I'm working on a password validation routine, and am surprised to find that VB does not consider '*' to be a symbol per the Char.IsSymbol() check.
Here is the output from the QuickWatch:
char.IsSymbol("*") False Boolean
The MS documentation does not specify what characters are matched by IsSymbol, but does imply that standard mathematical symbols are included here.
Does anyone have any good ideas for matching all standard US special characters?
Characters that are symbols in this context: UnicodeCategory.MathSymbol, UnicodeCategory.CurrencySymbol, UnicodeCategory.ModifierSymbol and UnicodeCategory.OtherSymbol from the System.Globalization namespace. These are the Unicode characters designated Sm, Sc, Sk and So, respectively. All other characters return False.
From the .Net source:
internal static bool CheckSymbol(UnicodeCategory uc)
{
switch (uc)
{
case UnicodeCategory.MathSymbol:
case UnicodeCategory.CurrencySymbol:
case UnicodeCategory.ModifierSymbol:
case UnicodeCategory.OtherSymbol:
return true;
default:
return false;
}
}
or converted to VB.Net:
Friend Shared Function CheckSymbol(uc As UnicodeCategory) As Boolean
Select Case uc
Case UnicodeCategory.MathSymbol, UnicodeCategory.CurrencySymbol, UnicodeCategory.ModifierSymbol, UnicodeCategory.OtherSymbol
Return True
Case Else
Return False
End Select
End Function
CheckSymbol is called by IsSymbol with the Unicode category of the given char.
Since the * is in the category OtherPunctuation (you can check this with char.GetUnicodeCategory()), it is not considered a symbol, and the method correctly returns False.
To answer your question: use char.GetUnicodeCategory() to check which category the character falls in, and decide to include it or not in your own logic.
If you simply need to know that character is something else than digit or letter,
use just
!char.IsLetterOrDigit(c)
preferably with
&& !char.IsControl(c)
Maybe you have the compiler option "strict" of, because with
Char.IsSymbol("*")
I get a compiler error
BC30512: Option Strict On disallows implicit conversions from 'String' to 'Char'.
To define a Character literal in VB.NET, you must add a c to the string, like this:
Char.IsSymbol("*"c)
IsPunctuation(x) is what you are looking for.
This worked for me in C#:
string Password = "";
ConsoleKeyInfo key;
do
{
key = Console.ReadKey(true);
// Ignore any key out of range.
if (char.IsPunctuation(key.KeyChar) ||char.IsLetterOrDigit(key.KeyChar) || char.IsSymbol(key.KeyChar))
{
// Append the character to the password.
Password += key.KeyChar;
Console.Write("*");
}
// Exit if Enter key is pressed.
} while (key.Key != ConsoleKey.Enter);
Related
In this Hangman game how can I give the condition to check if input !=Char? It says that Kotlin: Operator '!=' cannot be applied to 'String?' and 'Char.Companion'
How can I solve this issue?
while (letters != correctGuesses) {
printExploredWord(word, correctGuesses)
println("\n#Wrong guesses: $fails\n\n")
print("Guess letter:")
val input = readLine()
if (input == null) {
continue
} else if (input.length != 1) {
println("Please enter one letter")
continue
} else if (input != Char) {
println("Please enter a character")
}
if (word.toLowerCase().contains(input.toLowerCase())) {
correctGuesses.add(input[0].toLowerCase())
} else {
++fails
}
Sounds from your comments that what you want to check is if the input String? has a single alphabetic character. You need to be precise with your terminology. Numbers and punctuation are also made up of characters. Char is a class representing any element of a String, so it doesn't make sense to be asking if something in a String is a Char because the answer is true no matter what.
The question you need to be asking is whether the first character in the given String is a letter. There's a function for that: Char.isLetter(). And since we're checking the content of the first character of the String, we need to get its value with input[0] because it doesn't make sense to ask if a whole String is a letter character. A String is never a Char because these are different classes. So in your case you would use:
if (input == null) {
continue
} else if (input.length != 1) {
println("Please enter one letter")
continue
} else if (!input[0].isLetter()) {
println("Please enter a character")
continue
}
But again, the terminology is wrong here. You should be reminding the user to enter a letter, not a character.
Your input variable is a type String?, so there's no reason to check if it's a char or not. To check for instanceof in Kotlin you use the "is" operator. So to compare if your input is a Char you can do,
else if(!(input is Char)) {
//This is unnecessary though since you're already checking if the length of the input isn't 1
}
How can I implement functions that take an optional flag in Perl6? For example, say that I want to invoke my function like this:
format 'a b c';
or like this:
format :pretty 'a b c';
How can I do this? Thanks
It's just a named argument, in case of flags a boolean one. This all works out because :pretty is syntactic sugar for the pair :pretty(True) aka pretty => True.
You can either use the boolean value
sub format($arg, Bool :$pretty = False) {
if $pretty { ... }
else { ... }
}
or use its presence for multi-dispatch
multi format($arg) { ... }
multi format($arg, Bool :$pretty!) { ... }
In the first example, we provided a default value (which isn't really necessary as the undefined value boolifies to False, but it's arguably the 'right thing to do' semantically), in the second one we made it a required parameter by adding the !.
Also note that named arguments still have to be separated by commas, ie you'd use it as
format :pretty, 'a b c';
If you really want that odd syntax, you can use an operator and some subsignature magic. The Bool method is optional and the class Pretty can be empty. It's just there to provide something for the MMD-dispatcher to hold onto.
class Pretty { method Bool {True} };
sub prefix:<:pretty>(|c){ Pretty.new, c };
multi sub format((Pretty:D $pretty, |a)){ dd $pretty.Bool, a };
multi sub format(|c){ dd c };
format :pretty 'a b c'; format 'a b c';
# OUTPUT«Bool::True\(\("a b c"))\("a b c")»
I'm using list labels to gather tokens and semantic predicates to validate sequences in my parser grammar. E.g.
line
:
(text+=WORD | text+=NUMBER)+ ((BLANK | SKIP)+ (text+=WORD | text+=NUMBER)+)+
{Parser.validateContext(_localctx)}?
(BLANK | SKIP)*
;
where
WORD: [\u0021-\u002F\u003A-\u007E]+; // printable ASCII characters (excluding SP and numbers)
NUMBER: [\u0030-\u0039]+; // printable ASCII number characters
BLANK: '\u0020';
SKIP: '\u0020\u0020' | '\t'; // two SPs or a HT symbol
The part of Parser.validateContext used to validate the line rule would be implemented like this
private static final boolean validateContext(ParserRuleContext context) {
//.. other contexts
if(context instanceof LineContext)
return "<reference-sequence>".equals(Parser.joinTokens(((LineContext) context).text, " "));
return false;}
where Parser.joinTokens is defined as
private static String joinTokens(java.util.List<org.antlr.v4.runtime.Token> tokens, String delimiter) {
StringBuilder builder = new StringBuilder();
int i = 0, n;
if((n = tokens.size()) == 0) return "";
builder.append(tokens.get(0).getText());
while(++i < n) builder.append(delimiter + tokens.get(i).getText());
return builder.toString();}
Both are put in a #parser::members clause a the beginning of the grammar file.
My problem is this: sometimes the _localctx reference is null and I receive "no viable alternative" errors. These are probably caused because the failing predicate guards the respective rule and no alternative is given.
Is there a reason–potentially an error on my part–why _localctx would be null?
UPDATE: The answer to this question seems to suggest that semantic predicates are also called during prediction. Maybe during prediction no context is created and _localctx is set to null.
The semantics of _localctx in a predicate are not defined. Allowable behavior includes, but is not limited to the following (and may change during any release):
Failing to compile (no identifier with that name)
Using the wrong context object
Not having a context object (null)
To reference the context of the current rule from within a predicate, you need to use $ctx instead.
Note that the same applies for rule parameters, locals, and/or return values which are used in a predicate. For example, the parameter a cannot be referenced as a, but must instead be $a.
I need some help here, I am currently making a game, but I got stuck somewhere. So, what I want is, if a Labels text is higher then the other labels text, then something will happen, I typed If Label26.Text > Label24.Text Then Label33.Visible = True which seems not to work, please, I need some help here, thanks. And yes, the labels text is NUMBERS.
The Text property of a label is a string. As far as computers go, you can't do math (using comparison operators like > will not return the result you are expecting) with strings because they are just a sequence of characters.
Even if the string only contains a number, the computer still sees it as a sequence of characters and not a number ("5" is a string literal with the character 5 in it, while 5 is an integer that can be used in a mathematic expression).
As some of the other commenters mentioned, you need to cast the Text property to an Integer or Double (or some other numeric data type). To do so, you'd want to use Int32.Parse to change the strings to integers.
If Int32.Parse(Label26.Text) > Int32.Parse(Label24.Text) Then Label33.Visible = True
You can use the int.tryParse to check if the content of the variable is a number or not. The output of the TryParse is a boolean, see the example below:
int num1 = 0;
bool num1_ = false;
num1_ = int.TryParse(txt1.Text.ToString(), out num1);
if (num1_)
{
// Is a number/integer
//Do something
}
else
{
//Is a string
//Do something else
}
Here's what I'm trying to do. A user can type in a search string, which can include '*' or '?' wildcard characters. I'm finding this works with regular strings but not with ones including numeric characters.
e.g:
414D512052524D2E535441524B2E4E45298B8751202AE908
1208
if I look for a section of that hex string, it returns false. If I look for "120" or "208" in the "1208" string it fails.
Right now, my regular expression pattern ends up looking like this when a user enters, say "w?f": '\bw.?f\b'
I'm (obviously) not well-versed in regular expressions at the moment, but would appreciate any pointers someone may have to handle numeric characters in the way I need to - thanks!
Code in question:
/**
*
* #param searchString
* #param strToBeSearched
* #return
*/
public boolean findString(String searchString, String strToBeSearched) {
Pattern pattern = Pattern.compile(wildcardToRegex(searchString));
return pattern.matcher(strToBeSearched).find();
}
private String wildcardToRegex(String wildcard){
StringBuffer s = new StringBuffer(wildcard.length());
s.append("\\b");
for (int i = 0, is = wildcard.length(); i < is; i++) {
char c = wildcard.charAt(i);
switch(c) {
case '*':
s.append(".*");
break;
case '?':
s.append(".?");
break;
default:
s.append(c);
break;
}
}
s.append("\\b");
return(s.toString());
}
Let's assume your string to search in is
1208
The search "term" the user enters is
120
The pattern then is
\b120\b
The \b (word boundary) meta-character matches beginning and end of "words".
In our example, this can't work because 120 != 1208
The pattern has to be
\b.*120.*\b
where .* means match a variable number of characters (including null).
Solution:
either add the .*s to your wildcardToRegex(...) method to make this functionality work out-of-the-box,
or tell your users to search for *120*, because your * wildcard character does exactly the same.
This is, in fact, my preference because the user can then define whether to search for entries starting with something (search for something*), including something (*something*), ending with something (*something), or exactly something (something).