About Question enthuware.ocajp.i.v7.2.1240 :

[AndaRO]

If I am multiplied : char * char result int?
byte * byte result int?

Is it true?

Thanks Paul

[admin]

Yes, that is correct.

[lenalena]

Hi, Paul,

First of all, I want to thank you for taking such good care of the board. I found your explanations excellent.

However, I found the following statement from a long time ago:
===

FYI,
If any variable is a long, then the result is a long.
If any variable is a float or a double, then the result is a double.

===

I found today (accidentally), that it's not exactly accurate. Maybe it was 7 years ago. But if any variable is a float, and the rest are not doubles, the result will be a float, not a double.

Code: Select all

float f = 10.0f;
int x = 10;
System.out.println( (f*x) instanceof float);

This fails to compile, but the compiler gives the following error:
"required reference, found float". So clearly, it gets evaluated to a float, not a double.
If you change type of f to a double, the compile changes the complaint to "required reference, found double".
So it will evaluate to double only if double is present. If only float is present, it will evaluate to float.

I might be missing something, though...

[admin]

You are right. As per section 4.2.4 of JLS 8:

If at least one of the operands to a numerical operator is of type double, then the operation is carried out using 64-bit floating-point arithmetic, and the result of the numerical operator is a value of type double. If the other operand is not a double, it is first widened (§5.1.5) to type double by numeric promotion (§5.6).

Otherwise, the operation is carried out using 32-bit floating-point arithmetic, and the result of the numerical operator is a value of type float. (If the other operand is not a float, it is first widened to type float by numeric promotion.)

thank you!
Paul.

[elias86]

Why in:

s = b * b ;

b * b returns an int???

and in:

c = c + b;

c + b returns an int???

[admin]

Because that is how the Java language designers designed the mathematical operators to work.

[elias86]

Because that is how the Java language designers designed the mathematical operators to work.

Are there specifications where i can find all of the rules that indicates the implicit and explicit cast for primitive variables?

Thanks, Elias.

[admin]

Yes, of course. You can find all the rules here: https://docs.oracle.com/javase/specs/jl ... index.html

[flex567]

From the explanation:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once. Note that the implied cast to type T may be either an identity conversion or a narrowing primitive conversion. For example, the following code is correct:

What does it mean,
evaluated only once
identity conversion,
narrowing primitive conversion ?

[admin]

1. Example: ia[0] += 1; is same as ia[0] = ia[0] + 1; but in the second case, the exact array element needs to figure out twice - once for evaluating the RHS and once for assigning that value to the LHS.

2. Identity conversion
3. narrowing primitive conversion

[javabean68]

Hi,

I don't understand what the sentence in bold means:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once. Note that the implied cast to type T may be either an identity conversion or a narrowing primitive conversion.

Can you please explain it to me perhaps by means of an example?

Thanks
Regards
Fabio

[admin]

Sorry, I am not sure what is the need for the bold part.