how to get two decimal places in java

DecimalFormat ("#.00"); 3 String str = Df.format (d); 4 System.out.println (str);2.%.2f means that after two bits are retained, it is handled in a way that is also directly truncated and not rounded.1 double d = 114.145; 2 String.Format ("%.2f", D);3.roundingmode.half_down six, negative number first to take the absolute value and then five six into the negative, roundingmode.half_up: for rounding, negative number first to take the absolute value of five six into the negative.1 double d = 114.14

Computer Question (intermediate)-output the frequency of occurrence of the specified symbol in decimal form (Java) The questions are as follows: The Code is as follows: Package huawei; import java. math. bigDecimal; public final class Demo {/** function: count the frequency of occurrence of a specified character in the input benchmark string. The string on

Excerpt from Http://irobot.iteye.com/blog/285537Java two decimal places (four methods) oneLong is the length of the type, how to have a decimal, is double barJava.text.DecimalFormat df=new Java.text.DecimalFormat ("#.##");Double d=3.14159;System.out.println (Df.format (d));TwoJava.math.BigDecimalBigDecimal bd = new BigDecimal ("3.14159265");BD = Bd.setscale (2,BI

Java. text. DecimalFormat class decimal format, The specific code of the java. text. DecimalFormat class in decimal format is for your reference. The specific content is as follows: Import java. text. decimalFormat; class FormatDemo {public void format1 (String pattern, doub

() { theSystem.out.println ("Mathutil.scale (\" 2.23956\ ", 3) \ t" + mathutil.scale ("2.23956", 3)); -System.out.println ("Mathutil.scale (\" 2.23956\ ", 0) \ t" + mathutil.scale ("2.23956", 0)); -System.out.println ("Mathutil.scale" (\ "2.23956\",-3) \ t "+ mathutil.scale (" 2.23956 ",-3)); - } +}At this point, Java Learning -047-numeric formatting and the number of decimal rounding successfully comp

The following two methods are recommended: The first method is to use the java. Text. decimalformat class: For example, decimalformat df = new decimalformat ("#.##"); Double D = 1.100000001; System. Out. println (DF. Format (d )); Result: 1.10 The second method is to use Java. Math. bigdecimal Bigdecimal BD = new bigdecimal ("1.00100010 "); BD = BD. setscale (2, bigdecimal. round_half_up); /

Because of the needs of personal applications, it is easy to write. I hope everyone will give suggestions and learn together. For more information, seeCopy codeThe Code is as follows: Import java. text .*;Import java. math .*;Public class Test{Public Test (){Double a = saveNumber (15.12312312, 6 );System. out. println ("a..." + );}// Retain the number of decimal

) >>> 16); // move the 8-bit height to 0, and then shift the 16-bit value to the right. Sb. append ("."); Sb. append (string. valueof (longip 0x0000ffff) >>> 8 )); Sb. append ("."); Sb. append (string. valueof (longip 0x000000ff )); Sb. append ("."); Returnsb. tostring (); } Publicstaticvoidmain (string [] args) { System. out. println ("ip address representation: rn "); System. out. print ("32-bit binary format :"); System. out. println (long. tobinarystring (3366362403l )); System. out. print

. valueof (longip 0x00ffffff) >>> 16); // Move the 8-bit height to 0, and then shift the 16-bit value to the right.Sb. append (".");Sb. append (string. valueof (longip 0x0000ffff) >>> 8 ));Sb. append (".");Sb. append (string. valueof (longip 0x000000ff ));Sb. append (".");Return sb. tostring ();} Public static void main (string [] args){ System. out. println ("IP address representation: rn ");System. out. print ("32-bit binary format :");Syste

height to 0, and then shift the 16-bit value to the right.Sb. append (".");Sb. append (string. valueof (longip 0x0000ffff) >>> 8 ));Sb. append (".");Sb. append (string. valueof (longip 0x000000ff ));Sb. append (".");Return sb. tostring ();}Public static void main (string [] args){System. out. println ("ip address representation: rn ");System. out. print ("32-bit binary format :");System. out. println (long. tobinarystring (3366362403l ));System. out. print ("

Some small algorithms are of the Java version, and a large number of questions on the network are aimed at C ++. Therefore, Java implementation is rare, but they are the basis of the test, The implementation is the same. You can broaden your thinking and be helpful. /*** 1 hop steps* Question: there are N levels in a step. If you can skip 1 level at a time, you can also skip 2 levels.* Calculate the total

split () is a string parameter in parentheses, so be sure to conform to this: Split (".") form, that is, the point '. ' Enclose them in double quotation marks """." is already defined as a special character in Java, so you want to use "." In Split (). It must be escaped in order to avoid ambiguity, i.e. in the use of "." "\ \" is required for string segmentation.such as: String myString = "Myblog.cnblogs";Mystring.split ("\ \");

split () is a string parameter in parentheses, so be sure to conform to this: Split (".") form, that is, the point '. ' Enclose them in double quotation marks """." is already defined as a special character in Java, so you want to use "." In Split (). It must be escaped in order to avoid ambiguity, i.e. in the use of "." "\ \" is required for string segmentation.such as:String myString = "Myblog.cnblogs";Mystring.split ("\ \");

For example: 5 of the binary conversion method is 5 divided by 2 is the remainder 1 and then the quotient value 2 continues to divide by 2 the remainder 0 and then the quotient value is 1, so the last to get the binary value is 101Public static void Main (string[] args) {StringBuffer SBF = ToBin (ten);String Str=sbf.reverse (). toString ();System.out.println (str); }static StringBuffer toBin (int x) {stringbuffer result=new stringbuffer ();do{resul

This paper mainly introduces 3 kinds of methods of obtaining random number in Java, mainly using random () function to realize Method 1 (data type) (Minimum +math.random () * (maximum-min +1)) Example: (int) (1+math.random () * (10-1+1)) from 1 to 10 int type with number Method 2 Get random numbers for (int i=0;i Through Java. The ra

First understand, ClassPath: refers to: web-inf\classes This directory, usually put the file in the SRC folder, or create a new file and add to the build path. The IDE automatically places files under classes when synchronizing files to Tomcat.Get the code for the properties:Public Properties loadProperties () {Properties Properties = new Properties (); InputStream stream = This.getclass (). getClassLoader (). getResourceAsStream ("/xxx.properties");

are correctly recognized as numerals, including decimal form "3.0 ". English and messy symbols (including the invisible line break, where did you find it ?) It is classified as one type --/nx! (Because I Don't Know What ICTCLAS calls it. It's invalid or invalid, or other characters. Can I name it by myself) There are two exclamation points in the test text. One is half-width English !, One is full of Chinese Characters !, Both are correctly recogniz

properties through an entity? The answer is--reflex! Public List querybyentity (Object model,string cont) throws Nosuchmethodexception, Illegalaccessexception, IllegalArgumentException, invocationtargetexception{mapwith this code, we have been able to get all the non-null attributes in the entity, and have it in the map. After we get to these, we can make the search very convenient. Summarize:The reflectio

quickly. When control is returned from a method call, the virtual machine has tried its best to reclaim space from all discarded objects . That is, the gc Java Virtual machine must be garbage collected when it is called. So why is it not necessarily done on the Internet or even in many textbooks gc ? Personal speculation may be that after running the gc method, memory consumption may not be significantly reduced, Because at this time there may n

Method 1(data type) (min. +math.random () * (max.-min. +1))Cases:(int) (1+math.random () * (10-1+1))int type with number from 1 to 10Method 2Get random numbersfor (int i=0;i{System.out.println ((int) (1+math.random () *10));}(int) (1+math.random () *10)Through Java. The random method of the math package gets a 1-10 int numberThe formula is: the minimum---The maximum (integer) random number(type) Minimum value +math.random () * Maximum valueMethod 3Ran