Is ++i really faster than i++ in Java

Recently there was a question whether ++j is better than j++  in terms of performance. Typically this suggestion is applicable for C++ programming.

In below java code, let’s assume a value of 10 is passed as input to both the functions. After executing the line, in func1() the value of k will be 10 whereas value of j will be 11 and in func2(), the value of both k, j will be 11.

In C++, in order to return a value of 10 for variable k in func1(), the value of j is copied to a temp location, j is incremented and then the value of temp location is returned to k. For func2() this temp location is not necessary and hence one instruction is less for ++j when compared to j++.

Code

Java byte code

package com.test; public class Test {         public static void func1(int j) {                 int k = j++;         }         public static void func2(int j) {                 int k = ++j;         } }

public class com.test.Test extends java.lang.Object{ public com.test.Test();   Code:    0:   aload_0    1:   invokespecial   #1; //Method java/lang/Object."<init>":()V    4:   return public static void func1(int);   Code:    0:   iload_0    1:   iinc    0, 1    4:   istore_1    5:   return public static void func2(int);   Code:    0:   iinc    0, 1    3:   iload_0    4:   istore_1    5:   return }

To verify this in java, let’s examine the byte code and how this is handled.

func1():

                The passed input is stored on to the stack using iload and using iinc the value of j is incremented (so j= 11 now) and k is populated with value on the stack (which is 10) using istore

func2():

                First the value of j is incremented using iinc instruction and then incremented value is stored on to stack (which is 11) and then the same is returned to k using iload and istore respectively.

Hence the number of instructions is same in both the cases.

Now if the argument is that if no assignment is required then is ++j better? Let’s write a simple code and examine its byte code as shown in the below table:

Code	Java byte code
package com.test; public class Test {         public static void func1(int j) {                 j++;         }         public static void func2(int j) {                 ++j;         } }	public class com.test.Test extends java.lang.Object{ public com.test.Test();   Code:    0:   aload_0    1:   invokespecial   #1; //Method java/lang/Object."<init>":()V    4:   return public static void func1(int);   Code:    0:   iinc    0, 1    3:   return public static void func2(int);   Code:    0:   iinc    0, 1    3:   return }

Compiler is smart to understand that the variable needs to be incremented irrespective of its position.

So what’s the conclusion? In case of java the number of instructions is same for the both the cases and hence we are free to choose our option while coding :-)