java - Binary, hex, decimal comparison -

i playing bit numbers, , interesting came upon me, don't quite understand.

 public static void main(string[] args) {          int hexnumber = 0x7a;//decimal: 122  binary:0111 1010          int decnumber = 122;          int binnumber = 1111010;     system.out.println(hexnumber);//122    system.out.println(integer.tostring(hexnumber, 16)); //7a    system.out.println(integer.tohexstring(hexnumber));  //7a    system.out.println(integer.tostring(hexnumber, 2)); //   1111010    system.out.println(integer.tobinarystring(hexnumber)); //1111010     system.out.println(hexnumber==binnumber);//false    system.out.println(hexnumber==decnumber);//true    system.out.println(decnumber==binnumber);//false     } 

why "false" @ #1 , #3? doesn't change if binnumber = 01111010;

well, can't directly store binary values in java without prefix.

binnumber isn't stored binary number 1111010; instead, it's stored decimal number 1111010. have store int binnumber = integer.parseint("1111010", 2); or better yet int binnumber = 0b1111010;.

for octal:

int octalno = 0177; //'0' prefix

or

int octalno = integer.parseint("0177", 8); //leading '0's ignored

for hexadecimal:

int hexno = 0x177; //'0x' prefix

or

int hexno = integer.parseint("0177", 16); //leading '0's ignored

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for more info, have @ this.