Hexa / Octa Conversions.

[Shivani Chandna]

On don's sites questions like these puzzle me....
Are we supposed to do the conversions and check the decimal value each time such questions are posed :roll: ( is there a shorter method ) - And same goes for unicode values assigned to a char> how do we ascertain if its assignable or not?
===========
<Find errors in the lines below >

 char c1 = 0xffff;    // 1 OK
 char c2 = 0xfffff; // 2 ERROR
 byte b1 = 0xffff;  // 3 ERROR
 byte b2 = 0x7f;    // 4 ERROR
 byte b3 = 0xff;    // 5 OK
 byte b4 = -0x80;   // 6 OK

or below:

 char a = 061;      // 1
   char b = '\61';    // 2
   char c = '\061';   // 3
   char d = 0x0031;   // 4
   char e = '\u0031'; // 5

Thank you so much..
Shivani

[Roy Simon]

hi Shivani

well before doing for the conversion to decimal you can check for the size of the assigned operand
//first case
char c1 = 0xffff;
a character is an unsigned 16 bit integer
every hexadecimal bit represents 4 binary bit's (f==1111)
hence char c1=0xffff is valid
//second case
char c2 = 0xfffff;
the value being assigned is 20 bit long and a char can accept on 16 bits
hence u get the error
//third case
byte b1 = 0xffff;
byte is a 8 bit number and the above code tries to assign a 16 bit number to it hence you get the error
//forth case
well this is tricky and i am still working on as to why did u get an error
cos 0x7f is 0111 1111 = 127
are you absolutely sure you got an error???
the other two are fine

you can use this to eliminate options and to be sure with the remaining i guess you have to do it the hard way


Regards
Simon

[Michael Ernest]

char c1 = 0xffff; // 1 OK
char c2 = 0xfffff; // 2 ERROR
byte b1 = 0xffff; // 3 ERROR
byte b2 = 0x7f; // 4 ERROR
byte b3 = 0xff; // 5 OK
byte b4 = -0x80; // 6 OK

I can see how c2, b1, and b3 (the ones in boldface) have problems; the rest are fine.

In each problem case, the value assigned would lose precision if the compiler allowed it. The char type is 16 bits wide. A hex number requires 4 bits for each column in the value. That is, 0xfffff requires 20 bits; the char type can only hold 16 of them. It's a bad fit.

The byte type can only hold 7 bits of value, since the most significant, or high-order, bit is used to express positive or negative value. So you can fit in 0x7f but not 0xff.
[ September 06, 2005: Message edited by: Michael Ernest ]

[Bert Bates]

From the broken record dept:

Remember - this topic applies only to the SCJP 1.4 exam... this topic is NOT on the SCJP 5.0 exam...

carry on

[shanthisri mocherla]

Hi Shivani
byte XX = 0xff would result in compiler error.
because 0xff =255. 0x7f = 127.
127 is within the range.but 255 is not within the range of byte i.e -128 to 127.
0xff = 1111 1111;
0x7f = 0111 1111;

Can Anyone explain why -1 and 255 have same binary number 1111 1111???

Shanthi