Handling of Chracter in Java -


this question has answer here:

i came across case in java prints character ascii. if try print 2 character prints sum of there ascii value.

system.out.println('c'); =>> c system.out.println('a'+'b'); =>> 195 (97+98)

just want know why in second case prints sum of there ascii value

behind scenes, compiler acting smart , replacing char + char int constant value.

in first case println(char) called , in second case println(int) called

sample code :

public static void main(string[] args) {     system.out.println('a');     system.out.println('a' + 'b'); // compiler first "resolves" expression 'a'+'b' (since char cannot added added ints) , tries call correct `println(int)` method. }

byte code:

public static void main(java.lang.string[]);    descriptor: ([ljava/lang/string;)v    flags: acc_public, acc_static    code:      stack=2, locals=1, args_size=1         0: getstatic     #16                 // field java/lang/system.out:ljav /io/printstream;         3: bipush        97                 // single char. (pushed byte int)         5: invokevirtual #22                 // method java/io/printstream.prin ln:(c)v         8: getstatic     #16                 // field java/lang/system.out:ljav /io/printstream;        11: sipush        195                // push sum of 2 chars short        14: invokevirtual #28                 // method java/io/printstream.prin ln:(i)v        17: return

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