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I am new to java, and was running some code last night, and this really bothered me. I was building a simple program to display every X outputs in a for loop, and I noticed a MASSIVE decrease in performance, when I used modulus as variable % variable vs variable % 5000 or whatnot. Can someone explain to me why this is and what is causing it? So I can be better...

Here is the "efficient" code (sorry if I get a bit of syntax wrong I'm not on the computer with the code right now)

long startNum = 0;
long stopNum = 1000000000L;

for (long i = startNum; i <= stopNum; i++){
    if (i % 50000 == 0) {
        System.out.println(i);
    }
}

Here is the "inefficient code"

long startNum = 0;
long stopNum = 1000000000L;
long progressCheck = 50000;

for (long i = startNum; i <= stopNum; i++){
    if (i % progressCheck == 0) {
        System.out.println(i);
    }
}

Mind you I had a date variable to measure the differences, and once it became long enough, the first one took 50ms while the other took 12 seconds or something like that. You may have to increase the stopNum or decrease the progressCheck if your PC is more efficient than mine or what not.

I looked for this question around the web, but I can't find an answer, maybe I'm just not asking it right.

EDIT: I did not expect my question to be so popular, I appreciate all the answers. I did perform a benchmark on each half in time taken, and the inefficient code took considerably longer, 1/4 of a second vs 10 seconds give or take. Granted they are using println, but they are both doing the same amount, so I wouldn't imagine that would skew it much, especially since the discrepancy is repeatable. As for the answers, since I am new to Java, I will let votes decide for now which answer is best. I will try to pick one by Wednesday.

EDIT2: I am going to make another test tonight, where instead of modulus, it just increments a variable, and when it reaches progressCheck, it will perform one, and then reset that variable to 0. for a 3rd option.

EDIT3.5:

I used this code, and below I'll show my results.. Thank you ALL for the wonderful help! I also tried comparing the short value of the long to 0, so all my new checks happen ever "65536" times making it equal in repeats.

public class Main {


    public static void main(String[] args) {

        long startNum = 0;
        long stopNum = 1000000000L;
        long progressCheck = 65536;
        final long finalProgressCheck = 50000;
        long date;

        // using a fixed value
        date = System.currentTimeMillis();
        for (long i = startNum; i <= stopNum; i++) {
            if (i % 65536 == 0) {
                System.out.println(i);
            }
        }
        long final1 = System.currentTimeMillis() - date;
        date = System.currentTimeMillis();
        //using a variable
        for (long i = startNum; i <= stopNum; i++) {
            if (i % progressCheck == 0) {
                System.out.println(i);
            }
        }
        long final2 = System.currentTimeMillis() - date;
        date = System.currentTimeMillis();

        // using a final declared variable
        for (long i = startNum; i <= stopNum; i++) {
            if (i % finalProgressCheck == 0) {
                System.out.println(i);
            }
        }
        long final3 = System.currentTimeMillis() - date;
        date = System.currentTimeMillis();
        // using increments to determine progressCheck
        int increment = 0;
        for (long i = startNum; i <= stopNum; i++) {
            if (increment == 65536) {
                System.out.println(i);
                increment = 0;
            }
            increment++;

        }

        //using a short conversion
        long final4 = System.currentTimeMillis() - date;
        date = System.currentTimeMillis();
        for (long i = startNum; i <= stopNum; i++) {
            if ((short)i == 0) {
                System.out.println(i);
            }
        }
        long final5 = System.currentTimeMillis() - date;

                System.out.println(
                "\nfixed = " + final1 + " ms " + "\nvariable = " + final2 + " ms " + "\nfinal variable = " + final3 + " ms " + "\nincrement = " + final4 + " ms" + "\nShort Conversion = " + final5 + " ms");
    }
}

Results:

  • fixed = 874 ms (normally around 1000ms, but faster due to it being a power of 2)
  • variable = 8590 ms
  • final variable = 1944 ms (Was ~1000ms when using 50000)
  • increment = 1904 ms
  • Short Conversion = 679 ms

Not surprising enough, due to a lack of division, the Short Conversion was 23% faster than the "fast" way. This is interesting to note. If you need to show or compare something every 256 times (or about there) you can do this, and use

if ((byte)integer == 0) {'Perform progress check code here'}

ONE FINAL INTERESTING NOTE, using modulus on the "Final declared Variable" with 65536 (not a pretty number) was half the speed of (slower) than the fixed value. Where before it was benchmarking near the same speed.

Robert Cotterman
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  • I would be extremely surprised if those two snippets don't compile to the same bytecode. – Thomas Jungblut Jan 28 '19 at 16:05
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    I got the same result actually. On my machine, the first loop runs in about 1,5 seconds and the second runs in about 9 seconds. If I add `final` in front of the `progressCheck` variable, both run at the same speed again. That leads me to believe that the compiler or the JIT manages to optimize the loop when it knows that `progressCheck` is constant. – marstran Jan 28 '19 at 16:07
  • so adding `final` to a variable declares it as constant, and I will see more efficiency? – Robert Cotterman Jan 28 '19 at 16:17
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    Related: [How do I write a correct micro-benchmark in Java?](https://stackoverflow.com/q/504103/1371329) – jaco0646 Jan 28 '19 at 16:21
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    Division by a constant [can be easily converted to a multiplication by the multiplicative inverse](https://stackoverflow.com/q/41183935/995714). Division by a variable can't. And a 32-bit division is faster than a 64-bit division on x86 – phuclv Jan 28 '19 at 16:27
  • @ThomasJungblut hard to believe that the same bytecode produces just different results... and I doubt the compiler checks if `println` is eventually changing the variables (however that could happen) - if the varaible was defined `final` then it should be the same/similar bytecode [CONFIRMED not the same bytecode unless variable is final] – user85421 Jan 28 '19 at 16:28
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    @phuclv note 32-bit division is not a issue here, it is a 64-bit remainder operation in both cases – user85421 Jan 28 '19 at 16:36
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    BTW both of those variables can be declared not just final but static final. But this is a very interesting observation. I am up-oting the question – Michael Gantman Jan 28 '19 at 16:38
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    @RobertCotterman if you declare the variable as final, the compiler creates the same bytecode as using the constant (eclipse/Java 11) ((despite using one more memory slot for the variable)) – user85421 Jan 28 '19 at 16:39
  • Thank you for the final info. The reason i was using long over int was due to a longer bit of code i was trying to produce, but it was so strange/foreign to me that it didn't just work, i only know powershell at the moment. So my view of programming is skewed – Robert Cotterman Jan 28 '19 at 16:53
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    I'd look at [Do java finals help the compiler create more efficient bytecode?](https://stackoverflow.com/questions/8354412/do-java-finals-help-the-compiler-create-more-efficient-bytecode) – GBlodgett Jan 28 '19 at 16:54
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    I tried to isolate the crucial part here and created a simple microbenchmark. (Not one with JMH or so, just calling the loop several times, with some black hole thrown in, and omitting the `println` which would otherwise distort *all* results), I did not notice any difference in performance. You should try to create an example where the issue can be reproduced. – Marco13 Jan 28 '19 at 18:42
  • I think @phuclv actually holds the answer. The compiler will apply some magic to generate efficient code (because division is slow), but that can only be done if the value is known at forehand. If the value is constant (either a literal or a `final` variable), the compiler can perform the magic, but if the value is unknown (i.e.variable), then the compiler must fall back to integer division. – MC Emperor Jan 28 '19 at 20:15
  • Everyone keeps saying division is slow, and I get that. AND @phuclv shared a link, but I am not keen to understand how to replace modulo with multiplication. Any light? – Robert Cotterman Jan 28 '19 at 21:53
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    @CarlosHeuberger: A smart-enough compiler would see that `i` can't become greater than 1 billion, so it fits in a 32-bit register. Thus it could optimize by using 32-bit operand-size for everything. And it's non-negative, so implicit zero-extension to 64-bit (on x86-64) still makes it free to pass a 64-bit value to functions that are expecting it. – Peter Cordes Jan 29 '19 at 00:32
  • @CarlosHeuberger: Out of curiosity, what would Java do if the variable were final, but were set in the constructor? From a performance standpoint, the optimal approach would probably be to create a second backing field to hold an approximate reciprocal, but I don't know if a JIT would be able to do such a thing. – supercat Jan 29 '19 at 23:37
  • The second will be as fast as the first if declare the variable final: `final long progressCheck = 50000;` –  Feb 14 '19 at 09:17

4 Answers4

141

You are measuring the OSR (on-stack replacement) stub.

OSR stub is a special version of compiled method intended specifically for transferring execution from interpreted mode to compiled code while the method is running.

OSR stubs are not as optimized as regular methods, because they need a frame layout compatible with interpreted frame. I showed this already in the following answers: 1, 2, 3.

A similar thing happens here, too. While "inefficient code" is running a long loop, the method is compiled specially for the on-stack replacement right inside the loop. The state is transferred from the interpreted frame to OSR-compiled method, and this state includes progressCheck local variable. At this point JIT cannot replace the variable with the constant, and thus cannot apply certain optimizations like strength reduction.

In particular this means JIT does not replace integer division with multiplication. (See Why does GCC use multiplication by a strange number in implementing integer division? for the asm trick from an ahead-of-time compiler, when the value is a compile-time constant after inlining / constant-propagation, if those optimizations are enabled. An integer literal right in the % expression also gets optimized by gcc -O0, similar to here where it's optimized by the JITer even in an OSR stub.)

However, if you run the same method several times, the second and the subsequent runs will execute the regular (non-OSR) code, which is fully optimized. Here is a benchmark to prove the theory (benchmarked using JMH):

@State(Scope.Benchmark)
public class Div {

    @Benchmark
    public void divConst(Blackhole blackhole) {
        long startNum = 0;
        long stopNum = 100000000L;

        for (long i = startNum; i <= stopNum; i++) {
            if (i % 50000 == 0) {
                blackhole.consume(i);
            }
        }
    }

    @Benchmark
    public void divVar(Blackhole blackhole) {
        long startNum = 0;
        long stopNum = 100000000L;
        long progressCheck = 50000;

        for (long i = startNum; i <= stopNum; i++) {
            if (i % progressCheck == 0) {
                blackhole.consume(i);
            }
        }
    }
}

And the results:

# Benchmark: bench.Div.divConst

# Run progress: 0,00% complete, ETA 00:00:16
# Fork: 1 of 1
# Warmup Iteration   1: 126,967 ms/op
# Warmup Iteration   2: 105,660 ms/op
# Warmup Iteration   3: 106,205 ms/op
Iteration   1: 105,620 ms/op
Iteration   2: 105,789 ms/op
Iteration   3: 105,915 ms/op
Iteration   4: 105,629 ms/op
Iteration   5: 105,632 ms/op


# Benchmark: bench.Div.divVar

# Run progress: 50,00% complete, ETA 00:00:09
# Fork: 1 of 1
# Warmup Iteration   1: 844,708 ms/op          <-- much slower!
# Warmup Iteration   2: 105,893 ms/op          <-- as fast as divConst
# Warmup Iteration   3: 105,601 ms/op
Iteration   1: 105,570 ms/op
Iteration   2: 105,475 ms/op
Iteration   3: 105,702 ms/op
Iteration   4: 105,535 ms/op
Iteration   5: 105,766 ms/op

The very first iteration of divVar is indeed much slower, because of inefficiently compiled OSR stub. But as soon as the method reruns from the beginning, the new unconstrained version is executed which leverages all the available compiler optimizations.

Ian Kemp
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apangin
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    I hesitate to vote on this. On the one hand, it sounds like an elaborate way of saying "You messed up your benchmark, read something about JIT". On the other hand, I wonder why you seem to be so sure that OSR was the main relevant point here. I mean, doing a (micro) benchmark that involves `System.out.println` will nearly *necessarily* produce garbage results, and the fact that both versions are equally fast does not have to do anything with OSR in *particular*, as far as I can tell.. – Marco13 Jan 29 '19 at 00:31
  • @Marco13: the takeaway here for me is that first-pass JITing (in an OSR) is a bit like gcc without optimization: it doesn't propagate constants from local merely variables that happen not to change, so the first warmup with the compile-time constant "hidden" in a local variable is much slower than with a numeric literal in the `%` expression. Other answers say that making it `final` would also enable better optimization right away, even in the OSR, because the compiler knows the value is compile-time constant without checking the rest of the data-flow graph or w/e, like with a numeric literal. – Peter Cordes Jan 29 '19 at 07:04
  • @Marco13: it prints infrequently enough that `println` may not make this unusable as a benchmark. – Peter Cordes Jan 29 '19 at 07:06
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    @Marco13 The answer refers to the similar issues ([1](https://stackoverflow.com/questions/33842116/java-for-loop-optimization/33858960#33858960), [2](https://stackoverflow.com/questions/42570135/java-method-call-performance/42590152#42590152)) where I have proven that the reason is OSR compilation. This can be also verified by running the benchmark with `-XX:+PrintCompilation`. – apangin Jan 29 '19 at 07:47
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    (I'm curious and like to understand this. I hope the comments are not disturbing, might delete them later, but: ) The link `1` is a bit dubious - the empty loop could also be optimized away completely. The second is more similar to that one. But again, it's not clear why you attribute the difference to the OSR *specifically*. I'd just say: At some point, the method is JITed and becomes faster. To my understanding, the OSR only causes the usage of the final, optimized code to (roughly) be ~"deferred to the next optimization pass". (continued...) – Marco13 Jan 29 '19 at 10:32
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    (continued:) Unless you are specifically analyzing the hotspot logs, you cannot say whether the difference is caused by comparing JITed and un-JITed code, or by comparing JITed and OSR-stub-code. And you *certainly* cannot say that for sure when the question does not contain the real code or a complete JMH benchmark. So arguing that the difference is caused by OSR sounds, for me, inappropriately specific (and "unjustified") compared to saying that it is caused by the JIT in general. (No offense - I'm just wondering...) – Marco13 Jan 29 '19 at 10:34
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    @Marco13 there’s a simple heuristic: without the JIT’s activity, each `%` operation would have the same weight, as an optimized execution is only possible, well, if an optimizer did actual work. So the fact that one loop variant is significantly faster than the other proves the presence of an optimizer and further proves that it failed to optimize one of the loops to the same degree as the other (within the same method!). As this answer proves the capability of optimizing both loops to the same degree, there must be something that hindered the optimization. And that’s OSR in 99.9% of all cases – Holger Jan 29 '19 at 13:19
  • @Holger I see, but the original code did not indicate that they had been located in the same method (in fact, they couldn't have been, due to the variable names). The EDITed question now contains a snippet where multiple variants *are* contained in one method, and there, different timings can be explained by OSR. To put it that way: Unless the *exact* test conditions are stated, the OSR just seems like a guess for me, and the difference could *originally* have been caused by an awful lot of other flaws in the test code. There are just too many possible microbenchmarking pitfalls... – Marco13 Jan 29 '19 at 13:28
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    @Marco13 That was an "educated guess" based on the knowledge of HotSpot Runtime and the experience of analyzing similar issues before. Such a long loop could hardly be compiled in a way other than OSR, especially in a simple hand-made benchmark. Now, when OP has posted the complete code, I can only confirm the reasoning once again by running the code with `-XX:+PrintCompilation -XX:+TraceNMethodInstalls`. – apangin Jan 29 '19 at 21:12
  • While the guess was educated and reasonable, and was "confirmed" to be right in hindsight, it still feels like going out on a limb, considering the initial version of the question. A clarification (e.g. asking for the code that was added with the last EDIT) could have been worthwhile and appropriate. But the most important thing is that the guess - and thus the answer - was "right", so we can leave it at that. – Marco13 Jan 30 '19 at 13:37
40

In follow-up to @phuclv comment, I checked the code generated by JIT1, the results are as follows:

for variable % 5000 (division by constant):

mov     rax,29f16b11c6d1e109h
imul    rbx
mov     r10,rbx
sar     r10,3fh
sar     rdx,0dh
sub     rdx,r10
imul    r10,rdx,0c350h    ; <-- imul
mov     r11,rbx
sub     r11,r10
test    r11,r11
jne     1d707ad14a0h

for variable % variable:

mov     rax,r14
mov     rdx,8000000000000000h
cmp     rax,rdx
jne     22ccce218edh
xor     edx,edx
cmp     rbx,0ffffffffffffffffh
je      22ccce218f2h
cqo
idiv    rax,rbx           ; <-- idiv
test    rdx,rdx
jne     22ccce218c0h

Because division always takes longer than multiplication, the last code snippet is less performant.

Java version:

java version "11" 2018-09-25
Java(TM) SE Runtime Environment 18.9 (build 11+28)
Java HotSpot(TM) 64-Bit Server VM 18.9 (build 11+28, mixed mode)

1 - VM options used: -XX:+UnlockDiagnosticVMOptions -XX:CompileCommand=print,src/java/Main.main

Oleksandr Pyrohov
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    To give an order of magnitude on "slower", for x86_64: `imul` is 3 cycles, `idiv` is between 30 and 90 cycles. So integer division is between 10x and 30x slower than integer multiplication. – Matthieu M. Jan 29 '19 at 13:19
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    Could you explain what all of that means for readers who are interested, but don't speak assembler? – Nico Haase Jan 29 '19 at 14:00
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    @NicoHaase The two commented lines are the only important ones. In the first section, the code is performing an integer multiplication, whereas the second section, the code is performing an integer division. If you think about doing multiplication and division by hand, when you multiply you're usually doing a bunch of small multiplications and then one big set of additions, but division is a small division, a small multiplication, a subtraction, and repeat. Division is slow because you're essentially doing a bunch of multiplications. – MBraedley Jan 29 '19 at 15:05
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    @MBraedley thanks for your input, but such explanation should be added to the answer itself and not be hidden in the comment section – Nico Haase Jan 29 '19 at 15:16
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    @MBraedley: More to the point, multiplication in a modern CPU is fast because the partial products are independent and can thus be computed separately, while each stage of a division is dependent upon the preceding stages. – supercat Jan 29 '19 at 20:34
  • @supercat I wonder if this means that multiplication could be theoretically vectorized, at least in theory – Eugene Jan 29 '19 at 20:54
  • @Eugene: It is, routinely. Division is harder. Given an approximate reciprocal of a the divisor, one can multiply the dividend by that, multiply that result by the divisor, and subtract the result from the dividend, and then process the residue as one did the original number. Depending upon the quality of the original estimated reciprocal, the residue may diminish quite quickly, thus allowing division to be reduced to a reasonable-length sequence of very large multiplications which must be done sequentially, but the parts of each multiply could be done in parallel. – supercat Jan 29 '19 at 21:05
  • The OP is reporting a factor of 200 difference though. This seems like it could explain a factor of 50 difference at the absolute most. – David Schwartz Jan 31 '19 at 07:17
  • The question is, why does it choose div over mul. And when i thought about it, i bet it's because during compilation it's not sure if the check is gonna change, so it uses div, but since it's a static integer for the other, it can always mul to get the right answer. Compilers are weird. But it is doing it's best to be as optimal as possible. – Robert Cotterman May 24 '20 at 04:37
24

As others have noted, the general modulus operation requires a division to be done. In some cases, the division can be replaced (by the compiler) by a multiplication. But both can be slow compared to addition/subtraction. Hence, the best performance can be expected by something along these lines:

long progressCheck = 50000;

long counter = progressCheck;

for (long i = startNum; i <= stopNum; i++){
    if (--counter == 0) {
        System.out.println(i);
        counter = progressCheck;
    }
}

(As a minor optmiziation attempt we use a pre-decrement down-counter here because on many architectures comparing to 0 immediately after an arithmetic operation costs exactly 0 instructions/CPU cycles because the ALU's flags are already set appropriately by the preceeding operation. A decent optimizing compiler will, however, do that optimization automatically even if you write if (counter++ == 50000) { ... counter = 0; }.)

Notice that often you don't really want/need modulus, because you know that your loop counter (i) or whatever is only ever incremented by 1, and you really don't care about the actual remainder the modulus will give you, just see if the incrementing-by-one counter hits some value.

Another 'trick' is to use power-of-two values/limits, e.g. progressCheck = 1024;. Modulus a power of two can be quickly calculated via bitwise and, i.e. if ( (i & (1024-1)) == 0 ) {...}. This should be pretty fast too, and may on some architectures outperform the explicit counter above.

JimmyB
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    A smart compiler would invert the loops here. Or you could do that in the source. The `if()` body becomes an outer-loop body, and the stuff outside the `if()` becomes an inner loop body that runs for `min(progressCheck, stopNum-i)` iterations. So at the start, and every time `counter` reaches 0, you do `long next_stop = i + min(progressCheck, stopNum-i);` to set up for a `for(; i< next_stop; i++) {}` loop. In this case that inner loop is empty and should hopefully optimize away entirely, of you can do that in the source and make it easy for the JITer, reducing your loop to i+=50k. – Peter Cordes Jan 29 '19 at 00:26
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    But yes, in general a down-counter is a good efficient technique for fizzbuzz / progresscheck type stuff. – Peter Cordes Jan 29 '19 at 00:26
  • I added to my question, and did an increment, the `--counter` is just as fast as my increment version, but less code.also it was 1 lower than it should be, i'm curious if it should be `counter--` to get the exact number you want, not that it's much of a difference – Robert Cotterman Jan 29 '19 at 03:36
  • @PeterCordes A *smart* compiler would just print the numbers, no loop at all. (I think some only slightly more trivial benchmarks started to fail that way maybe 10 years ago.) – Peter - Reinstate Monica Jan 29 '19 at 08:19
  • @PeterA.Schneider: well yes, of course, in this case where there's no loop body outside the `if()`. But depending how long it runs, it's not necessarily a win to fully unroll into a potentially-huge straight-line block of code. (Or into a very long string, for a theoretical compiler that optimizes multiple `println` calls into one call with a long string containing newlines, after constant propagation). – Peter Cordes Jan 29 '19 at 09:18
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    @RobertCotterman Yes, `--counter` is off by one. `counter--` will give you exactly the `progressCheck` number of iterations (or you could set `progressCheck = 50001;` of course). – JimmyB Jan 29 '19 at 14:33
1

I am also surprised by seeing the performance of the above codes. It's all about the time taken by the compiler for executing the program as per the declared variable. In the second (inefficient) example:

for (long i = startNum; i <= stopNum; i++) {
    if (i % progressCheck == 0) {
        System.out.println(i)
    }
}

You are performing the modulus operation between two variables. Here, compiler has to check the value of stopNum and progressCheck to go to the specific memory block located for these variables every time after each iteration because it is a variable and its value might be change.

That's why after each iteration compiler went to the memory location to check the latest value of the variables. Therefore at the compile time the compiler was not able to create efficient byte code.

In the first code example, you are performing modulus operator between a variable and a constant numeric value which is not going to change within execution and compiler no need to check the value of that numeric value from the memory location. That's why compiler was able to create efficient byte code. If you declare progressCheck as a final or as a final static variable then at the time of run-time/compile-time compiler know that it's a final variable and its value not going to change then compiler replace the progressCheck with 50000 in code:

for (long i = startNum; i <= stopNum; i++) {
    if (i % 50000== 0) {
        System.out.println(i)
    }
}

Now you can see that this code also looks like the first (efficient) code example. The performance of first code and as we mentioned above both code will work efficiently. There will not be much difference in execution time of either code example.

double-beep
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Bishal Dubey
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    There is a HUGE difference, albeit i was doing the operation a trillion times, so over 1 trillion operations it saved 89% time to do the "efficient" code. mind you if you're only doing it a few thousand times, were talking such a tiny difference, it's probably not a big deal. i mean over 1000 operations it would save you 1 millionth of 7 seconds. – Robert Cotterman Jan 29 '19 at 03:38
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    @Bishal Dubey "There will not be much difference in execution time of both code." Did you read the question? – Grant Foster Jan 29 '19 at 16:27
  • "That's why after each iteration compiler went to the memory location to check the latest value of the variables" - Unless the variable was declared `volatile` the 'compiler' will *not* read its value from RAM over and over again. – JimmyB Feb 10 '19 at 10:07