How java convert long to float or int to float?
441040
Because a double can represent a much wider range of magnitudes that a float, the cast/conversion can produce some enexpected results.\n\nTo understand the conversion fully, you must format the number being converted into binary format. That is when you see how the bits are altered when converting from long (64 bits) to float (32 bits).
1008669
sizeof doesnt give the length of what you actually have, its gives the size of how big the variable has been declared to be . . .\n\nsizeof will return the number of bytes reserved for a variable or data type. \n\nThe following code shows sizeof returning the length of a data type. \n\n\n        /* How big is an int? expect an answer of 4. */\n\nmain()\n {\n   printf("%d &#92;n", sizeof(int));\n        }\n\n \n\n\nsizeof will also return the number of bytes reserved for a structure. \n\n\n        /* Will print 8 on most machines. */\n\n        main()\n{\n  struct \n  {\n    int a;\n    int b;\n  } TwoInts;\n\nprintf("%d &#92;n", sizeof(TwoInts));\n}\n\n\n \n\n\nFinally, sizeof will return the length of a variable. \n\n\nmain()\n{\n  char String[20];\n  \n  printf ("%d &#92;n", sizeof String);\n   printf ("%d &#92;n", sizeof (String));\n        }\n\n\n \n\n\nIn the example above I have printed the size of 'String' twice. This is to show that when dealing with variables, the brackets are optional. I recommend that you always place the brackets around the sizeof argument.
257370
There are a number places in the C++ spec that say that the method of implementation is left up to the designer of the compiler as long as they meet certain requirement.\n\nFor example, the rules require that the storage size of a long be at least as large as that for an int.\nMicrosoft C++ makes both ints and longs 32 bits. A program that works with 32 bit longs could very easily fail with 16 bit longs and vice versa. \n\nIt's also possible that you are breaking a rule that you can get away with on with one compiler, but not the other. For example, if you are running a program built with MS C++ using debug heap allocation, when you free the memory, it gets overwritten with a specific pattern to aid in debugging. If you use a stale pointer, you'll be pointing to trash. The same code built using another compiler might still find valid values in the memory, even though you released it.
114514
I have the answer.\n\nYou should change the numbers to be "400.0" instead of "400". This is your primary problem.\n\nThe deeper reason that this is happening is that 400*400/400 is an INTEGER expression. It will be evaluated as an integer, and only assigned to a floating point result.\n\nI believe that your compiler is a 16 bit compiler, this is the only answer that makes sense.\n\nNow let's just see what the compiler is doing. Its first step is to evaluate 400*400. Now if you get out your calculator, you will get the answer 160,000. NOT SO FAST!!!! An integer in your compiler is only 16 bits long, and can only express numbers between -32768 and 32767. The answer that it comes up with is 28,928 (160,000 - 2*65,536). It will next divide 28,928 by 400 to give 72 (again it drops the fractional result because it is integer).\n\nIn both C and C++, the result of an integer expression is an integer. It does not matter if it gets assigned to a float or not. If you want float accuracy, then the numbers in the expression must be floats.
779217
There are built-in functions in a programming language to perform that task.For instance in C++ we have: cint("string value") for converting the string value into integer.And in Visual Basic we have the val() function to do the same thing.
936111
2 different ways:\n\nint n = 5;\nchar buffer[10]; /* long enough to hold your number + null */\nsprintf(buffer, "%i", n);\n\n\nor\n\nitoa(number,string,radix) \n\nradix - 10 for decimal, 16 for hex...
872841
double min(double* s, int length)\n{\ndouble min = *s;\nint i;\nfor(i=1, i< length; ++i)\nif((*s) < min ) {\n    min = *s;\n}\nreturn min;\n}
565428
According to the docs in java sun, \nDecimalFormat is an object...\n\nSo when you use a method within that object, say Format in your case,\n\n StringBuffer format(double number, StringBuffer result, FieldPosition fieldPosition)\n          Formats a double to produce a string.\n StringBuffer format(long number, StringBuffer result, FieldPosition fieldPosition)\n          Format a long to produce a string.\n\nThose are its entities, so it returns a StringBuffer. ( http://java.sun.com/j2se/1.5.0/docs/api/java/lang/StringBuffer.html )
272231
You can format the output of cout like this:\ncout << setprecision(2) << yourVar << endl;\n\nAnd if you want it to show the decimal even when it is .00, use cout << fixed << ...\n\nAnd while we're at it, if you want to make your output even prettier, you might want to make use of setw(), setfill(), and left/right...\n\nOh, and to use any of those, you need to #include <iomanip>
642129
ok, here is one declaration, of many;\n\nfloat Doc(int arr[100], int NumCt)\n{\nfloat sum = 0.0;\nfor (int i = 0; i < NumCt; i++)\nsum += arr[i];\n\nif (!NumCt)\nreturn sum;\n\nreturn sum / NumCt;\n}
83391
parseInt Method of Integer Wrapper class is used to parse integer from a string.\nsyntax:\nInteger.parseInt(string)\nit returns a integer value so an integer is used the store the return value at the left hand side of assignment statement.
985771
You can generate this in one line of Perl, Unix scripting, or a small C/C++ program.  Simply write one number per line from your program counter.  You may have to get tricky and use two long ints, but it's not hard.\n\n Realize that your document may be very large, and for most purposes, you don't need a document, just use the variable(s) in your final program.
91787
Unfortunately, no. You'll need to parse the integer from the String entered in the text field. You can use JTextArea or JTextPane or even JPasswordField. To compare the values you'd either want to use a JButton at the bottom (for the user to click once he/she is finished entering text) and then parse an integer out of it.  View the sources below (especially the API) to view the different subclasses of the different types of text components in Swing. If you need help parsing the integer out of it, then let me know and I'll tell you how to do that (if I'm on, so don't count on it ;). Good luck!
339901
To validate text input converted to a int try this:\n\ntry {\n    String s = new JTextField().getText(); // The user enters text in the text field\n    int value = Integer.parseInt(s);\n}\ncatch(NumberFormatException ex) { // Invalid input, not a number\n    ex.printStackTrace();\n}\n\nHope this helps
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.
918056
There is probably a better way, or at least a more streamline way, of doing this, but off the top of my head here is what I would do:\n\n1. Create a loop to run through all the characters in your string.\n2. Test each character to see if it falls between ASCII values 47 - 58 (i.e. 0-9). If it does then the character is a number.\n3. If the first test fails then you know that it isn't a number; therefore, you know that it must be a character, so now test to see if it is a character between 'a' and 'd' (i.e. ASCII 97-100).\n\nHere is some sample code (hopefully it can help):\npublic static void main(String[] args){\nString teststr = "d3";\nchar x;\n\nSystem.out.println("Test String:" + teststr);\n\nfor(int i = 0; i < teststr.length(); i++){\nx = teststr.charAt(i);\nif(x >= 47 && x <= 58){\n// Store it however\nSystem.out.println(Integer.parseInt(""+ x));\n}\nelse if(x >= 97 && x <= 100){\nif(x == 'a')\nSystem.out.println("a=" + 1);\nelse if(x == 'b')\nSystem.out.println("b=" + 2);\nelse if(x == 'c')\nSystem.out.println("c=" + 3);\nelse if(x == 'd')\nSystem.out.println("d=" + 4);\n}\n}\n}
80562
Relatively easy way\n\n0.008895478\n0.012222453\n\nmultiply both by 100 (ie, move decimal point two spaces to the right):\n\n0.8895478\n1.2222453\n\nObviously first one is less.