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Java Forum / General / June 2007char[]
Hi, if I do: char cop[] = new char[30]; and then System.out.print(cop[0]+" +++ "); why is the output an empty string and not +++ ? Thanks > Hi, if I do: > [quoted text clipped - 7 lines] > > Thanks Hint: What are the initial values of the elements of cop? What about the character with Unicode code 0? Maybe some research about "null terminated strings" can help. >> Hi, if I do: >> [quoted text clipped - 11 lines] > What about the character with Unicode code 0? > Maybe some research about "null terminated strings" can help. Null terminated strings are not relevant here. This is Java, not C. I get the result that Josh expected, not the one he observed. /gordon -- > Null terminated strings are not relevant here. This is Java, not C. > > I get the result that Josh expected, not the one he observed. > > /gordon What is relevant here is the fact that mull characters are not expected to be handled by a printing device. Obviously, in OP's case it takes such a character for a string terminator. Faton Berisha > > Hi, if I do: > [quoted text clipped - 11 lines] > What about the character with Unicode code 0? > Maybe some research about "null terminated strings" can help. if cop is an array of char when I create it with 'new' it should have a reference to an object of array of char with 30 cells with empty chars. But I don't understand that output! | > > Hi, if I do: | > [quoted text clipped - 15 lines] | should have a reference to an object of array of char with 30 | cells with empty chars. But I don't understand that output! Depending on how your system handles I/O, you may also need to write out a line terminator (cr / lf) or otherwise flush the output buffer. See what println gives instead of print. Also, are you sure you recompiled? -- this exercise works for me. Matt Humphrey matth@ivizNOSPAM.com http://www.iviz.com/ > | > > Hi, if I do: > | > [quoted text clipped - 22 lines] > > Matt Humphrey m...@ivizNOSPAM.comhttp://www.iviz.com/ also if I write: System.out.println(x[0] + " ----- "); in the stdout is only print the '\n' char but nothing of concatenating x[0] with " ----- " !!! when java meet a + sign it should create a StringBuffer tmp object like: new StringBuffer(x[0]).concat(" ----- " ).toString() and than print it. But it print nothing! why? > also if I write: > System.out.println(x[0] + " ----- "); [quoted text clipped - 5 lines] > new StringBuffer(x[0]).concat(" ----- " ).toString() > and than print it. But it print nothing! why? It prints something, but whatever you're using to display the output is mishandling the nul character. public static void main( String [] args ) { char cop[] = new char[30]; System.out.println( ">>>" + cop[0] +" +++ " ); } If I run this in a console, I get: >>> +++ If I run it in Eclipse, I get: If I run it in a console and pipe the output into a file, and then look at it with vi[m], I see: >>>^@ +++ where the ^@ is a representation of the nul character. I shall silently pass over the issue of why you were deliberately generating output with nul characters in and then frothing and foaming about the result rather than doing three extra minutes of investigation on it.  Signature"He's dead, Jim, but not as we know it." Unsaid /Trek/ Hewlett-Packard Limited registered office: Cain Road, Bracknell, registered no: 690597 England Berks RG12 1HN > if cop is an array of char when I create it with 'new' it > should have a reference to an object of array of char with 30 > cells with empty chars. But I don't understand that output! There is no such thing like an empty char. The single elements of the char-array have the value 0 and the question is what happens if you give that out to stdout. Regards, Lothar SignatureLothar Kimmeringer E-Mail: spamfang@kimmeringer.de PGP-encrypted mails preferred (Key-ID: 0x8BC3CD81) Always remember: The answer is forty-two, there can only be wrong questions! > if cop is an array of char when I create it with 'new' it > should have a reference to an object of array of char with 30 > cells with empty chars. But I don't understand that output! It would create an array with null characters, which are different form empty char. Faton Berisha > Hi, if I do: > [quoted text clipped - 7 lines] > > Thanks according to the java language specification, an array of char is not a string. i checked the api and i think you should try this and see if it works: String[] strCop = String.copyValueOf(cop); then System.out.print(strCop[0]+" +++ "); am sure it will now. >> Hi, if I do: >> [quoted text clipped - 12 lines] > it works: > String[] strCop = String.copyValueOf(cop); Invalid syntax. copyValueOf() does not return an array. > then > System.out.print(strCop[0]+" +++ "); > am sure it will now. No, it won't, whatever verb you left out. It will result in compiler error. However, the result of "arbitraryTypeOfObject + someString" is a String, which is what applies here, so the whole "strCop" dodge is irrelevant. Even if you did declare "strCop" as a String and used the correct method and syntax to get its first character, it would be literally the same type of expression: char plus String equals String. SignatureLew > >> Hi, if I do: > [quoted text clipped - 31 lines] > -- > Lew i think i was really in a hurry. String[] references String objects in the array varaible, strCop. so i guessed he should should read in each reference singly, i asked him to read it as a whole. read singly like this: String[] strCop = new String[copy.length]; for (int i=0; i<cop.length; i++){ //do the valueOfCopy here, component by component. strCopy[i] = String.valueOfCopy(cop[i]); /....everything should be okay from here } my mistake. i just know it'll work. an IDE not within reach though. it will this time > > >> Hi, if I do: > [quoted text clipped - 47 lines] > my mistake. i just know it'll work. an IDE not within reach though. > it will this time got to hide my head in shame. i tried my code above at home... well, the asker didn't post wrong code then all he just didn't understsand was the result syso returned. default for char values is \u0000 which has a code point of -1 and doesn't map to string the same way return key and tab key do not map to string. hmphhh... char[] cop = new char[30]; Character y = new Character('k'); System.out.println(y.getNumericValue('k')); System.out.println(y.getNumericValue(cop[0])); //unicode maps from code point zero System.out.println(y.getNumericValue('\u0000')); if (y.isDefined(cop[0])) System.out.println("yes"); //yes } //\u0000 is the first unicode character, rather a symbol, like \u0009 is the tab character or your '\t'. > got to hide my head in shame. i tried my code above at home... > well, the asker didn't post wrong code then all he just didn't > understsand was the result syso returned. default for char values is > \u0000 which has a code point of -1 and doesn't map to string the same > way return key and tab key do not map to string. > hmphhh... Chars don't "map to" Strings, Strings comprise chars. Unicode '\u0000' is a perfectly valid char, even in a String. Not all display devices handle it the same way. Many posters have made that point to the OP. SignatureLew Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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