Here is my codeļ¼
int a = 0x451998a0;
float b = *((float *)&a);
printf("coverto float: %f, %.10lf\n", b, b);
In windows the output is:
coverto float: 2457.539063, 2457.5390625000
In linux the output is:
coverto float: 2457.539062, 2457.5390625000
Is there any way to make sure the output is the same?
The behavior you're seeing is just a consequence of the fact that Windows' printf() function is implemented differently from Linux's printf() function. Most likely the difference is in how printf() implements number rounding.
How printf() works under the hood in either system is an implementation detail; thus the system is not likely to provide such fine-grained control on how printf() displays the floating point values.
There are two ways that may work to keep them the same:
Use more precision during calculation than while displaying it. For example, some scientific and graphing calculators use double precision for all internal calculations, but display the results with only float precision.
Use a cross-platform printf() library. Such libraries would most likely have the same behavior on all platforms, as the calculations required to determine what digits to display are usually platform-agnostic.
However, this really isn't as big of a problem as you think it is. The difference between the outputs is 0.000001. That is a ~0.0000000004% difference from either the two values. The display error is really quite negligible.
Consider this: the distance between Los Angeles and New York is 2464 miles, which is of the same order of magnitude as the numbers in your display outputs. A difference of 0.000001 miles is 1.61 millimeters. We of course don't measure distances between cities with anywhere near that kind of precision. :-)
If you use the same printf() implementation, there's a good chance they'll show the same output. Depending on what you're up to, it may be easier to use GNU GCC on both OSes, or to get printf() source code and add it to your project (you should have no trouble googling one).
BTW - have you actually checked what that hex number encodes? Should it round up or down? The 625 thing is likely itself rounded, so you shouldn't assume it should round to 63....
The obvious answer is to use less precision in your output. In general,
if there's any calculation involved, you can't even be sure that the
actual floating point values are identical. And how printf and
ostream round is implementation defined, even if the floating point
values are equal.
In general, C++ doesn't guarantee that two implementations produce the
same results. In this particular case, if it's important, you can do
the rounding by hand, before doing the conversion, but you'll still have
occasional problems because the actual floating point values will be
different. This may, in fact, occur even with different levels of
optimization with the same compiler. So anything you try (other than
writing the entire program in assembler) is bound to be a loosing battle
in the end.
Related
I'm in the process of converting a program from Scilab code to C++. One loop in particular is producing a slightly different result than the original Scilab code (it's a long piece of code so I'm not going to include it in the question but I'll try my best to summarise the issue below).
The problem is, each step of the loop uses calculations from the previous step. Additionally, the difference between calculations only becomes apparent around the 100,000th iteration (out of approximately 300,000).
Note: I'm comparing the output of my C++ program with the outputs of Scilab 5.5.2 using the "format(25);" command. Meaning I'm comparing 25 significant digits. I'd also like to point out I understand how precision cannot be guaranteed after a certain number of bits but read the sections below before commenting. So far, all calculations have been identical up to 25 digits between the two languages.
In attempts to get to the bottom of this issue, so far I've tried:
Examining the data type being used:
I've managed to confirm that Scilab is using IEEE 754 doubles (according to the language documentation). Also, according to Wikipedia, C++ isn't required to use IEEE 754 for doubles, but from what I can tell, everywhere I use a double in C++ it has perfectly match Scilab's results.
Examining the use of transcendental functions:
I've also read from What Every Computer Scientist Should Know About Floating-Point Arithmetic that IEEE does not require transcendental functions to be exactly rounded. With that in mind, I've compared the results of these functions (sin(), cos(), exp()) in both languages and again, the results appear to be the same (up to 25 digits).
The use of other functions and predefined values:
I repeated the above steps for the use of sqrt() and pow(). As well as the value of Pi (I'm using M_PI in C++ and %pi in Scilab). Again, the results were the same.
Lastly, I've rewritten the loop (very carefully) in order to ensure that the code is identical between the two languages.
Note: Interestingly, I noticed that for all the above calculations the results between the two languages match farther than the actual result of the calculations (outside of floating point arithmetic). For example:
Value of sin(x) using Wolfram Alpha = 0.123456789.....
Value of sin(x) using Scilab & C++ = 0.12345yyyyy.....
Where even once the value computed using Scilab or C++ started to differ from the actual result (from Wolfram). Each language's result still matched each other. This leads me to believe that most of the values are being calculated (between the two languages) in the same way. Even though they're not required to by IEEE 754.
My original thinking was one of the first three points above are implemented differently between the two languages. But from what I can tell everything seems to produce identical results.
Is it possible that even though all the inputs to these loops are identical, the results can be different? Possibly because a very small error (past what I can see with 25 digits) is occurring that accumulates over time? If so, how can I go about fixing this issue?
No, the format of the numbering system does not guarantee equivalent answers from functions in different languages.
Functions, such as sin(x), can be implemented in different ways, using the same language (as well as different languages). The sin(x) function is an excellent example. Many implementations will use a look-up table or look-up table with interpolation. This has speed advantages. However, some implementations may use a Taylor Series to evaluate the function. Some implementations may use polynomials to come up with a close approximation.
Having the same numeric format is one hurdle to solve between languages. Function implementation is another.
Remember, you need to consider the platform as well. A program that uses an 80-bit floating point processor will have different results than a program that uses a 64-bit floating point software implementation.
Some architectures provide the capability of using extended precision floating point registers (e.g. 80 bits internally, versus 64-bit values in RAM). So, it's possible to get slightly different results for the same calculation, depending on how the computations are structured, and the optimization level used to compile the code.
Yes, it's possible to have a different results. It's possible even if you are using exactly the same source code in the same programming language for the same platform. Sometimes it's enough to have a different compiler switch; for example -ffastmath would lead the compiler to optimize your code for speed rather than accuracy, and, if your computational problem is not well-conditioned to begin with, the result may be significantly different.
For example, suppose you have this code:
x_8th = x*x*x*x*x*x*x*x;
One way to compute this is to perform 7 multiplications. This would be the default behavior for most compilers. However, you may want to speed this up by specifying compiler option -ffastmath and the resulting code would have only 3 multiplications:
temp1 = x*x; temp2 = temp1*temp1; x_8th = temp2*temp2;
The result would be slightly different because finite precision arithmetic is not associative, but sufficiently close for most applications and much faster. However, if your computation is not well-conditioned that small error can quickly get amplified into a large one.
Note that it is possible that the Scilab and C++ are not using the exact same instruction sequence, or that one uses FPU and the other uses SSE, so there may not be a way to get them to be exactly the same.
As commented by IInspectable, if your compiler has _control87() or something similar, you can use it to change the precision and/or rounding settings. You could try combinations of this to see if it has any effect, but again, even you manage to get the settings identical for Scilab and C++, differences in the actual instruction sequences may be the issue.
http://msdn.microsoft.com/en-us/library/e9b52ceh.aspx
If SSE is used, I"m not sure what can be adjusted as I don't think SSE has an 80 bit precision mode.
In the case of using FPU in 32 bit mode, and if your compiler doesn't have something like _control87, you could use assembly code. If inline assembly is not allowed, you would need to call an assembly function. This example is from an old test program:
static short fcw; /* 16 bit floating point control word */
/* ... */
/* set precision control to extended precision */
__asm{
fnstcw fcw
or fcw,0300h
fldcw fcw
}
I am porting some program from Matlab to C++ for efficiency. It is important for the output of both programs to be exactly the same (**).
I am facing different results for this operation:
std::sin(0.497418836818383950) = 0.477158760259608410 (C++)
sin(0.497418836818383950) = 0.47715876025960846000 (Matlab)
N[Sin[0.497418836818383950], 20] = 0.477158760259608433 (Mathematica)
So, as far as I know both C++ and Matlab are using IEEE754 defined double arithmetic. I think I have read somewhere that IEEE754 allows differents results in the last bit. Using mathematica to decide, seems like C++ is more close to the result. How can I force Matlab to compute the sin with precision to the last bit included, so that the results are the same?
In my program this behaviour leads to big errors because the numerical differential equation solver keeps increasing this error in the last bit. However I am not sure that C++ ported version is correct. I am guessing that even if the IEEE754 allows the last bit to be different, somehow guarantees that this error does not get bigger when using the result in more IEEE754 defined double operations (because otherwise, two different programs correct according to the IEEE754 standard could produce completely different outputs). So the other question is Am I right about this?
I would like get an answer to both bolded questions. Edit: The first question is being quite controversial, but is the less important, can someone comment about the second one?
Note: This is not an error in the printing, just in case you want to check, this is how I obtained these results:
http://i.imgur.com/cy5ToYy.png
Note (**): What I mean by this is that the final output, which are the results of some calculations showing some real numbers with 4 decimal places, need to be exactly the same. The error I talk about in the question gets bigger (because of more operations, each of one is different in Matlab and in C++) so the final differences are huge) (If you are curious enough to see how the difference start getting bigger, here is the full output [link soon], but this has nothing to do with the question)
Firstly, if your numerical method depends on the accuracy of sin to the last bit, then you probably need to use an arbitrary precision library, such as MPFR.
The IEEE754 2008 standard doesn't require that the functions be correctly rounded (it does "recommend" it though). Some C libms do provide correctly rounded trigonometric functions: I believe that the glibc libm does (typically used on most linux distributions), as does CRlibm. Most other modern libms will provide trig functions that are within 1 ulp (i.e. one of the two floating point values either side of the true value), often termed faithfully rounded, which is much quicker to compute.
None of those values you printed could actually arise as IEEE 64bit floating point values (even if rounded): the 3 nearest (printed to full precision) are:
0.477158760259608 405451814405751065351068973541259765625
0.477158760259608 46096296563700889237225055694580078125
0.477158760259608 516474116868266719393432140350341796875
The possible values you could want are:
The exact sin of the decimal .497418836818383950, which is
0.477158760259608 433132061388630377105954125778369485736356219...
(this appears to be what Mathematica gives).
The exact sin of the 64-bit float nearest .497418836818383950:
0.477158760259608 430531153841011107415427334794384396325832953...
In both cases, the first of the above list is the nearest (though only barely in the case of 1).
The sine of the double constant you wrote is about 0x1.e89c4e59427b173a8753edbcb95p-2, whose nearest double is 0x1.e89c4e59427b1p-2. To 20 decimal places, the two closest doubles are 0.47715876025960840545 and 0.47715876025960846096.
Perhaps Matlab is displaying a truncated value? (EDIT: I now see that the fourth-last digit is a 6, not a 0. Matlab is giving you a result that's still faithfully-rounded, but it's the farther of the two closest doubles to the desired result. And it's still printing out the wrong number.
I should also point out that Mathematica is probably trying to solve a different problem---compute the sine of the decimal number 0.497418836818383950 to 20 decimal places. You should not expect this to match either the C++ code's result or Matlab's result.
I am writing a simulation program that proceeds in discrete steps. The simulation consists of many nodes, each of which has a floating-point value associated with it that is re-calculated on every step. The result can be positive, negative or zero.
In the case where the result is zero or less something happens. So far this seems straightforward - I can just do something like this for each node:
if (value <= 0.0f) something_happens();
A problem has arisen, however, after some recent changes I made to the program in which I re-arranged the order in which certain calculations are done. In a perfect world the values would still come out the same after this re-arrangement, but because of the imprecision of floating point representation they come out very slightly different. Since the calculations for each step depend on the results of the previous step, these slight variations in the results can accumulate into larger variations as the simulation proceeds.
Here's a simple example program that demonstrates the phenomena I'm describing:
float f1 = 0.000001f, f2 = 0.000002f;
f1 += 0.000004f; // This part happens first here
f1 += (f2 * 0.000003f);
printf("%.16f\n", f1);
f1 = 0.000001f, f2 = 0.000002f;
f1 += (f2 * 0.000003f);
f1 += 0.000004f; // This time this happens second
printf("%.16f\n", f1);
The output of this program is
0.0000050000057854
0.0000050000062402
even though addition is commutative so both results should be the same. Note: I understand perfectly well why this is happening - that's not the issue. The problem is that these variations can mean that sometimes a value that used to come out negative on step N, triggering something_happens(), now may come out negative a step or two earlier or later, which can lead to very different overall simulation results because something_happens() has a large effect.
What I want to know is whether there is a good way to decide when something_happens() should be triggered that is not going to be affected by the tiny variations in calculation results that result from re-ordering operations so that the behavior of newer versions of my program will be consistent with the older versions.
The only solution I've so far been able to think of is to use some value epsilon like this:
if (value < epsilon) something_happens();
but because the tiny variations in the results accumulate over time I need to make epsilon quite large (relatively speaking) to ensure that the variations don't result in something_happens() being triggered on a different step. Is there a better way?
I've read this excellent article on floating point comparison, but I don't see how any of the comparison methods described could help me in this situation.
Note: Using integer values instead is not an option.
Edit the possibility of using doubles instead of floats has been raised. This wouldn't solve my problem since the variations would still be there, they'd just be of a smaller magnitude.
I've worked with simulation models for 2 years and the epsilon approach is the sanest way to compare your floats.
Generally, using suitable epsilon values is the way to go if you need to use floating point numbers. Here are a few things which may help:
If your values are in a known range you and you don't need divisions you may be able to scale the problem and use exact operations on integers. In general, the conditions don't apply.
A variation is to use rational numbers to do exact computations. This still has restrictions on the operations available and it typically has severe performance implications: you trade performance for accuracy.
The rounding mode can be changed. This can be use to compute an interval rather than an individual value (possibly with 3 values resulting from round up, round down, and round closest). Again, it won't work for everything but you may get an error estimate out of this.
Keeping track of the value and a number of operations (possible multiple counters) may also be used to estimate the current size of the error.
To possibly experiment with different numeric representations (float, double, interval, etc.) you might want to implement your simulation as templates parameterized for the numeric type.
There are many books written on estimating and minimizing errors when using floating point arithmetic. This is the topic of numerical mathematics.
Most cases I'm aware of experiment briefly with some of the methods mentioned above and conclude that the model is imprecise anyway and don't bother with the effort. Also, doing something else than using float may yield better result but is just too slow, even using double due to the doubled memory footprint and the smaller opportunity of using SIMD operations.
I recommend that you single step - preferably in assembly mode - through the calculations while doing the same arithmetic on a calculator. You should be able to determine which calculation orderings yield results of lesser quality than you expect and which that work. You will learn from this and probably write better-ordered calculations in the future.
In the end - given the examples of numbers you use - you will probably need to accept the fact that you won't be able to do equality comparisons.
As to the epsilon approach you usually need one epsilon for every possible exponent. For the single-precision floating point format you would need 256 single precision floating point values as the exponent is 8 bits wide. Some exponents will be the result of exceptions but for simplicity it is better to have a 256 member vector than to do a lot of testing as well.
One way to do this could be to determine your base epsilon in the case where the exponent is 0 i e the value to be compared against is in the range 1.0 <= x < 2.0. Preferably the epsilon should be chosen to be base 2 adapted i e a value that can be exactly represented in a single precision floating point format - that way you know exactly what you are testing against and won't have to think about rounding problems in the epsilon as well. For exponent -1 you would use your base epsilon divided by two, for -2 divided by 4 and so on. As you approach the lowest and the highest parts of the exponent range you gradually run out of precision - bit by bit - so you need to be aware that extreme values can cause the epsilon method to fail.
If it absolutely has to be floats then using an epsilon value may help but may not eliminate all problems. I would recommend using doubles for the spots in the code you know for sure will have variation.
Another way is to use floats to emulate doubles, there are many techniques out there and the most basic one is to use 2 floats and do a little bit of math to save most of the number in one float and the remainder in the other (saw a great guide on this, if I find it I'll link it).
Certainly you should be using doubles instead of floats. This will probably reduce the number of flipped nodes significantly.
Generally, using an epsilon threshold is only useful when you are comparing two floating-point number for equality, not when you are comparing them to see which is bigger. So (for most models, at least) using epsilon won't gain you anything at all -- it will just change the set of flipped nodes, it wont make that set smaller. If your model itself is chaotic, then it's chaotic.
i am running a program in c++ on windows and on linux.
the output is meant to be identical.
i am trying to make sure that the only differences are real differences oppose to working inviorment differences.
so far i have taken care of all the differences that can be caused by \r\n differences
but there is one thing that i can't seem to figure out.
in the windows out put there is a 0.000 and in linux it is -0.000
does any one know what can it be that is making the difference?
thanx
Probably it comes from differences in how the optimizer optimizes some FP calculations (that can be configurable - see e.g. here); in one case you get a value slightly less than 0, in the other slightly more. Both in output are rounded to a 0.000, but they keep their "real" sign.
Since in the IEEE floating point format the sign bit is separate from the value, you have two different values of 0, a positive and a negative one. In most cases it doesn't make a difference; both zeros will compare equal, and they indeed describe the same mathematical value (mathematically, 0 and -0 are the same). Where the difference can be significant is when you have underflow and need to know whether the underflow occurred from a positive or from a negative value. Also if you divide by 0, the sign of the infinity you get depends on the sign of the 0 (i.e. 1/+0.0 give +Inf, but 1/-0.0 gives -Inf). In other words, most probably it won't make a difference for you.
Note however that the different output does not necessarily mean that the number itself is different. It could well be that the value in Windows is also -0.0, but the output routine on Windows doesn't distinguish between +0.0 and -0.0 (they compare equal, after all).
Unless using (unsafe) flags like -ffast-math, the compiler is limited in the assumptions it can make when 'optimizing' IEEE-754 arithmetic. First check that both platforms are using the same rounding.
Also, if possible, check they are using the same floating-point unit. i.e., SSE vs FPU on x86. The latter might be an issue with math library function implementations - especially trigonometric / transcendental functions.
I was just reading about rounding errors in C++. So, if I'm making a math intense program (or any important calculations) should I just drop floats all together and use only doubles or is there an easier way to prevent rounding errors?
Obligatory lecture: What Every Programmer Should Know About Floating-Point Arithmetic.
Also, try reading IEEE Floating Point standard.
You'll always get rounding errors. Unless you use an infinite arbitrary precision library, like gmplib. You have to decide if your application really needs this kind of effort.
Or, you could use integer arithmetic, converting to floats only when needed. This is still hard to do, you have to decide if it's worth it.
Lastly, you can use float or double taking care not to make assumption about values at the limit of representation's precision. I'd wish this Valgrind plugin was implemented (grep for float)...
The rounding errors are normally very insignificant, even using floats. Mathematically-intense programs like games, which do very large numbers of floating-point computations, often still use single-precision.
This might work if your highest number is less than 10 billion and you're using C++ double precision.
if ( ceil(10000*(x + 0.00001)) > ceil(100000*(x - 0.00001))) {
x = ceil(10000*(x + 0.00004)) / 10000;
}
This should allow at least the last digit to be off +/- 9. I'm assuming dividing by 1000 will always just move a decimal place. If not, then maybe it could be done in binary.
You would have to apply it after every operation that is not +, -, *, or a comparison. For example, you can't do two divisions in the same formula because you'd have to apply it to each division.
If that doesn't work, you could work in integers by scaling the numbers up and always use integer division. If you need advanced functions maybe there is a package that does deterministic integer math. Integer division is required in a lot of financial settings because of round off error being subject to exploit like in the movie "The Office".