Assignment of hex and binary literals

[nick woodward]

I'm a bit confused as to why some of these assignments are allowed, if anyone minds straightening me out please!

float f = 0b1000; // line 1
float f1 = 0b1000f; // line 2
float f2 = 0x1000f; // line 3

int f3 = 0x1000f; // line 4
                         

line 1: allowed as it's a binary representation of an integer literal widened to float
line 2: binary can't represent float literals? (enthuware)
line 3: hexadecimal can - so the f is a suffix, not a hexadecimal value?
line 4: but this line makes me think it's a hexadecimal value - how can a float literal be assigned to an int implicitly?

hopefullly this is outside the scope of the exam and i can go back to my revision

Nick

[Ramya R Subramanian]

I havent seen the enthuware question or answer.
I think line 2 doesnt work because, you cannot have values like' f' in a binary representation. you can have only 0 or 1.If you remove 'f' it should work.

float f1 = 0b1000;

and I think in line 3 and line 4,'f' must be hexadecimal value only. If we calculate the hexadecimal value for 0x1000f

hexadecimal 'f' represents 15. then ( here ^ represents power and not bitwise exclusive OR)

1*(16^4)+0*(16^3)+0*(16^2)+0*(16^1)+15*(16^0) = 65536+15=65551

If you print these values in line 3 and line 4. it gives

float f2 = 0x1000f; // line 3
		int f3 = 0x1000f; // line 4
		System.out.println("f2---"+f2);//65551.0 (because it is float)
		System.out.println("f3---"+f3);//65551 (because it is int)

[Roel De Nijs]

nick woodward wrote:I'm a bit confused as to why some of these assignments are allowed, if anyone minds straightening me out please!

I think you get a bit carried away with the additional complexity of the hex and binary literals and you are simply forgetting the basics

nick woodward wrote:line 1: allowed as it's a binary representation of an integer literal widened to float

Correct!

nick woodward wrote:line 2: binary can't represent float literals? (enthuware)

The suffix 'f' (or 'F') is not an allowed character in the binary representation. That's why you'll get a compiler error.

nick woodward wrote:line 3: hexadecimal can - so the f is a suffix, not a hexadecimal value?

It's similar with line1: an int value which is widened to a float. If you print the values, you'll notice that f is part of the value and it's not the suffix to indicate a float

float f2a = 0x10000;
float f2b = 0x1000f;
System.out.printf("%f %f%n", f2a, f2b); // 65536.000000 65551.000000

nick woodward wrote:line 4: but this line makes me think it's a hexadecimal value - how can a float literal be assigned to an int implicitly?

For exactly the same reason as line3: f is part of the value and it's not the suffix to indicate a float

float f2b = 0x1000f;
int f3 = 0x1000f;
System.out.printf("%f %d%n", f2b, f3);

Hope it helps!
Kind regards,
Roel

[Roel De Nijs]

Ramya Subraamanian wrote:I think line 2 doesnt work because, you cannot have values like' f' in a binary representation. you can have only 0 or 1.If you remove 'f' it should work.
...
and I think in line 3 and line 4,'f' must be hexadecimal value only.

On both cases you are spot-on! Have a cow for not only providing an excellent answer but also faster than me

[Joe Bishara]

Ramya Subraamanian wrote:...you can have only 0 or 1

Not quite right. An l or L suffix is allowed.

In Java SE 7, the integer types (byte, short, int, and long) can also be expressed using the binary number system.

byte b = 0b1000;
short s = 0b1000;
int i = 0b1000;
long l = 0b1000L;

[Roel De Nijs]

And I like to mention one very important tip: you can only express integer literals with the number systems binary, octal and hexadecimal (so no floating-point literals).

It seems the suffix 'l' (or 'L') is allowed in any of the number systems

long lng1 = 0xCaFeBaBeFL;
long lng2 = 0312775352766L;
long lng3 = 0b1100101011111110101110101011111000001L;

Without the suffix 'L' all three lines would fail to compile because the value is outside the range of int (but not outside the range of long).

But for the suffix 'f' (or 'F') it's a different story

float flt1 = 0xCaFeBaBeFF;
float flt2 = 0312775352766F;
float flt3 = 0b1100101011111110101110101011111000001F;

Only flt2 compiles successfully, both flt1 and flt3 give at least one compiler error. The value of flt1 is outside the range of int (so the suffix 'F' is part of the value). For flt3 you'll get two (related) compiler errors: value outside the range of int and syntax error on token 'F'.

And since we are having so much fun, let's try the same for the suffix 'd' (or 'D')

double dbl1 = 0xCaFeBaBeFD;
double dbl2 = 0312775352766D;
double dbl3 = 0b1100101011111110101110101011111000001D;

The results are exactly the same as with the suffix 'f' (or 'F'). Yay, consistency!

Hope it helps!
Kind regards,
Roel

PS. I think this will definitely be outside the scope of the OCAJP exam. I even can't remember if I got a question on the actual exam using different number systems.

[Roel De Nijs]

Joe Bishara wrote:

Ramya Subraamanian wrote:...you can have only 0 or 1

Not quite right. An l or L suffix is allowed.

That's indeed spot-on! And you truely deserves a cow for posting that at exactly the same time as me. I think you have better odds for winning the lottery

[Ramya R Subramanian]

Not quite right. An l or L suffix is allowed.

oh yes, it does work with suffix L or l.

it is like widening primitive conversion. from long to float and long to double.

         float f1 = 0b1000L;
		 float f2= 0b1000l;
		 double d1=0b1000L;

and Narrowing primitive conversion

        byte b1 = (byte)0b1000L;//narrowing conversion - long to byte
		short s1 = (short)0b1000L;//narrowing conversion -long to short
		int i1=(int)0b1000L;//narrowing conversion -long to int

and long with L suffix

long l1 = 0b1000L;

[nick woodward]

yeah, it seems that i somehow got confused by the answer:

Test 5, question 30

float x = 0b10_000f; This is invalid because the floating point suffices f, F or d, D are used only when using decimal system or hexadecimal and not while using binary.

to me this suggests that the f as a suffix can be used with hexadecimals. but the answer is not distinguishing between being a suffix, and just being allowed at the end of the hex number as a normal value.

however, from what you're saying Roel, it seems the distinction is when the literal being assigned is greater than the max range of an integer, and that this isn't possible with binary numbers, but is with others.

think i'm going to have to search for a thread on the basics again, just to make sure i've got them sorted.

thanks for your help, both of you!

Nick

[Roel De Nijs]

nick woodward wrote:to me this suggests that the f as a suffix can be used with hexadecimals. but the answer is not distinguishing between being a suffix, and just being allowed at the end of the hex number as a normal value.

You definitely can use 'f' (or 'F') and 'd' (or 'D') in a hexadecimal number, because those are valid characters in the hexadecimal number system. But you can not use them as the suffix to convert the default type of an integer literal (which is - as you definitely know - int) to float or double. So if your hexadecimal number is outside the range of an int, you can only use the suffix 'l' (or 'L') to convert the int literal to a long literal. Illustrated in the following code snippet

int int1 = 0xCaFeBaBe; // ok
int int2 = 0xCaFeBaBe0; // out of range
double dbl1 = 0xCaFeBaBe0D; // out of range
float flt1 = 0xCaFeBaBe0F; // out of range
long lng1 = 0xCaFeBaBe0L; // ok

nick woodward wrote:however, from what you're saying Roel, it seems the distinction is when the literal being assigned is greater than the max range of an integer, and that this isn't possible with binary numbers, but is with others.

For binary numbers it's different: 'f' (or 'F') and 'd' (or 'D') or not valid characters in the binary number system. So strictly speaking you could use them as the suffix to convert an int literal to a float or double. But it's not allowed for binary numbers! When you use one of these characters in a binary literal, you'll get a compiler error. So if your binary number is outside the range of an int (which is possible), you can only use the suffix 'l' (or 'L') to convert the int literal to a long literal. illustrated in the following code snippet

int int3 = 0b11111111111111111111111111111110; // ok
int int4 = 0b111111111111111111111111111111100; // out of range
double dbl2 = 0b111111111111111111111111111111100D; // syntax error on 'D'
float flt2 = 0b111111111111111111111111111111100F; // syntax error on 'F'
long lng2 = 0b111111111111111111111111111111100L; // ok

Hope it helps!
Kind regards,
Roel

[nick woodward]

oh ok, yes it does, thanks!

do you see where my confusion came from in that enthuware answer though? it does seem to suggest that 'f' and 'd' can be used as a suffix in hex.

anyway, good to know about converting an int literal to a long literal. apreciated!

Nick

[Roel De Nijs]

nick woodward wrote:do you see where my confusion came from in that enthuware answer though? it does seem to suggest that 'f' and 'd' can be used as a suffix in hex.

Yeah! I can definitely see your confusing. And for some reason the octal system is missing too in that statement (which allows using 'f'/'F' and 'd'/'D' as a suffix. Illustrative code snippet

float flt1 = 0312775352766; // out of int range
float flt2 = 0312775352766F; // ok
double dbl2 = 0312775352766D; // ok

So they probably should rephrase that statement like "This is invalid because the floating point suffices f, F or d, D are used only when using decimal or octal system, not while using binary. And f, F or d, D can (of course) be used in hexadecimal system too, but then they are interpreted as a valid character (value 15 and 13 respectively) not as a suffix." (or something similar of course).

Hope it helps and that statement is less confusing!
Kind regards,
Roel

[Paul Anilprem]

nick woodward wrote:yeah, it seems that i somehow got confused by the answer:

Test 5, question 30

float x = 0b10_000f; This is invalid because the floating point suffices f, F or d, D are used only when using decimal system or hexadecimal and not while using binary.

to me this suggests that the f as a suffix can be used with hexadecimals. but the answer is not distinguishing between being a suffix, and just being allowed at the end of the hex number as a normal value.
Nick

Nick,
I just checked this question (qid enthuware.ocajp.i.v7.2.1377 ) and the complete explanation that is provided with this option is as follows:

This is invalid because the floating point suffices ///f, F, d, and D/// are used only when using decimal system and not while using binary. However, since f is a valid digit in hexadecimal system, a hex number may end with an f although it will not be interpreted as float but as the digit f. Thus, float x = 0x10_000f; and float x = 10_000f; are valid because they are written in hex and decimal respectively but float x = 0b10_000f; is invalid because is written in binary.

Note that a floating point number cannot be written in Octal. Therefore, float x = 010_000f; is valid but it is not octal even though it starts with a 0. It is interpreted in decimal.

So it does make it quite clear that an f at the end of a hex number is interpreted as a digit and not as a floating point suffix.

HTH,
Paul.

[Roel De Nijs]

Paul Anilprem wrote:

Note that a floating point number cannot be written in Octal. Therefore, float x = 010_000f; is valid but it is not octal even though it starts with a 0. It is interpreted in decimal.

That's a really nice note Have a cow! I didn't know that (since I didn't print these values). But now I have and that's indeed true! Not that I doubted about correctness of Enthuware's explanation, but it's always a (very) good idea to try it out yourself

short shrt = 010_000;
float flt = 010_000F;
double dbl = 010_000D;
System.out.printf("%d %.2f %.2f%n", shrt, flt, dbl);

Output: 4096 10000.00 10000.00

Kind regards,
Roel

[nick woodward]

Thanks Paul, I think this is a case of me needing to update the software, because the line

However, since f is a valid digit in hexadecimal system, a hex number may end with an f although it will not be interpreted as float but as the digit f

does not appear in my answer (it is otherwise exactly the same).

Thanks for the reply though!

Nick