为什么如果我使用 int 会打印 ASCII 字符，而如果我使用变量则不会？

Question

The following code prints A to the console:

char ch = (char)65.25;

System.out.println(ch);

However I don't understand why the next piece of code doesn't print the ASCII character?

int rand = (int)currentSeconds % 26;

System.out.println(" the number is " + rand);

char randN = (char)rand;

System.out.println(randN);

Below follows the full code:

System.out.println(System.currentTimeMillis());
System.out.println(" using current second to  generate random Uppercase letter");

long totalsecs = System.currentTimeMillis()/1000;
long currentSeconds = totalsecs % 60;
long totalMinute = totalsecs/60;
long currentMinute = totalMinute%60;
long totalHours = totalMinute/60;
long currentHours = totalHours%24;

System.out.println("current time is " +currentHours +":"+ currentMinute +":"+currentSeconds);


int rand = (int)currentSeconds % 26;
System.out.println(" the number is " + rand);
char randN = (char)rand;

System.out.println(randN);

Answer 1

There are many versions of the println method, with different parameter types. In other words, if s is a String , i is an int and c is a char , then System.out.println(i) is actually calling a different method from System.out.println(c) , which is calling a different method from System.out.println(s) . It's the compiler's job to figure out which method you're trying to call, and it takes into account the type of the expression that you pass in. So

• System.out.println(rand) calls the int version of the method, which prints the number.
• System.out.println(randN) calls the char version of the method, which prints the character, rather than its numeric equivalent.
• System.out.println("the number is " + rand) calls the String version of the method, which prints a String , and in this case, the String is made by concatenating the number to the end of a String literal.

If you're trying to print a character value in the range A to Z in place of a number from 0 to 25, then writing char randN = (char) rand; isn't enough - that won't actually convert the number to the range you want, because the numeric values of the characters from A to Z are not actually 0 to 25.

You need to write char randN = 'A' + rand; to effect the desired conversion.

Answer 2

tl;dr

Character.toString
( 
    "A".codePointAt( 0 ) 
    +
    ThreadLocalRandom.current().nextInt( 0 , 27 )
)

Or:

Character.toString
( 
    ThreadLocalRandom.current().nextInt( 65 , ( 65 + 27 ) )
)

Start with A

As others pointed out, to get a random character from AZ, you need to add 0-25 to a starting number, that number being the code for A which is 65.

Avoid char

Another problem with your code is its use of the char . The char type in Java is legacy, essentially broken. As a 16-bit value, that type cannot represent most characters.

While your particular case of AZ works with char , using char is a bad habit, and an unnecessary habit.

Code points

Instead learn to use Unicode code point integer numbers. Unicode is a superset of US-ASCII, so code points 0-127 represent the same characters in both.

int codePointForA = "A".codePointAt( 0 ) ;  // 65
int addend = ThreadLocalRandom.current().nextInt( 0 , 27 ) ;  // ( inclusive , exclusive )
int randomCodePoint = ( codePointForA + addend ) ;
String randomCharacter = Character.toString( randomCodePoint ) ;

See code run live at IdeOne.com .

Z