Team,
I am not able to use the Java 7 Underscores in Numeric Literals feature for getting the input from user and printing out in same format as declared. Please help in doing that? OR Is this feature is incomplete?
Scanner input = new Scanner( System.in );
int x = 1_00_000;
System.out.print( "Enter numeric literals with underscores: " ); //2_00_000
x = input.nextInt(); //java.util.InputMismatchException
System.out.println(x); // Prints in normal format, but want to be in 2_00_000.
NOTE: In Eclipse; I am able to change the value of numeric literal with Underscored numeric literal in runtime. This may be hack, but this is needed feature to input Underscored numeric literal in runtime rit?.
http://www.eclipse.org/jdt/ui/r3_8/Java7news/whats-new-java-7.html#miscellaneous
if you want maintain the underscores you can use String:
Scanner input = new Scanner( System.in );
System.out.print( "Enter numeric literals with underscores: " ); //2_00_000
String stringLiterals = input.nextLine();
System.out.println(stringLiterals); // Prints 2_00_000.
Related
I'm using Visual FoxPro and I need to convert currency amount into numeric. The 2 columns in the table are tranamt(numeric,12,2) and tranamt2(character)
Here's my example:
tranamt2=-$710,000.99
I've tried
replace all tranamt with val(tranamt2)
and
replace all tranamt with val(strtran(tranamt2, ",",""))
both results give me zero. I know it has something to do with the negative sign but I can't figure it out. Any help is appreciated.
Try this:
replace all tranamt with VAL(STRTRAN(STRTRAN(tranamt2, "$", ""), ",", ""))
This removes the dollar sign and comma in one shot.
need to convert currency amount into numeric
tranamt(numeric,12,2) and tranamt2(character)
First of all a neither a Character Field Type nor a Numeric Field type (tranamt2) are Not a VFP Currency Field type
You may be using the value of a Character field to represent currency, but that does not make it a currency value - just a String value.
And typically when that is done, you do NOT store the Dollar Sign '$' and Comma ',' as part of the data.
Instead, you store the 'raw' value (in this case: "-710000.99") and merely format how that 'raw' value is displayed when needed.
So in your Character field you have a value of: -$710,000.99
Do you have the Dollar Sign '$' and the Comma ',' as part of the field data?
If so, to convert it to a Numeric, you will first have to eliminate those extraneous characters prior to the the conversion.
If they are not stored as part of your field value, then you can use the VAL() 'as is'.
Example:
cStr = "-710000.99" && The '$' and ',' are NOT stored as part of Character value
nStr = VAL(cStr)
?nStr
However if you have the Dollar Sign and the Comma as part of the field data itself, then you can use STRTRAN() to eliminate them during the conversion.
Example:
cStr = "-$710,000.99" && Note both '$' and ',' are part of data value
* --- Remove both '$' and ',' and convert with VAL() ---
nStr = VAL(STRTRAN(STRTRAN(cStr,",",""),"$",""))
?nStr
Maybe something like:
REPLACE tranamt WITH VAL(STRTRAN(STRTRAN(tranamt2,",",""),"$",""))
EDIT: Another alternative would be to use CHRTRAN() to remove the '$' and ','
Something like:
cRemoveChar = "$," && Characters to be removed from String
REPLACE tranamt WITH VAL(CHRTRAN(tranamt2,cRemoveChar,""))
Good Luck
A little late but I use this function call
function MoneyToDecimal
LPARAMETER tnAmount
LOCAL lnAmount
IF VARTYPE(tnAmount) = "Y"
lnAmount = VAL(STRTRAN(TRANSFORM(tnAmount), "$", ""))
ELSE
lnAmount = tnAmount
ENDIF
return lnAmount
endfunc
And can be tested with these calls:
wait wind MoneyToDecimal($112.50)
wait wind MoneyToDecimal($-112.50)
Use the built-in MTON() function to convert a currency value into a numeric value:
replace all tranamt with mton(tranamt2)
I am trying to go through a string one character at a time and change those characters into their ASCII numbers. I have it working wine except for the spaces in the text turned into '' (with no space) instead of ' ' (with space).
Scanner in = new Scanner(original_file);
PrintWriter out = new PrintWriter("encoded_message.txt");
text_file = "";
while (in.hasNextLine()) {
text_file += in.nextLine() + "\n";
}
char[] text_file_array = text_file.toCharArray();
for (char c : text_file_array) {
int code = Character.getNumericValue(c); //<-- The problem is here.
out.println(code);
}
in.close();
out.close();
When I run this, the c becomes '' which makes code have a value of -1. I later use code in an array. How do I make it return ' '?
The Character.getNumericValue() method returns the unicode numeric value not the ASCII value.
e.g. for 'e' ASCII is 101, unicode is 14.
ASCII is relatively simple:
int ascii = (int) character;
Convert character to ASCII numeric value in java
The javadoc for getNumericValue discusses this but sheds little light to the matter:
https://docs.oracle.com/javase/7/docs/api/java/lang/Character.html#getNumericValue%28char%29
The letters A-Z in their uppercase ('\u0041' through '\u005A'), lowercase ('\u0061' through '\u007A'), and full width variant ('\uFF21' through '\uFF3A' and '\uFF41' through '\uFF5A') forms have numeric values from 10 through 35. This is independent of the Unicode specification, which does not assign numeric values to these char values.
If the character does not have a numeric value, then -1 is returned. If the character has a numeric value that cannot be represented as a nonnegative integer (for example, a fractional value), then -2 is returned.
The c doesn't in fact change to become '' (with no space). If you change your out.println method to this (debugging, yes? you can clean it up later), you can get more info:
out.println("'" + c + "'" + " " + code + " " + (int)c);
I am wondering how to make something where if X=5 and Y=2, then have it output something like
Hello 2 World 5.
In Java I would do
String a = "Hello " + Y + " World " + X;
System.out.println(a);
So how would I do that in TI-BASIC?
You have two issues to work out, concatenating strings and converting integers to a string representation.
String concatenation is very straightforward and utilizes the + operator. In your example:
"Hello " + "World"
Will yield the string "Hello World'.
Converting numbers to strings is not as easy in TI-BASIC, but a method for doing so compatible with the TI-83+/84+ series is available here. The following code and explanation are quoted from the linked page:
:"?
:For(X,1,1+log(N
:sub("0123456789",ipart(10fpart(N10^(-X)))+1,1)+Ans
:End
:sub(Ans,1,length(Ans)-1?Str1
With our number stored in N, we loop through each digit of N and store
the numeric character to our string that is at the matching position
in our substring. You access the individual digit in the number by
using iPart(10fPart(A/10^(X, and then locate where it is in the string
"0123456789". The reason you need to add 1 is so that it works with
the 0 digit.
In order to construct a string with all of the digits of the number, we first create a dummy string. This is what the "? is used
for. Each time through the For( loop, we concatenate the string from
before (which is still stored in the Ans variable) to the next numeric
character that is found in N. Using Ans allows us to not have to use
another string variable, since Ans can act like a string and it gets
updated accordingly, and Ans is also faster than a string variable.
By the time we are done with the For( loop, all of our numeric characters are put together in Ans. However, because we stored a dummy
character to the string initially, we now need to remove it, which we
do by getting the substring from the first character to the second to
last character of the string. Finally, we store the string to a more
permanent variable (in this case, Str1) for future use.
Once converted to a string, you can simply use the + operator to concatenate your string literals with the converted number strings.
You should also take a look at a similar Stack Overflow question which addresses a similar issue.
For this issue you can use the toString( function which was introduced in version 5.2.0. This function translates a number to a string which you can use to display numbers and strings together easily. It would end up like this:
Disp "Hello "+toString(Y)+" World "+toString(X)
If you know the length of "Hello" and "World," then you can simply use Output() because Disp creates a new line after every statement.
I need to actually print a Dollar sign in Dart, ahead of a variable. For example:
void main()
{
int dollars=42;
print("I have $dollars."); // I have 42.
}
I want the output to be: I have $42. How can I do this? Thanks.
Dart strings can be either raw or ... not raw (normal? cooked? interpreted? There isn't a formal name). I'll go with "interpreted" here, because it describes the problem you have.
In a raw string, "$" and "\" mean nothing special, they are just characters like any other.
In an interpreted string, "$" starts an interpolation and "\" starts an escape.
Since you want the interpolation for "$dollars", you can't use "$" literally, so you need to escape it:
int dollars = 42;
print("I have \$$dollars.");
If you don't want to use an escape, you can combine the string from raw and interpreted parts:
int dollars = 42;
print(r"I have $" "$dollars.");
Two adjacent string literals are combined into one string, even if they are different types of string.
You can use a backslash to escape:
int dollars=42;
print("I have \$$dollars."); // I have $42.
When you are using literals instead of variables you can also use raw strings:
print(r"I have $42."); // I have $42.
I am currently learning c++/cli and I want to convert a character to its ASCII code decimal and vice versa( example 'A' = 65 ).
In JAVA, this can be achieved by a simple type casting:
char ascci = 'A';
char retrieveASCII =' ';
int decimalValue;
decimalValue = (int)ascci;
retrieveASCII = (char)decimalValue;
Apparently this method does not work in c++/cli, here is my code:
String^ words = "ABCDEFG";
String^ getChars;
String^ retrieveASCII;
int decimalValue;
getChars = words->Substring(0, 1);
decimalValue = Int32:: Parse(getChars);
retrieveASCII = decimalValue.ToString();
I am getting this error:
A first chance exception of type 'System.ArgumentOutOfRangeException' occurred in mscorlib.dll
Additional information: Input string was not in a correct format.
Any Idea on how to solve this problem?
Characters in a TextBox::Text property are in a System::String type. Therefore, they are Unicode characters. By design, the Unicode character set includes all of the ASCII characters. So, if the string only has those characters, you can convert to an ASCII encoding without losing any of them. Otherwise, you'd have to have a strategy of omitting or substituting characters or throwing an exception.
The ASCII character set has one encoding in current use. It represents all of its characters in one byte each.
// using ::System::Text;
const auto asciiBytes = Encoding::ASCII->GetBytes(words->Substring(0,1));
const auto decimalValue = asciiBytes[0]; // the length is 1 as explained above
const auto retrieveASCII = Encoding::ASCII->GetString(asciiBytes);
Decimal is, of course, a representation of a number. I don't see where you are using decimal except in your explanation. If you did want to use it in code, it could be like this:
const auto explanation = "The encoding (in decimal) "
+ "for the first character in ASCII is "
+ decimalValue;
Note the use of auto. I have omitted the types of the variables because the compiler can figure them out. It allows the code to be more focused on concepts rather than boilerplate. Also, I used const because I don't believe the value of "variables" should be varied. Neither of these is required.
BTW- All of this applies to Java, too. If your Java code works, it is just out of coincidence. If it had been written properly, it would have been easy to translate to .NET. Java's String and Charset classes have very similar functionality as .NET String and Encoding classes. (Encoding to the proper term, though.) They both use the Unicode character set and UTF-16 encoding for strings.
More like Java than you think
String^ words = "ABCDEFG";
Char first = words [0];
String^ retrieveASCII;
int decimalValue = ( int)first;
retrieveASCII = decimalValue.ToString();