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B = "a" + "true";System.out.println ((A = = b)); result = TrueString a = "a3.4";String B = "a" + 3.4;System.out.println ((A = = b)); result = TrueAnalysis: JVM for string constant "+" number connection, the program compile period, the JVM will be the constant string "+" connection optimization to the concatenated value, take "a" + 1, the compiler is optimized in class is already A1. The value of its string constants is determined at compile time, so the final result of the above program is true

Today's content:Knowledge Point one: RegularWhat is a regular:is to use a series of characters with special meanings to form a set of rules that are used to describe a character string.The regular is used to remove a small string of rules in a large string.Why use the Regular:1. User Registration2. Crawler programsHow to use:Re.findallRe.findall module:\w letters, numbers, underscoresPrint (Re.findall (' \w ', ' yangzz:age_18 '))\w no letters, numbers, underscoresPrint (Re.findall (' \w ', ' yan

reference "value" (that is, the memory address), which forces the reference to point only to the object that was initially pointed to, and changes its point to cause a compile-time error.Final is not responsible for the change in the object it points to. several examples of string constant pool problems Here are a few common examples of comparative analysis and understanding:String A = "A1";String B = "a" + 1;System.out.println ((a = = b));//result = TrueString a = "atrue";String B = "a" + "tru

= 1000000000000000L; System.out.println(l); //定义float类型的变量 float f = 12.34F; System.out.println(f); //定义double类型的变量 double d = 12.34; System.out.println(d); //定义char类型的变量 char c = ‘a‘; System.out.println(c); //定义boolean类型的变量 boolean bb = false; System.out.println(bb); }}Considerations for 1.1.1 Variable Definition

PHP does not seem to be the same as the ASP "/" to represent the root directory, replaced by $_server[' Document_root ', others are the same: ... /represents an upward layer./represents the current layer. If a/b/c/s.php now wants to invoke the/bb/s2.txt under the root directory, then: $RootDir = $_server[' Document_root '];$fireDir = "$RootDir/bb/s2.txt"; Or: ". /.. /.. /

: JVM for string constant "+" connection, the program compile period, the JVM will be the constant string of "+" connection optimization to the value after the connection, take "a" + 1, after the compiler optimization in class is already A1. The value of the string constant at compile time is determined, so the final result of the above program is true. Example 4: Java code String a = "AB"; String BB = "B"; String B = "a" +

First, awk uses ordinary variables in the shell Method 1: Use the quotation mark method Aa= "Hello" awk ' BEGIN {print ' $aa '} ' Output $AA awk ' BEGIN {print ' $aa '} ' Output Hello awk ' BEGIN {print ' ' $aa '} ' Output empty Aa= "hello! Coder " awk ' BEGIN {print ' $aa '} ' Error, double quotes do not match, because space appears in the middle awk ' BEGIN {print ' ' $aa ' '} ' Hello! can be displayed normally Coder Method 2: Using the-V method Awk-v varaa= "$aa" ' BEGIN {print Varaa}

assign it to the string variable S2. Judge S1+s2= "A" in the character table, then s1=s1+s2= "a" (see table 2, second row). 4) Read the next character a in the image data stream and assign it to the string variable S2. Judge S1+s2= "AA" is not in the string table, output s1= "a" in the word list of index 0H, and at the end of the string table for s1+s2= "AA" to add index 4H, and s1=s2= "a" (see table 2, the third row). 5) Read the next character B assigned to S2. Judge S1+s2= "AB" is not in the

= "A1"; String B = "a" + 1; System.out.println ((A = = b)); result = True String a = "atrue"; String B = "a" + "true"; System.out.println ((A = = b)); result = True String a = "a3.4"; String B = "a" + 3.4; System.out.println ((A = = b)); result = True Analysis: JVM for string constant "+" number connection, the program compile period, the JVM will be the constant string "+" connection optimization to the concatenated value, take "a" + 1, the compiler is optimized in class is already A1. The val

Some may ask questions like this. the expected result of the table's data in the following example is idaabb201734562506643898920.00002. of course, the environment is SQL Server's answer and analysis as follows: 1, many friends often ask questions similar to this, The data in the table is as follows: ID AA BB 1 123 456 1 4535 54 1 60 6564 1 60 656 2 50 664 2 60 6 3 89 4 40 4242 The expected result is: ID AA BB