i++ vs i=i+1 in Java

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Often, i++ and i=i+1 are considered to be same statements but in Java, both statements work differently internally. We will take a look at:

The compiler instructions used
Typecasting along with the operations
benchmarking the two operations

We eventually see that i=i+1 is always faster than i++.

Compiler Instructions

As both i = i+1 and i++ are different commands, both uses different JVM instructions in bytecode.

i+1 uses iadd instruction

i++ uses iinc instruction

The exact command depends on the type of compiler being used.

Typecasting

In the expression i=i+1, as we are adding 1 an integer, if i is of lower precision such as byte or short, it will give error: incompatible types: possible lossy conversion.

Example (with the error):

class opengenus 
{
	public static void main (String[] args) 
	{
	    short i = 1;
	    
	    i = i + 1; // gives compilation error
	    System.out.println(i);
	}
}

Output:

error: incompatible types: possible lossy conversion from int to short
	    i = i + 1;
	          ^
1 error

Correct code with explicit type casting:

class opengenus 
{
	public static void main (String[] args) 
	{
	    short i = 1;
	    
	    i = (short)(i + 1); // compiles correctly
	    System.out.println(i);
	}
}

Output:

This problem does not take place in the expression i++ as typecasting is handled internally by the compiler

Example:

class opengenus 
{
	public static void main (String[] args) 
	{
	    short i = 1;
	    
	    i++; // compiles correctly
	    System.out.println(i);
	}
}

Output

Hence, while replacing i=i+1 by i++, we need to consider the type of i as well.

Question

Depending upon above points, which should be faster?

i=i+1

i++

depends on platform

depends on code

As i++ does automatic typecasting and uses a compiler instruction which internally uses iadd instruction, i=i+1 is faster than i++.

Benchmark

To measure the performance of the two similar statements, we used the following Java code:

import java.lang.Math;
class benchmark_opengenus 
{
	public static void main (String[] args) 
	{
	    int i=0, limit = 1000000;
	    long t1_1 = System.nanoTime();
		for(int j=0; j<=limit; j++)
		{
		    i++;
		}
		long t1_2 = System.nanoTime();
		
		long t2_1 = System.nanoTime();
		for(int j=0; j<=limit; j++)
		{
		    i=i+1;
		}
		long t2_2 = System.nanoTime();
		
		System.out.println("i++ took "+(t1_2-t1_1)+ " nanoseconds");
		System.out.println("i=i+1 took "+(t2_2-t2_1)+ " nanoseconds");
		
		System.out.println("i=i+1 took "+ String.format("%.2f",(double)(t2_2-t2_1)*100/(t1_2-t1_1))+" % of time taken by i++");
	}
}

Output:

i++ took 4041400 nanoseconds
i=i+1 took 2676134 nanoseconds
i=i+1 took 66.22 % of time taken by i++

Hence, we see i = i+1 is nearly 40% faster than i++.

Compiler Instructions

Typecasting

Question

Depending upon above points, which should be faster?

Benchmark

Always use i=i+1 instead of i++