This question already has answers here:
are all instructions in jvm atomic?
(1 answer)
Can num++ be atomic for 'int num'?
(13 answers)
Closed 16 days ago.
Single JVM instruction is not atomic but single assemble instruction is atomic.
Is this correct?
Demo
public class DemoInstruction {
public static void main(String[] args) {
int i=0;
i++;
}
}
i++ translate into JVM instruction:
IINC 1 1
i++ translates into local assembly instructions consisting of three assembly instructions.
Related
This question already has answers here:
Is Java "pass-by-reference" or "pass-by-value"?
(93 answers)
Assigning in Java?
(5 answers)
Closed 3 years ago.
PS: This question is different than others because I need to know exactly what code to write in order to prevent the variable from changing.
I have this example of code to demonstrate my problem:
float[][] one = {{0}};
float[][] two = one;
System.out.println(two[0][0]);
one[0][0] ++;
System.out.println(two[0][0]);
int three = 0;
int four = three;
System.out.println(four);
three ++;
System.out.println(four);
The output is:
0.0
1.0
0
0
For some reason, when I change float[][] 'one', 'two' automatically changes as well.
However, when I performed the same task but for a different data type int, then the code works as intended.
What is causing this variable to drag the other one with it?
Why doesn't it happen for the int type?
How do I fix it?
This question already has answers here:
Is ++x more efficient than x++ in Java?
(3 answers)
Difference between pre-increment and post-increment in a loop?
(22 answers)
Closed 4 years ago.
I've read recently that ++i is much more efficient in C++ than its counterpart i++ because the first (++i) does the increment and returns the incremented value, whereas the latter (i++) creates a copy of the value, return it and performs an increment.
Does the same happen in Java?
Both are not atomic operations composed of the multiple steps. Unlike in C++, these operators can't be overloaded. So there is no difference in Java in the matter of performance.
The only and only difference you should mind between x++ and ++x is that x++ returns the value before it's incremented. And ++x does the same but after the incrementation.
This answer provides a bytecode example.
This question already has answers here:
Confusion on NaN in Java
(4 answers)
Are the bit patterns of NaNs really hardware-dependent?
(4 answers)
Closed 5 years ago.
I had an issue with some NaNs that came from 0/0 not being Canonical in a system I was making and I reduced the problem to this case.
public class Test {
public static void main(String[] args){
Float zero = Float.intBitsToFloat(0);
// Passes
assert Float.floatToRawIntBits(zero) == Float.floatToRawIntBits(0f);
// Fails
assert Float.floatToRawIntBits(zero / zero) == Float.floatToRawIntBits(0f / 0f);
}
}
This is quite bizarre to me. Additionally when debugging with Intellij, watch expressions respond like you would expect; both assertions pass.
I am using openjdk version "1.8.0_131".
This question already has answers here:
What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it?
(26 answers)
Closed 5 years ago.
I wrote a Java program below:
int[] tar = {1,2,5};
for(int i=0 ; i<tar.length ; i++)
{
if(tar[i] - tar[i-1] > 2)
{
System.out.print("true");
}
}
why "tar[i] - tar[i-1]" doesn't mean any error？ Isn't an error of ArrayIndexOutBoundsException?
It is an Exception and not error (if you mean it by compiler), cause Java knows about your array values only at run time.
Compiler cannot run your code and see that. What if at runtime you provided some good values at right indices ? That is the reason, compiler only checks the compiler level rules. It won't hint you about runtime behaviour.
This question already has answers here:
++someVariable vs. someVariable++ in JavaScript
(7 answers)
Closed 7 years ago.
I don't consider myself to be bad at programming, but there's been something troubling me since the past few days.
int counter = 3;
++counter;
Is the following code above the same as counter++;.
It is similar, but not the same.
In your expression it doesn't matter, but if you had something more complicated, like System.out.println(counter++), it would make a big difference.
For example:
int counter = 3;
System.out.println(counter++)
This will print 3, then increment counter to 4.
However, if you do
int counter = 3;
System.out.println(++counter)
it will print 4 because it increments prior to giving the value as a parameter to the print function.
It's a question of when the increment is performed, the prefix performs it before other operations, postfix performs it after. They have different precedences.