Add "."(dot) in a Numbers

Add "."(dot) in a Numbers

By : Kurt
Date : November 22 2020, 10:56 AM
help you fix your problem See the number_format function,
code :
$number = number_format($number, 0, '', '.');
// change 70000000 to 70.000.000 etc.

// ex. with a number set 
$numbers = array(70000000, 75000000, 300000);
foreach ($numbers as $number) {
    echo number_format($number, 0, '', '.') . '<br>';

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Bizarre issue with printf in bash script:"09" and "08" are invalid numbers, "07" and "

Bizarre issue with printf in bash script:"09" and "08" are invalid numbers, "07" and "

By : Eric Lee
Date : March 29 2020, 07:55 AM
To fix this issue This is my bash script - I just want to left-pad a set of numbers with zeroes: , If you have your "09" in a variable, you can do
code :
echo "$a"
echo "${a#0}"
printf "%04d" "${a#0}"
How restrict textbox in C# to only receive numbers and (dot "." or comma ","), after "." o

How restrict textbox in C# to only receive numbers and (dot "." or comma ","), after "." o

By : Kazek Ciaś
Date : March 29 2020, 07:55 AM
I hope this helps . Try this code ! I hope this helps. Let me know if I can assist you further.
This is auxiliary function i have written
code :
private bool alreadyExist(string _text , ref char KeyChar)
            if (_text.IndexOf('.')>-1)
                KeyChar = '.';
                return true;
            if (_text.IndexOf(',') > -1)
                KeyChar = ',';
                return true;
            return false;
 private void txtValormetrocubico_KeyPress(object sender, KeyPressEventArgs e)
            if (!char.IsControl(e.KeyChar)
                    && !char.IsDigit(e.KeyChar)
                    && e.KeyChar != '.' && e.KeyChar != ',')
                e.Handled = true;

            //check if '.' , ',' pressed
            char sepratorChar='s';
            if (e.KeyChar == '.' || e.KeyChar == ',')
                // check if it's in the beginning of text not accept
                if (txtValormetrocubico.Text.Length == 0) e.Handled = true;
                // check if it's in the beginning of text not accept
                if (txtValormetrocubico.SelectionStart== 0 ) e.Handled = true;
                // check if there is already exist a '.' , ','
                if (alreadyExist(txtValormetrocubico.Text , ref sepratorChar)) e.Handled = true;
                //check if '.' or ',' is in middle of a number and after it is not a number greater than 99
                if (txtValormetrocubico.SelectionStart != txtValormetrocubico.Text.Length && e.Handled ==false)
                    // '.' or ',' is in the middle
                    string AfterDotString = txtValormetrocubico.Text.Substring(txtValormetrocubico.SelectionStart);

                    if (AfterDotString.Length> 2)
                        e.Handled = true;
            //check if a number pressed

            if (Char.IsDigit(e.KeyChar))
                //check if a coma or dot exist
                if (alreadyExist(txtValormetrocubico.Text ,ref sepratorChar))
                    int sepratorPosition = txtValormetrocubico.Text.IndexOf(sepratorChar);
                    string afterSepratorString = txtValormetrocubico.Text.Substring(sepratorPosition + 1 );
                    if (txtValormetrocubico.SelectionStart > sepratorPosition && afterSepratorString.Length >1)
                        e.Handled = true;


printf with "%d" of numbers starting with 0 (ex "0102") giving unexpected answer (ex '"66"

printf with "%d" of numbers starting with 0 (ex "0102") giving unexpected answer (ex '"66"

By : Janete Sales de Oliv
Date : March 29 2020, 07:55 AM
I hope this helps you . This is because when the first digit of a number (integer constant) is 0 (and second must not be x or X), the compiler interprets it as an octal number. Printing it with %d will give you a decimal value.
To print octal value you should use %o specifier
code :
   printf("%o", n);  
 12  125  3546  
 012 0125 03546  
 0xf 0xff 0X5fff   
Why does Array("1","2","3").join() convert strings to numbers in JavaScript?

Why does Array("1","2","3").join() convert strings to numbers in JavaScript?

By : evenagainstthewrold
Date : November 22 2020, 10:33 AM
it helps some times It doesn't. Quotes around strings are a notation thing, telling the system that you want to create a string; they're not part of the string itself. So when you join ["1", "2", "3"], you get "123", because that's the content of the three strings. It happens to look the same as if they'd been numbers, but this is only a coincidence.
There are a couple of ways to add quotes to strings. In general, you need to escape them in some way. The easiest way to do this in JavaScript is to put a backslash (\) in front of the character you want to escape, like this:
code :
Text to speech pronouncing numbers like "4th", "8ths", or "2nd"

Text to speech pronouncing numbers like "4th", "8ths", or "2nd"

By : user1803328
Date : March 29 2020, 07:55 AM
I hope this helps you . This sort of behavior is going to vary between TextToSpeech Engines -- the Google TTS engine, for example, will behave differently than, say, the SVOX PICO (emulator < API 24) engine... so it's not your fault how each engine behaves slightly differently... and if there are any pronunciation controls, then the engine is responsible for supplying them directly to the end user via settings.
You're probably just testing on a different engine than you were before... or even an update to the same engine.
code :
private ArrayList<String> whatEnginesAreInstalled(Context context) {
    final Intent ttsIntent = new Intent();
    final PackageManager pm = context.getPackageManager();
    final List<ResolveInfo> list = pm.queryIntentActivities(ttsIntent, PackageManager.GET_META_DATA);
    ArrayList<String> installedEngineNames = new ArrayList<>();
    for (ResolveInfo r : list) {
        String engineName = r.activityInfo.applicationInfo.packageName;

        // just logging the version number out of interest
        String version = "null";
        try {
            version = pm.getPackageInfo(engineName,
            } catch (Exception e) {
                Log.i("XXX", "try catch error");
        Log.i("XXX", "we found an engine: " + engineName);
        Log.i("XXX", "version: " + version);
    return installedEngineNames;
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