npublic class s2i {\n\n   public static void main (String[] args) {\n\n      // String s = "fred";    // do this if you want an exception\n\n      String s = "100";\n\n      try {\n         int i = Integer.parseInt(s.trim());\n         System.out.println("int i = " + i);\n      } catch (NumberFormatException nfe) {\n         System.out.println("NumberFormatException: " + nfe.getMessage());\n      }\n\n   }\n\n}\n\n\n\nOther notes\n\nInteger.toString(int i) is used to convert in the other direction, from int to String. \n  \nIf you're interested in converting a String to an Integer object, use the valueOf() method of the Integer class instead of the parseInt() method.
76546
How about using the char array that is the input string, converting to uppercase, then using the chars as indexes into your substitution array of chars? Use the ASCII table to help you.\n\nchar subst_array[] = {'Z', 'Y', 'X', ...};\n\nconversion to uppercase: if (char > 90) char -= 32;\n\nfor (i = 0; i < strlen(input_string); i++)\n  input_string[i] = subst_array[input_string[i] - 65];\n\ncout << input_string;
868581
Java doesn't allow operator overloading.
876620
Calling the toCharArray() method on the String will return an array of characters.  To get just one character in the string, call the charAt(int position) method.
642877
int substitute(char c1, char c2, char *s) {\n// init. counter\nint num = 0;\n\nwhile (*s != '&#92;0') {\n\n// compare string char with 1st char\nif (*s == c1) {\n// if equal substitute with 2nd char and incr. counter\n*s = c2;\nnum++;\n} \n// incr. string pointer\ns++;\n}\n\n// return number of occurrences\nreturn num;\n}
293980
an Integer and an int are not the same object.  try int x = Integer.intValue();  where Integer is the name of your Integer
790827
main () {char *s, b[100]; while(s=gets(b)) printf (atoi(s)%2 ? "nope&#92;n" : "yep&#92;n");}
780385
Try it out - you need a semicolon at the end of the first print statement for it to compile.\nIn java,\nit prints out:\n0000000000\n0123456789\n0246802468\n0369258147\n0482604826\n0505050505\n0628406284\n0741852963\n0864208642\n0987654321
717813
In JavaScript you can do it. In Java you cannot, unless perhaps you typeset a char to a byte to an int. But that's not what you mean, is it?\n\nYou need to use at least one method.\n\nEdit:\nOf course, what the people who answered below me don't realize is that everything is in a method ("function" in other languages). Perhaps you want to write your own method to convert strings to numbers... it's still part of a method.\n\nBut why would you want to do that when there are parseInt and valueOf?\nRead down from http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Integer.html#parseInt(java.lang.String)
6190
alternative 1)\nUse int/short instead. Make sure you mask potential values with "& 0xff" so they don't overflow.\n\nalternative 2)\nUse (signed) byte for storage, but when doing comparisons, assignment to other integer types, cast to int/short and mask with 0xff
1069483
Try using String.Format()\n\nFor example:\n\nString.Format("Hello {0}, you are {1}",\n      value1,\n      value2)
220005
Use a string. Calculate the first 16 digits, then put them into a string and subtract that number from your value and then use it as though its scientific notation. For example\n3.141592653589793 is the first 16 digits. So convert that to a string and then multiply your deriving value by 1,000,000,000,000,000 and start again. Then strip out the decimal and append to the string\n"2384626433832795"\nand then\n"0288419716939937", "5105820974944592", "3078164062862089",\n"9862803482534211", etc. \n\nThis should allow you a very large item.\n\n\nThe other option I can think of is declare an array 5 or 10 digits each and then store the numbers in there. That might work.\n\n\nEither way, good luck!
890450
char *Months[] = { "January", "February", "March", ... "December"};\n\nchar *GetMonth( int Index )\n{\n  return( Months[Index - 1] ); // Note -1 because first month is 1 not 0\n}
717749
its because "int" is only a whole number, no fractions allowed. you need to change "int n = 2.05*100;" to \n"float n = 2.05*100;".
515624
You can start off by declaring a new integer variable "low", initialize it to 0, and then after any input compare the new value to the value in low and if the new value is lower then set low equal to the new value.\nThe following sites should also be of use.
623394
IsNumeric will tell you if a value being passed into it can be converted into a number, were you to try to run Decimal.Parse on it. You might use it like this:\n\nFunction Translate(Foo as String) As Decimal\n    if IsNumeric(Foo) = True then\n      Return Decimal.Parse(Foo)\n    Else\n      Return 0\n    End if\nEnd Function
98590
import java.util.*;\npublic class HelloWorld {   \n   public static Stack stack=new Stack(); \npublic static void main(String[] args) {\n   try {\n   String s="this is java string reverse ";\n   reverseString(s);\n   System.out.println("the reversed string value is ");\n   for(int i=stack.size()-1;i>=0;i--){\n   System.out.print(stack.get(i));\n   }\n       } \n   catch (Exception e) {\n            e.printStackTrace();\n       }\n    }\n   public static void reverseString(String s) throws Exception{\n      StringTokenizer temp=new StringTokenizer(s);\n   if(temp.hasMoreTokens()){\n   String s1=temp.nextToken();\n   stack.push(s1+" ");  \n   reverseString(s.substring(s1.length()+1));\n   }\n   else{\n   return;\n   }     \n   }  \n   \n}