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Firstly, consider an object with date value.Object arrObj[] = { "Date", Calendar.getInstance() };After that, use the String.format() method to format Calendar and display the date.The following is an example.Example Live Demoimport java.util.Calendar; public class Demo { public static void main(String []args){ Object arrObj[] = { "Date", Calendar.getInstance() }; System.out.println("Formatting Date..."); System.out.println(String.format("%1$s = %2$tY %2$tm %2$te", arrObj)); } }OutputFormatting Date... Date = 2018 11 17
The following are the conversion characters for date-time.CharacterDescriptioncComplete date and timeFISO 8601 dateDU.S. formatted date (month/day/year)T24-hour timer12-hour timeR24-hour time, no secondsYFour-digit year (with leading zeroes)yLast two digits of the year (with leading zeroes)CFirst two digits of the year (with leading zeroes)BFull month namebAbbreviated month namemTwo-digit month (with leading zeroes)dTwo-digit day (with leading zeroes)eTwo-digit day (without leading zeroes)AFull weekday nameaAbbreviated weekday namejThree-digit day of year (with leading zeroes)HTwo-digit hour (with leading zeroes), between 00 and 23kTwo-digit hour (without leading zeroes), between 0 and 23ITwo-digit hour (with leading zeroes), between 01 and 12lTwo-digit hour (without leading zeroes), between 1 and 12MTwo-digit minutes ... Read More
Let’s say the following is our string with special characters.String str = "test*$demo";Check for the special characters.Pattern pattern = Pattern.compile("[^A-Za-z0-9]"); Matcher match = pattern.matcher(str); boolean val = match.find();Now, if the bool value “val” is true, that would mean the special characters are in the string.if (val == true) System.out.println("Special characters are in the string.");Example Live Demoimport java.util.regex.Matcher; import java.util.regex.Pattern; public class Demo { public static void main(String []args) { String str = "test*$demo"; System.out.println("String: "+str); Pattern pattern = Pattern.compile("[^A-Za-z0-9]"); Matcher match = pattern.matcher(str); boolean val ... Read More
To delete all whitespaces from a string, use the replaceAll() method and replace every whitespace with empty.Let’s say the following is our string.String str = "This is it!";Now, let us replace the whitespaces that will eventually delete them.String res = str.replaceAll("\s+","");Example Live Demopublic class Demo { public static void main(String []args) { String str = "This is it!"; System.out.println("String: "+str); String res = str.replaceAll("\s+",""); System.out.println("String after deleting whitespace: "+res); } }OutputString: This is it! String after deleting whitespace: Thisisit!
INTEL 8085 has a very enriched Instruction Set. Varieties of instructions it can execute. Instructions will have different Byte counts, ranging from 1-Byte to 3-Bytes. Opcode always occupies 1-Byte in the memory. As we know that, with 8 bits for the opcode, 28 = 256 distinct opcodes are possible. In hexadecimal notation, the opcodes can range from 00H to FFH. Each opcode will correspond to an instruction. Thus from the calculation, it is possible to have 256 instructions in the instruction set of 8085. However, only 246 opcodes are implemented in 8085. They can be classified under 66 types, which ... Read More
In 8085 Instruction set, and specially in its arithmetic group of instructions, we have only add and subtract instructions. 8085 does not have instructions to perform multiplication or division numbers. Now let us discuss the instructions to perform addition operations only.To perform addition of two numbers, 8085 imposes the restriction that one of the operands must be kept in the Accumulator. The other operand can be at any one of the following possible locationsClassificationsExamplesThe other operand can be kept in 8-bit immediate data in the instruction.ADI 43HADI FFHThe other 8-bit operand can be kept in a memory location and whose ... Read More
Argument indices allow programmers to reorder the output. Let us see an example.Example Live Demopublic class Demo { public static void main(String[] args) { System.out.printf("Before reordering = %s %s %s %s %s %s", "one", "two", "three", "four", "five", "six" ); System.out.printf("After reordering = %6$s %5$s %4$s %3$s %2$s %1$s", "one", "two", "three", "four", "five", "six" ); System.out.printf("Before reordering = %d %d %d", 100, 200, 300); System.out.printf("After reordering = %2$d %3$d %1$d", 100, 200, 300); } }OutputBefore reordering = one two three four five six After reordering = ... Read More
Here is our character array.char[] ch = { 'T', 'E', 'S', 'T'};To create string object from the above character array is quite easy. Add the array to the string parameter as shown below −String str = new String(ch);Example Live Demopublic class Demo { public static void main(String[] args) { char[] ch = { 'T', 'E', 'S', 'T'}; String str = new String(ch); System.out.println(str); } }OutputTEST
Here is our character array.char[] ch = { 'T', 'E', 'S', 'T', 'I', 'N', 'G'};Create string object from some part of a string using the following String constructor. Through this we are fetching substring “IN” from the character array.String str = new String(ch, 4, 2);Example Live Demopublic class Demo { public static void main(String[] args) { char[] ch = { 'T', 'E', 'S', 'T', 'I', 'N', 'G'}; String str = new String(ch, 4, 2); System.out.println(str); } }OutputIN
To get a string from a subset of the character array elements, use the copyValueOf() method. This method returns a String that represents the character sequence in the array specified.Here is our character array.char[] ch = { 'T', 'E', 'S', 'T', 'I', 'N', 'G'};Now, let us create a string from the subset of the above array elements.String str = String.copyValueOf(ch, 4, 2);Example Live Demopublic class Demo { public static void main(String[] args) { char[] ch = { 'T', 'E', 'S', 'T', 'I', 'N', 'G'}; String str = String.copyValueOf(ch, 4, 2); System.out.println(str); } }OutputIN