What's the Difference Between X = X++; VS X++;

Is there a difference between x++ and ++x in java?

++x is called preincrement while x++ is called postincrement.

int x = 5, y = 5;

System.out.println(++x); // outputs 6
System.out.println(x); // outputs 6

System.out.println(y++); // outputs 5
System.out.println(y); // outputs 6

difference between X x = new X(); and x = new X();

Assuming that X is a class with a no-arg constructor (for example a default constructor), the following works:

    // Declare and initialize x
    X x = new X();
    // Assign new value to x
    x = new X();

The first code line declares a variable x and assigns a reference to the new instance of X to it (a new X object). The second line assigns a new X instance to the already declared variable x (thus discarding the reference to the first object). We declare a variable by putting the type name (or the word var) before them. So since in the first line, we have X x, this is a declaration.

In Java, variables need to be declared before they are first used. So the first code line would not work without the type name X at the front. Java would complain that the variable x had not been declared.

On the other hand, we are only allowed to declare each variable once. So putting type name X before the second line would be an error too. My Eclipse says Duplicate local variable x because it “thinks” that I am trying to declare a second variable also named x, which is not allowed (for good reasons).

What's the difference between X = X++; vs X++;?

X++ will increment the value, but then return its old value.

So in this case:

static void Main(string[] args)
{
    int x = 10;
    x = x++;
    Console.WriteLine(x);
}

You have X at 11 just for a moment, then it gets back to 10 because 10 is the return value of (x++).

You could instead do this for the same result:

static int plusplus(ref int x)
{
  int xOld = x;
  x++;
  return xOld;
}

static void Main(string[] args)
{
    int x = 10;
    x = plusplus(x);
    Console.WriteLine(x);
}

It is also worth mentioning that you would have your expected result of 11 if you would have done:

static void Main(string[] args)
{
    int x = 10;
    x = ++x;
    Console.WriteLine(x);
}

What exactly is the difference between x++ and x+1?

x++ and ++x

The increment operator x++ will modify and usually returns a copy of the old x. On a side note the prefixed ++x will still modify x but will returns the new x.

In fact x++ can be seen as a sort of:

{
    int temp = x; 
    x = x + 1; 
    return temp;
}

while ++x will be more like:

{
    x = x + 1;
    return x;
}

x + 1

The x+1 operation will just return the value of the expression and will not modify x. And it can be seen as:

{
    return (x + 1);
}

difference between x += x and x = x + x in Python list

x += x is calling append under the hood, which mutates the original variable

x = x + x is creating a new variable local to test2 and setting that value, which doesn't affect the original x

Difference in implementation of x = x + 1 and x++

My professor recently said that although x = x + 1 and x++ will obviously give the same result

I guess your professor perhaps meant - the value of x after x = x + 1 and x++ will be same. Just to re-phrase, as it seems to be creating confusion in interpreting the question.

Well, although the value of x will be same, they are different operators, and use different JVM instructions in bytecode. x + 1 uses iadd instruction, whereas x++ uses iinc instruction. Although this is compiler dependent. A compiler is free to use a different set of instructions for a particular operation. I've checked this against javac compiler.

For eclipse compiler, from one of the comment below from @Holger:

I just tested it with my eclipse and it produced iinc for both expressions. So I found one compiler producing the same instructions

You can check the byte code using javap command. Let's consider the following class:

class Demo {
    public static void main(String[] args) {
        int x = 5;

        x = x + 1;
        System.out.println(x);

        x++;
        System.out.println(x);
    }
} 

Compile the above source file, and run the following command:

javap -c Demo

The code will be compiled to the following bytecode (just showing the main method):

 public static void main(java.lang.String[]);
   Code:
      0: iconst_5
      1: istore_1
      2: iload_1
      3: iconst_1
      4: iadd
      5: istore_1
      6: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
      9: iload_1
     10: invokevirtual #3                  // Method java/io/PrintStream.println:(I)V
     13: iinc          1, 1
     16: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
     19: iload_1
     20: invokevirtual #3                  // Method java/io/PrintStream.println:(I)V
     23: return

Difference between unary operators ( += , =+ , ++x , and x++)

They differ in how they change the value and how they return the result.

The first two += and =+ behave in the way that the first increments a variable, the other sets a variable. They are not related. Observe the following code:

// +=
x = 1;
printf( x += 1 ); // outputs 2, the same as x = x+1
printf( x );      // outputs 2

// =+
x = 1;
printf( x =+ 1 ); // outputs 1, the same as x = 1;
printf( x );      // outputs 1

The next two, ++x and x++, differ in the order their function. ++x will increment your variable by 1 and return the result. x++ will return the result and increment by 1.

// ++x
x = 1;
printf( ++x ); // outputs 2, the same as x = x+1
printf( x );   // outputs 2

// x++
x = 1;
printf( x++ ); // outputs 1
printf( x );   // outputs 2

They are mostly useful for for loops and while loops.

In terms of speed, ++x is considered a lot faster than x++ since x++ needs to create an internal temporary variable to store the value, increment the main variable, but return the temporary variable, basically more operations are used. I learned this a looong time ago, I don't know if it still applies

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