Is ++i really faster than i++ in for-loops in java

问题:

In java I usually make a for-loop like following:

for (int i = 0; i < max; i++) {
   something
}

But recently a colleague typed it so:

for (int i = 0; i < max; ++i) {
   something
}

He said the latter would be faster. Is that true?

回答1:

No, it's not true. You could measure the performance by timing each loop for a large number of iterations, but I'm fairly certain they will be the same.

The myth came from C where ++i was regarded as faster than i++ because the former can be implemented by incremeting i then returning it. The latter might be implemented by copying the value of i to a temporary variable, incrementing i, then returning the temporary. The first version doesn't need to make the temporary copy and so many people assume that it is faster. However if the expression is used as a statement modern C compilers can optimize the temporary copy away so that there will be no difference in practice.

回答2:

This question needed some Java byte code. Consider the following code:

public class PostPre {
    public static void main(String args[]) {
        int n = 5;
        loop1(n);
        loop2(n);
    }

    public static void loop1(int n) {
        for (int i = 0; i < n; i++) {}
    }

    public static void loop2(int n) {
        for (int i = 0; i < n; ++i) {}
    }
}

Now compile it and disassemble it:

$ javac PostPre.java; javap -c PostPre.class 
Compiled from "PostPre.java"
public class PostPre {
  public PostPre();
    Code:
       0: aload_0       
       1: invokespecial #1                  // Method java/lang/Object."<init>":()V
       4: return        

  public static void main(java.lang.String[]);
    Code:
       0: iconst_5      
       1: istore_1      
       2: iload_1       
       3: invokestatic  #2                  // Method loop1:(I)V
       6: iload_1       
       7: invokestatic  #3                  // Method loop2:(I)V
      10: return        

  public static void loop1(int);
    Code:
       0: iconst_0      
       1: istore_1      
       2: iload_1       
       3: iload_0       
       4: if_icmpge     13
       7: iinc          1, 1
      10: goto          2
      13: return        

  public static void loop2(int);
    Code:
       0: iconst_0      
       1: istore_1      
       2: iload_1       
       3: iload_0       
       4: if_icmpge     13
       7: iinc          1, 1
      10: goto          2
      13: return        
}

loop1() and loop2() have the same byte code.

回答3:

For any reasonably capable optimizer, they will be exactly the same. If you aren't sure, look at the output bytecode or profile it.

回答4:

Even if it is, which I very much doubt, your colleague should really have better things to spend his time learning than how to optimise a loop expression.

回答5:

Try this in your environment

public class IsOptmized {
    public static void main(String[] args) {

        long foo; //make sure the value of i is used inside the loop
        long now = 0; 
        long prefix = 0;
        long postfix = 0;

        for (;;) {
            foo = 0;
            now = System.currentTimeMillis();
            for (int i = 0; i < 1000000000; i++) {
                foo += i;
            }
            postfix = System.currentTimeMillis() - now;

            foo = 0;
            now = System.currentTimeMillis();
            for (int i = 0; i < 1000000000; ++i) {
                foo += i;
            }
            prefix = System.currentTimeMillis() - now;

            System.out.println("i++ " + postfix + " ++i " + prefix + " foo " + foo);
        }
    }
}

Mine gives me

i++ 1690 ++i 1610 foo 499999999500000000
i++ 1701 ++i 1610 foo 499999999500000000
i++ 1701 ++i 1600 foo 499999999500000000
i++ 1701 ++i 1610 foo 499999999500000000
i++ 1701 ++i 1611 foo 499999999500000000
i++ 1701 ++i 1610 foo 499999999500000000
i++ 1691 ++i 1610 foo 499999999500000000
i++ 1701 ++i 1600 foo 499999999500000000
i++ 1701 ++i 1610 foo 499999999500000000
i++ 1691 ++i 1610 foo 499999999500000000
i++ 1701 ++i 1610 foo 499999999500000000
i++ 1691 ++i 1610 foo 499999999500000000
i++ 1701 ++i 1610 foo 499999999500000000
i++ 1701 ++i 1610 foo 499999999500000000
i++ 1701 ++i 1610 foo 499999999500000000
i++ 1692 ++i 1610 foo 499999999500000000
i++ 1701 ++i 1610 foo 499999999500000000
i++ 1691 ++i 1610 foo 499999999500000000

So even if it is not that much, I asume there is a difference

回答6:

It will not be any faster. The compiler and JVM with the JIT will make mincemeat of such insignificant differences.

You can use the usual loop optimization techniques to get speed benefits, like unrolling, if applicable.

回答7:

No there will be no difference at all.

This came from C++ but even there there would be no difference at all in this case. Where there is a difference is where i is an object. i++ would have to make an additional copy of the object as it has to return the original unchanged value of the item whereas ++i can return the changed object so saves a copy.

In c++ with user defined object the cost of a copy can be significant so it's definatly worth remembering. And because of this people tend to use it for int variables too, as it's just as good anyway...

回答8:

Decompile with "javap -c YourClassName" and see the result and decide from that. This way you see what the compiler actually does at each case, not what you think it does. This way you also see WHY one way is faster than the other.

回答9:

In Java there should be no difference - any modern compiler^* should generate the same byte code (just an iinc) in both cases since the result of the increment expression is not being used directly.
There is a third option, still the same byte code^*:

for (int i = 0; i < max; i += 1) {
   something
}

^* tested with Eclipse's compiler

回答10:

In Java there is no such difference. Java machine interpertes code and no matter if you write ++i or i++, it will be converted to byte code to exact same instruction set.

But in C/C++ there is a huge difference and if you are not using any optimisation flags, then your loop can be slower up to 3 times.

Using optimisation flags such like -O/-O3 will force compiler to make output asembly code simplier (in most cases) and therefore faster (in most cases).

回答11:

Even it one would be faster, nobody cares in the days of HotSpot. The first thing the JIT does is to remove all optimizations that javac made. After that, everything is left to the JIT to make it fast.