To repeat what everyone is saying... The value of c is assigned the decimal value of 1011111. Not the binary value of 1011111. Even if you convert it to string, using Integer.toString(), you are converting the wrong value.
IS THERE ANY WAY TO INITIALISE THE NUMBERS AS BINARY ??(i think it is not there in Java)
I know this seems to be rubbing salt on the wound, but you don't seem to understand that this is more your issue, than Java's. There is no such thing as "INITIALISE THE NUMBERS AS BINARY" -- integers are in the integer format.
Decimal, Octal, Hexidecimal, are literal formats used to help the programmer specify that integer value. In your case, you want to specify a literal value -- and it is that, that you want in binary.
If there was a huge problem with mentally converting a binary (literal) value to hexidecimal or octal -- then programmers would have pushed to have binary literals added a long time ago.
Quite frankly, it takes seconds to mentally convert a binary literal value to its octal or hexidecimal counterpart. And you don't need a calculator.
110110101 --> 110 110 101 --> 0665
110110101 --> 1 1011 0101 --> 0x1B5
But... To answer your question. There is currently no easy way to specify a integer literal as a binary in Java. So... if you absolutely need to have it done as a binary integer literal, then you can't do it. Sorry.
finally i have got the solution for my problem ... bitWiseCheck = Integer.parseInt(Integer.toBinaryString((Integer.parseInt(Integer.toString(binVal[j]),2)) & (Integer.parseInt(Integer.toString(vsVal),2))));
I don't think so. Try your code for -3 and -2 for instance, or 16 and 15.