How does computer convert floating point numbers to decimal string? - c++

When I run the following code, the output is accurately the number 2500 in decimal.
(g++ 5.3.1 on ubuntu)
#include<iostream>
#include<cmath>
using namespace std;
int main(){
cout.precision(0);
cout << fixed << pow(2.0,500.0);
return 0;
}
I wonder how C++ converted this floating point number to its decimal string at such a high precision.
I know that 2500 can be accurately presented in IEEE 754 format. But I think mod 10 and divided by 10 can cause precision loss on floating point numbers. What algorithm is used when the conversion proceed?

Yes, there exists an exact double-precision floating-point representation for 2500. You should not assume that pow(2.0,500.0) produces this value, though. There is no guarantee of accuracy for the function pow, and you may find SO questions that arose from pow(10.0, 2.0) not producing 100.0, although the mathematical result was perfectly representable too.
But anyway, to answer your question, the conversion from the floating-point binary representation to decimal does not in general rely on floating-point operations, which indeed would be too inaccurate for the intended accuracy of the end result. In general, accurate conversion requires reliance on big integer arithmetics. In the case of 2500, for instance, the naïve algorithm would be to repeatedly divide the big integer written in binary 1000…<500 zeroes in total>… by ten.
There are some cases where floating-point arithmetic can be used, for instance taking advantage of the fact that powers of 10 up to 1023 are represented exactly in IEEE 754 double-precision. But correctly rounded conversion between binary floating-point and decimal floating-point always require big integer arithmetics in general, and this is particularly visible far away from 1.0.

Related

How to accurately multiply a COleCurrency by a double?

I have a COleCurrency object that represents a unit price. And I have a double value that represents a quantity. And I need to calculate the total dollar amount to the nearest penny.
Looks like COleCurrency has built in multiplication operators, but only for multiplication with a long value.
I can multiply COleCurrency.m_cur.int64 by the double, but that converts the double to __int64 so it wouldn't be accurate.
What is the best way to accurately multiply a COleCurrency by a double?
Finite binary floating point values form a proper subset of finite decimal values. While any given floating point value has an exact, finite representation in decimal, the opposite isn't true. In other words, not every decimal can be represented using a finite binary floating point value. A simple example is 0.1 that produces an infinite sequence of binary digits when converted to a binary floating point value.
The important point here is that if you are dealing with fractional values, using binary floating point values to represent them will in general introduce inaccuracies (with very few exceptions, such as 0.5). The only way to perform accurate multiplications with an integer value is to use an integer as the multiplicand.
Since you have opted to use a floating point value the only thing you can do is limit the inaccuracies. The proposed solution:
__int64 x = currency.m_cur.int64 * (__int64)dbl;
suffers from the same "possible loss of data" issue the compiler warned about. Since you're now using an explicit cast, this silences the compiler. The effect is still the same: The floating point value gets truncated.
A better approach would be to convert the 64-bit integer value to a double first. This produces an exact floating point representation of the integer value, provided that it is within range (roughly +/- 1e15). You can then multiply with another floating point value (which is subject to rounding errors), and finally round the result using, e.g., std::llround:
__int64 x = std::llround(static_cast<double>(currency.m_cur.int64) * dbl);

Printing exact values for floats

How does a program (MySQL is an example) store a float like 0.9 and then return it to me as 0.9? Why does it not return 0.899...?
The issue I am currently experiencing is retrieving floating point values from MySQL with C++ and then reprinting those values.
There are software libraries, like Gnu MP that implement arbitrary precision arithmetic, that calculate floating point numbers to specified precision. Using Gnu MP you can, for example, add 0.3 to 0.6, and get exactly 0.9. No more, no less.
Database servers do pretty much the same thing.
For normal, run of the mill applications, native floating point arithmetic is fast, and it's good enough. But database servers typically have plenty of spare CPU cycles. Their limiting factors will not be available CPU, but things like available I/O bandwidth. They'll have plenty of CPU cycles to burn on executing complicated arbitrary precision arithmetic calculations.
There are a number of algorithms for rounding floating point numbers in a way that will result in the same internal representation when read back in. For an overview of the subject, with links to papers with full details of the algorithms, see
Printing Floating-Point Numbers
What's happening, in a nutshell, is that the function which converts the floating-point approximation of 0.9 to decimal text is actually coming up with a value like 0.90000....0123 or 0.89999....9573. This gets rounded to 0.90000...0. And then these trailing zeros are trimmed off so you get a tidy looking 0.9.
Although floating-point numbers are inexact, and often do not use base 10 internally, they can in fact precisely save and recover a decimal representation. For instance, an IEEE 754 64 bit representation has enough precision to preserve 15 decimal digits. This is often mapped to the C language type double, and that language has the constant DBL_DIG, which will be 15 when double is this IEEE type.
If a decimal number with 15 digits or less is converted to double, it can be coverted back to exactly that number. The conversion routine just has to round it off at 15 digits; of course if the conversion routine uses, say, 40 digits, there will be messy trailing digits representing the error between the floating-point value and the original number. The more digits you print, the more accurately rendered is that error.
There is also the opposite problem: given a floating-point object, can it be printed into decimal such that the resulting decimal can be scanned back to reproduce that object? For an IEEE 64 bit double, the number of decimal digits required for that is 17.

Is hardcode float precise if it can be represented by binary format in IEEE 754?

for example, 0 , 0.5, 0.15625 , 1 , 2 , 3... are values converted from IEEE 754. Are their hardcode version precise?
for example:
is
float a=0;
if(a==0){
return true;
}
always return true? other example:
float a=0.5;
float b=0.25;
float c=0.125;
is a * b always equal to 0.125 and a * b==c always true? And one more example:
int a=123;
float b=0.5;
is a * b always be 61.5? or in general, is integer multiply by IEEE 754 binary float precise?
Or a more general question: if the value is hardcode and both the value and result can be represented by binary format in IEEE 754 (e.g.:0.5 - 0.125), is the value precise?
There is no inherent fuzzyness in floating-point numbers. It's just that some, but not all, real numbers can't be exactly represented.
Compare with a fixed-width decimal representation, let's say with three digits. The integer 1 can be represented, using 1.00, and 1/10 can be represented, using 0.10, but 1/3 can only be approximated, using 0.33.
If we instead use binary digits, the integer 1 would be represented as 1.00 (binary digits), 1/2 as 0.10, 1/4 as 0.01, but 1/3 can (again) only be approximated.
There are some things to remember, though:
It's not the same numbers as with decimal digits. 1/10 can be
written exactly as 0.1 using decimal digits, but not using binary
digits, no matter how many you use (short of infinity).
In practice, it is difficult to keep track of which numbers can be
represented and which can't. 0.5 can, but 0.4 can't. So when you need
exact numbers, such as (often) when working with money, you shouldn't
use floating-point numbers.
According to some sources, some processors do strange things
internally when performing floating-point calculations on numbers
that can't be exactly represented, causing results to vary in a way
that is, in practice, unpredictable.
(My opinion is that it's actually a reasonable first approximation to say that yes, floating-point numbers are inherently fuzzy, so unless you are sure your particular application can handle that, stay away from them.)
For more details than you probably need or want, read the famous What Every Computer Scientist Should Know About Floating-Point Arithmetic. Also, this somewhat more accessible website: The Floating-Point Guide.
No, but as Thomas Padron-McCarthy says, some numbers can be exactly represented using binary but not all of them can.
This is the way I explain it to non-developers who I work with (like Mahmut Ali I also work on an very old financial package): Imagine having a very large cake that is cut into 256 slices. Now you can give 1 person the whole cake, 2 people half of the slices but soon as you decide to split it between 3 you can't - it's either 85 or 86 - you can't split the cake any further. The same is with floating point. You can only get exact numbers on some representations - some numbers can only be closely approximated.
C++ does not require binary floating point representation. Built-in integers are required to have a binary representation, commonly two's complement, but one's complement and sign and magnitude are also supported. But floating point can be e.g. decimal.
This leaves open the question of whether C++ floating point can have a radix that does not have 2 as a prime factor, like 2 and 10. Are other radixes permitted? I don't know, and last time I tried to check that, I failed.
However, assuming that the radix must be 2 or 10, then all your examples involve values that are powers of 2 and therefore can be exactly represented.
This means that the single answer to most of your questions is “yes”. The exception is the question “is integer multiply by IEEE 754 binary float [exact]”. If the result exceeds the precision available, then it can't be exact, but otherwise it is.
See the classic “What Every Computer Scientist Should Know About Floating-Point Arithmetic” for background info about floating point representation & properties in general.
If a value can be exactly represented in 32-bit or 64-bit IEEE 754, then that doesn't mean that it can be exactly represented with some other floating point representation. That's because different 32-bit representations and different 64-bit representations use different number of bits to hold the mantissa and have different exponent ranges. So a number that can be exactly represented in one way, can be beyond the precision or range of some other representation.
You can use std::numeric_limits<T>::is_iec559 (where e.g. T is double) to check whether your implementation claims to be IEEE 754 compatible. However, when floating point optimizations are turned on, at least the g++ compiler (1)erroneously claims to be IEEE 754, while not treating e.g. NaN values correctly according to that standard. In effect, the is_iec559 only tells you whether the number representation is IEEE 754, and not whether the semantics conform.
(1) Essentially, instead of providing different types for different semantics, gcc and g++ try to accommodate different semantics via compiler options. And with separate compilation of parts of a program, that can't conform to the C++ standard.
In principle, this should be possible. If you restrict yourself to exactly this class of numbers with a finite 2-power representation.
But it is dangerous: what if someone takes your code and changes your 0.5 to 0.4 or your .0625 to .065 due to whatever reasons? Then your code is broken. And no, even excessive comments won't help about that - someone will always ignore them.

Should I worry about precision when I use C++ mathematical functions with integers?

For example, The code below will give undesirable result due to precision of floating point numbers.
double a = 1 / 3.0;
int b = a * 3; // b will be 0 here
I wonder whether similar problems will show up if I use mathematical functions. For example
int a = sqrt(4); // Do I have guarantee that I will always get 2 here?
int b = log2(8); // Do I have guarantee that I will always get 3 here?
If not, how to solve this problem?
Edit:
Actually, I came across this problem when I was programming for an algorithm task. There I want to get
the largest integer which is power of 2 and is less than or equal to integer N
So round function can not solve my problem. I know I can solve this problem through a loop, but it seems not very elegant.
I want to know if
int a = pow(2, static_cast<int>(log2(N)));
can always give correct result. For example if N==8, is it possible that log2(N) gives me something like 2.9999999999999 and the final result become 4 instead of 8?
Inaccurate operands vs inaccurate results
I wonder whether similar problems will show up if I use mathematical functions.
Actually, the problem that could prevent log2(8) to be 3 does not exist for basic operations (including *). But it exists for the log2 function.
You are confusing two different issues:
double a = 1 / 3.0;
int b = a * 3; // b will be 0 here
In the example above, a is not exactly 1/3, so it is possible that a*3 does not produce 1.0. The product could have happened to round to 1.0, it just doesn't. However, if a somehow had been exactly 1/3, the product of a by 3 would have been exactly 1.0, because this is how IEEE 754 floating-point works: the result of basic operations is the nearest representable value to the mathematical result of the same operation on the same operands. When the exact result is representable as a floating-point number, then that representation is what you get.
Accuracy of sqrt and log2
sqrt is part of the “basic operations”, so sqrt(4) is guaranteed always, with no exception, in an IEEE 754 system, to be 2.0.
log2 is not part of the basic operations. The result of an implementation of this function is not guaranteed by the IEEE 754 standard to be the closest to the mathematical result. It can be another representable number further away. So without more hypotheses on the log2 function that you use, it is impossible to tell what log2(8.0) can be.
However, most implementations of reasonable quality for elementary functions such as log2 guarantee that the result of the implementation is within 1 ULP of the mathematical result. When the mathematical result is not representable, this means either the representable value above or the one below (but not necessarily the closest one of the two). When the mathematical result is exactly representable (such as 3.0), then this representation is still the only one guaranteed to be returned.
So about log2(8), the answer is “if you have a reasonable quality implementation of log2, you can expect the result to be 3.0`”.
Unfortunately, not every implementation of every elementary function is a quality implementation. See this blog post, caused by a widely used implementation of pow being inaccurate by more than 1 ULP when computing pow(10.0, 2.0), and thus returning 99.0 instead of 100.0.
Rounding to the nearest integer
Next, in each case, you assign the floating-point to an int with an implicit conversion. This conversion is defined in the C++ standard as truncating the floating-point values (that is, rounding towards zero). If you expect the result of the floating-point computation to be an integer, you can round the floating-point value to the nearest integer before assigning it. It will help obtain the desired answer in all cases where the error does not accumulate to a value larger than 1/2:
int b = std::nearbyint(log2(8.0));
To conclude with a straightforward answer to the question the the title: yes, you should worry about accuracy when using floating-point functions for the purpose of producing an integral end-result. These functions do not come even with the guarantees that basic operations come with.
Unfortunately the default conversion from a floating point number to integer in C++ is really crazy as it works by dropping the decimal part.
This is bad for two reasons:
a floating point number really really close to a positive integer, but below it will be converted to the previous integer instead (e.g. 3-1×10-10 = 2.9999999999 will be converted to 2)
a floating point number really really close to a negative integer, but above it will be converted to the next integer instead (e.g. -3+1×10-10 = -2.9999999999 will be converted to -2)
The combination of (1) and (2) means also that using int(x + 0.5) will not work reasonably as it will round negative numbers up.
There is a reasonable round function, but unfortunately returns another floating point number, thus you need to write int(round(x)).
When working with C99 or C++11 you can use lround(x).
Note that the only numbers that can be represented correctly in floating point are quotients where the denominator is an integral power of 2.
For example 1/65536 = 0.0000152587890625 can be represented correctly, but even just 0.1 is impossible to represent correctly and thus any computation involving that quantity will be approximated.
Of course when using 0.1 approximations can cancel out leaving a correct result occasionally, but even just adding ten times 0.1 will not give 1.0 as result when doing the computation using IEEE754 double-precision floating point numbers.
Even worse the compilers are allowed to use higher precision for intermediate results. This means that adding 10 times 0.1 may give back 1 when converted to an integer if the compiler decides to use higher accuracy and round to closest double at the end.
This is "worse" because despite being the precision higher the results are compiler and compiler options dependent, making reasoning about the computations harder and making the exact result non portable among different systems (even if they use the same precision and format).
Most compilers have special options to avoid this specific problem.

Rounding a float upward to an integer, how reliable is that?

I've seen static_cast<int>(std::ceil(floatValue)); before.
My question though, is can I absolutely count on this not "needlessly" rounding up? I've read that some whole numbers can't be perfectly represented in floating point, so my worry is that the miniscule "error" will trick ceil() into rounding upwards when it logically shouldn't. Not only that, but once rounded up, I worry it may be possible for a small "error" in representation to cause the number to be slightly less than a whole number, causing the cast to int to truncate it.
Is this worry unfounded? I remember a while back, an example in python where printing a specific whole number would cause it to print something very slightly less (like x.999, though I can't remember the exact number)
The reason I need to make sure, is I'm writing a texture buffer. The common case is whole numbers as floating point, but it'll occasionally get between values that need to be rounded to the nearest integer width and height that contains them. It increments in steps of power of 2, so the cost of rounding up needlessly can cause what should've only took a 256x256 texture to need a 512x512 texture.
If floatValue is exact, then there is no problem with rounding in your code. The only possible problem is overflow (if the result doesn't fit inside an int). Of course with such large values, the float will typically not have enough precision to distinguish adjacent integers anyway.
However, the danger usually lies in floatValue itself not being exact. For example, if it is the result of some computation whose exact answer is a whole number, it may end up a tiny amount greater than a whole number due to floating point rounding errors in the computation.
So whether you have a problem depends on how you got floatValue.
can I absolutely count on this not "needlessly" rounding up? I've read that some whole numbers can't be perfectly represented in floating point, so my worry is that the miniscule "error" will trick ceil()
Yes, some large numbers are impossible to represent exactly as floating-point numbers. In the zone where this happens, all floating-point numbers are integers. The error is not minuscule: the error in representing an integer by a floating-point, if error there is, is at least one. And, obviously, in the zone where some integers cannot be represented as floating-point and where all floating-point numbers are integers, ceil(f) == f.
The zone in question is |f| > 224 (16*1024*1024) for IEEE 754 single-precision and |f| > 253 for IEEE 754 double-precision.
A problem you are more likely to come across does not come from the impossibility of representing integers in floating-point format but from the cumulative effects of rounding errors. If your compiler offers IEEE 754 (the floating-point standard implemented exactly by the SSE2 instructions of modern and not so modern Intel processors) semantics, then any +, -, *, / and sqrt operation that results in a number exactly representable as floating-point is guaranteed to produce that result, but if several of the operations you apply do not have exactly representable results, the floating-point computation may drift away from the mathematical computation, even when the final result is an integer and is exactly representable. Then you may end up with a floating-point result slightly above the target integer and cause ceil() to return something other than you would have obtained with exact mathematical computations.
There are ways to be confident that some floating-point operations are exact (because the result is always representable). For instance (double)float1 * (double)float2, where float1 and float2 are two single-precision variables, is always exact, because the mathematical result of the multiplication of two single-precision numbers is always representable as a double. By doing the computation the “right” way, it is possible to minimize or eliminate the error in the end result.
The range is 0.0 to ~1024.0
All integers in this range can be represented exactly as float, so you'll be fine.
You'll only start having issues once you stray beyond the 24 bits of mantissa afforded by float.