10

Consider the following two method calls that are nearly equivalent. Take note of the way the byte array is declared and allocated on both.

void Method1()
{
    byte [] bytearray = new byte[16];

    /* some code */

}

void Method2()
{
    byte [] bytearray = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

    /* some code */    
}

In both cases, assume that when Method1 and Method2 return, "bytearray" is a candidate for garbage collection since none of the code that manipulates the bytearray variable will be holding a reference past the end of the method itself.

Does Method2 operate any faster (or differently) by avoiding the call to "new"? Or are the two implementations above equivalent? In either case, can the Java compiler or runtime make an optimization to avoid the overhead of hitting the memory allocator for this short-lived temporary buffer?

Taylan Aydinli
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selbie
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7 Answers7

12

Which form is the fastest? Depends on the JIT - it's possible they're equivalent. Very few, if any, programs will ever notice the difference if there is any.

Which form is the best? Almost invariably, the one that leaves your program more readable.

But is there any actual difference, regardless of if we'll ever notice? Let's have a look!

class ArrayTest {

    public int[] withNew() {
        int[] arr = new int[4];
        return arr;
    }

    public int[] withInitializer() {
        int[] arr = {0, 0, 0, 0};
        return arr;
    }

}

We disassemble this with javap -c ArrayTest: 

Compiled from "ArrayTest.java"
class ArrayTest {
  ArrayTest();
    Code:
       0: aload_0       
       1: invokespecial #1                  // Method java/lang/Object."<init>":()V
       4: return        

  public int[] withNew();
    Code:
       0: iconst_4      
       1: newarray       int
       3: astore_1      
       4: aload_1       
       5: areturn       

  public int[] withInitializer();
    Code:
       0: iconst_4      
       1: newarray       int
       3: dup           
       4: iconst_0      
       5: iconst_0      
       6: iastore       
       7: dup           
       8: iconst_1      
       9: iconst_0      
      10: iastore       
      11: dup           
      12: iconst_2      
      13: iconst_0      
      14: iastore       
      15: dup           
      16: iconst_3      
      17: iconst_0      
      18: iastore       
      19: astore_1      
      20: aload_1       
      21: areturn       
}

Nope, they're not the same in this case - using the initializer form causes the slots to be individually set to 0, which is kind of pointless since they're already zeroed by the array allocation. So, it's basically equivalent to this:

public int[] withNewAndSettingExplicitly() {
    int[] arr = new int[4];
    arr[0] = 0;
    arr[1] = 0;
    arr[2] = 0;
    arr[3] = 0;
    return arr;
}

although that compiles to yet another set of byte code, which is mostly the same but not quite:

  public int[] withNewAndSettingExplicitly();
    Code:
       0: iconst_4      
       1: newarray       int
       3: astore_1      
       4: aload_1       
       5: iconst_0      
       6: iconst_0      
       7: iastore       
       8: aload_1       
       9: iconst_1      
      10: iconst_0      
      11: iastore       
      12: aload_1       
      13: iconst_2      
      14: iconst_0      
      15: iastore       
      16: aload_1       
      17: iconst_3      
      18: iconst_0      
      19: iastore       
      20: aload_1       
      21: areturn

Thus, the moral of the story is this: If you want all elements set to 0, you're generating less bytecode with new int[size] (which may or may not be faster), but you also have to type less (which imho is a big win). If you want to set the values in the array directly when you're allocating it, use whatever looks best in your code, because the generated code will be pretty much the same whatever form you choose.

Now, to answer your actual questions:

Does Method2 operate any faster (or differently) by avoiding the call to "new"?

As we saw, the new is just hidden behind the initializer syntax (look for the newarray op code). By the way, allocation is extremely cheap in the JVM (generational garbage collectors have that pleasant side effect).

Or are the two implementations above equivalent?

As we saw - not quite, but it's unlikely anyone will ever notice the difference.

In either case, can the Java compiler or runtime make an optimization to avoid the overhead of hitting the memory allocator for this short-lived temporary buffer?

Again - allocation is cheap, so don't worry. Nevertheless, recent JVMs have this little feature called escape analysis which may result in the array being stack allocated rather than heap allocated.

gustafc
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5

There is no difference between both the ways. It's just 'Syntactic sugar' in syntax.

user207421
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Suresh Atta
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3

They are the same.

new keyword creates an object. But in Java, arrays are.. objects. See Chapter 10. Arrays:

In the Java programming language, arrays are objects (§4.3.1), are dynamically created, and may be assigned to variables of type Object (§4.3.2). All methods of class Object may be invoked on an array.
..
An array is created by an array creation expression (§15.10) or an array initializer (§10.6).

Maroun
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2

These 2 methods are exactly equivalent. However second method is more verbose and less convenient for longer arrays or if you want to modify the array length.

AlexR
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0

Its better to use the first style.

On the Second style i try to find a non-0. Its guarantees more error-free, and therefore better readable.

This shorthand has a good reason to be exists!

Grim
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0

Both are same...as both will occupy the same space over heap...and bytecode would be the option to study it well

nitesh
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0

Both are the same but I would not recommend the second one as it is difficult to read to write and debug. Imagine you find this:

byte [] bytearray = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

Could you tell if the size of the array is correct? Yes, but imagine with is:

byte [] bytearray = new byte[25];

It is far more easy.

And imagine the case the array length is 100!

Loïc
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