I have a string "1613894376.500012077" and I want to use strtof in order to convert to floating point 1613894376.500012077. The problem is when I use strtof I get the following result with the decimal misplaced 1.61389e+09. Please help me determine how to use strof properly.
A typical float is 32-bit and can only represent exactly about 232 different values. "1613894376.500012077" is not one of those.
"1.61389e+09" is the same value as "1613890000.0" and represents a close value that float can represent.
The 2 closest floats are:
1613894272.0
1613894400.0 // slightly closer to 1613894376.500012077
Print with more precision to see more digits.
The decimal point is not misplaced. The notation “1.61389e+09” means 1.61389•109, which is 1,613,890,000., which has the decimal point in the correct place.
The actual result of strtof in your computer is probably 1,613,894,400. This is the closest value to 1613894376.500012077 that the IEEE-754 binary32 (“single”) format can represent, and that is the format commonly used for float. When you print it with %g, the default is to use just six significant digits. To see it with more precision, print it with %.999g.
The number 1613894376.500012077 is equivalent (the same number up to the precision of the machine as 1.61389e+09.) The e+09 suffix means that the decimal point is located nine decimal digits right the place it has been placed (or that the number is multiplied by 10 to the ninth power). This is a common notation in computer science called scientific notation.
Related
I am the beginner, but I think same important thinks I should learn as soon as it possible.
So I have a code:
float fl=8.28888888888888888888883E-5;
cout<<"The value = "<<fl<<endl;
But my .exe file after run show:
8.2888887845911086e-005
I suspected the numbers to limit of the type and rest will be the zero, but I saw digits, which are random. Maybe it gives digits from memory after varible?
Could you explain me how it does works?
I suspected the numbers to limit of the type and rest will be the zero
Yes, this is exactly what happens, but it happens in binary. This program will show it by using the hexadecimal printing format %a:
#include <stdio.h>
int main(int c, char *v[]) {
float fl = 8.28888888888888888888883E-5;
printf("%a\n%a\n", 8.28888888888888888888883E-5, fl);
}
It shows:
0x1.5ba94449649e2p-14
0x1.5ba944p-14
In these results, 0x1.5ba94449649e2p-14 is the hexadecimal representation of the double closest to 8.28888888888888888888883*10-5, and 0x1.5ba944p-14
is the representation of the conversion to float of that number. As you can see, the conversion simply truncated the last digits (in this case. The conversion is done according to the rounding mode, and when the rounding goes up instead of down, it changes one or more of the last digits).
When you observe what happens in decimal, the fact that float and double are binary floating-point formats on your computer means that there are extra digits in the representation of the value.
I suspected the numbers to limit of the type and rest will be the zero
That is what happens internally. Excess bits beyond what the type can store are lost.
But that's in the binary representation. When you convert it to decimal, you can get trailing non-zero digits.
Example:
0b0.00100 is 0.125 in decimal
What you're seeing is a result of the fact that you cannot exactly represent a floating-point number in memory. Because of this, floats will be stored as the nearest value that can be stored in memory. A float usually has 24 bits used to represent the mantissa, which translates to about 6 decimal places (this is implementation defined though, so you shouldn't rely on this). When printing more than 6 decimal digits, you'll notice your value isn't stored in memory as the value you intended, and you'll see random digits.
So to recap, the problem you encountered is caused by the fact that base-10 decimal numbers cannot be represented in memory, instead the closest number to it is stored and this number will then be used.
each data type has range after this range all number is from memory or rubbish so you have to know this ranges and deal with it when you write code.
you can know this ranges from here or here
I want my answer in some precision form but not for input output purposes.
float a = cos ( 90*(PI/180)) gives 1.794897E-09
where as I want up to 8 decimal places answer in my variable which will give
0.00000000
setprecision or other methods are not helping to store the value in a variable. How can it be stored? basically it may not even be 8 or 9 digits .. all i want is restriction of an exponentiol form in my answer
You are limited by the type you are using. A single precision float can only represent between 6 and 9 significant decimal digits.
https://en.wikipedia.org/wiki/Single-precision_floating-point_format
Remember, a float is not a decimal value. So what you're seeing is the decimal representation. If you want more digits in the decimal representation, use a double.
https://en.wikipedia.org/wiki/Double-precision_floating-point_format
Double precision floats can represent values to 15-17 digits of decimal precision. This should guarantee the minimum of 8 that you require.
The precision and encoding of floating point values is concretely defined by IEEE 754. Without defining your own implementation of of floating points should be stored in memory, you can't really change the internal precision for how it's encoded and stored in memory.
If you want better precision you can use doubles. All the math functions work well with doubles.
I've made a BOMDAS calculator in C++ that uses doubles. Whenever I input an expression like
1000000000000000000000*1000000000000000000000
I get a result like 1000000000000000000004341624882808674582528.000000. I suspect it has something to do with floating-point numbers.
Floating point number represent values with a fixed size representation. A double can represent 16 decimal digits in form where the decimal digits can be restored (internally, it normally stores the value using base 2 which means that it can accurately represent most fractional decimal values). If the number of digits is exceeded, the value will be rounded appropriately. Of course, the upshot is that you won't necessarily get back the digits you're hoping for: if you ask for more then 16 decimal digits either explicitly or implicitly (e.g. by setting the format to std::ios_base::fixed with numbers which are bigger than 1e16) the formatting will conjure up more digits: it will accurately represent the internally held binary values which may produce up to, I think, 54 non-zero digits.
If you want to compute with large values accurately, you'll need some variable sized representation. Since your values are integers a big integer representation might work. These will typically be a lot slower to compute with than double.
A double stores 53 bits of precision. This is about 15 decimal digits. Your problem is that a double cannot store the number of digits you are trying to store. Digits after the 15th decimal digit will not be accurate.
That's not an error. It's exactly because of how floating-point types are represented, as the result is precise to double precision.
Floating-point types in computers are written in the form (-1)sign * mantissa * 2exp so they only have broader ranges, not infinite precision. They're only accurate to the mantissa precision, and the result after every operation will be rounded as such. The double type is most commonly implemented as IEEE-754 64-bit double precision with 53 bits of mantissa so it can be correct to log(253) ≈ 15.955 decimal digits. Doing 1e21*1e21 produces 1e42 which when rounding to the closest value in double precision gives the value that you saw. If you round that to 16 digits it's exactly the same as 1e42.
If you need more range, use double or long double. If you only works with integer then int64_t (or __int128 with gcc and many other compilers on 64-bit platforms) has a much larger precision (64/128 bits compared to 53 bits). If you need even more precision, use an arbitrary-precision arithmetic library instead such as GMP
I mean, for example, I have the following number encoded in IEEE-754 single precision:
"0100 0001 1011 1110 1100 1100 1100 1100" (approximately 23.85 in decimal)
The binary number above is stored in literal string.
The question is, how can I convert this string into IEEE-754 double precision representation(somewhat like the following one, but the value is not the same), WITHOUT losing precision?
"0100 0000 0011 0111 1101 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1010"
which is the same number encoded in IEEE-754 double precision.
I have tried using the following algorithm to convert the first string back to decimal number first, but it loses precision.
num in decimal = (sign) * (1 + frac * 2^(-23)) * 2^(exp - 127)
I'm using Qt C++ Framework on Windows platform.
EDIT: I must apologize maybe I didn't get the question clearly expressed.
What I mean is that I don't know the true value 23.85, I only got the first string and I want to convert it to double precision representation without precision loss.
Well: keep the sign bit, rewrite the exponent (minus old bias, plus new bias), and pad the mantissa with zeros on the right...
(As #Mark says, you have to treat some special cases separately, namely when the biased exponent is either zero or max.)
IEEE-754 (and floating point in general) cannot represent periodic binary decimals with full precision. Not even when they, in fact, are rational numbers with relatively small integer numerator and denominator. Some languages provide a rational type that may do it (they are the languages that also support unbounded precision integers).
As a consequence those two numbers you posted are NOT the same number.
They in fact are:
10111.11011001100110011000000000000000000000000000000000000000 ...
10111.11011001100110011001100110011001100110011001101000000000 ...
where ... represent an infinite sequence of 0s.
Stephen Canon in a comment above gives you the corresponding decimal values (did not check them, but I have no reason to doubt he got them right).
Therefore the conversion you want to do cannot be done as the single precision number does not have the information you would need (you have NO WAY to know if the number is in fact periodic or simply looks like being because there happens to be a repetition).
First of all, +1 for identifying the input in binary.
Second, that number does not represent 23.85, but slightly less. If you flip its last binary digit from 0 to 1, the number will still not accurately represent 23.85, but slightly more. Those differences cannot be adequately captured in a float, but they can be approximately captured in a double.
Third, what you think you are losing is called accuracy, not precision. The precision of the number always grows by conversion from single precision to double precision, while the accuracy can never improve by a conversion (your inaccurate number remains inaccurate, but the additional precision makes it more obvious).
I recommend converting to a float or rounding or adding a very small value just before displaying (or logging) the number, because visual appearance is what you really lost by increasing the precision.
Resist the temptation to round right after the cast and to use the rounded value in subsequent computation - this is especially risky in loops. While this might appear to correct the issue in the debugger, the accummulated additional inaccuracies could distort the end result even more.
It might be easiest to convert the string into an actual float, convert that to a double, and convert it back to a string.
Binary floating points cannot, in general, represent decimal fraction values exactly. The conversion from a decimal fractional value to a binary floating point (see "Bellerophon" in "How to Read Floating-Point Numbers Accurately" by William D.Clinger) and from a binary floating point back to a decimal value (see "Dragon4" in "How to Print Floating-Point Numbers Accurately" by Guy L.Steele Jr. and Jon L.White) yield the expected results because one converts a decimal number to the closest representable binary floating point and the other controls the error to know which decimal value it came from (both algorithms are improved on and made more practical in David Gay's dtoa.c. The algorithms are the basis for restoring std::numeric_limits<T>::digits10 decimal digits (except, potentially, trailing zeros) from a floating point value stored in type T.
Unfortunately, expanding a float to a double wrecks havoc on the value: Trying to format the new number will in many cases not yield the decimal original because the float padded with zeros is different from the closest double Bellerophon would create and, thus, Dragon4 expects. There are basically two approaches which work reasonably well, however:
As someone suggested convert the float to a string and this string into a double. This isn't particularly efficient but can be proven to produce the correct results (assuming a correct implementation of the not entirely trivial algorithms, of course).
Assuming your value is in a reasonable range, you can multiply it by a power of 10 such that the least significant decimal digit is non-zero, convert this number to an integer, this integer to a double, and finally divide the resulting double by the original power of 10. I don't have a proof that this yields the correct number but for the range of value I'm interested in and which I want to store accurately in a float, this works.
One reasonable approach to avoid this entirely issue is to use decimal floating point values as described for C++ in the Decimal TR in the first place. Unfortunately, these are not, yet, part of the standard but I have submitted a proposal to the C++ standardization committee to get this changed.
In C++,
What are the random digits that are displayed after giving setprecision() for a floating point number?
Note: After setting the fixed flag.
example:
float f1=3.14;
cout < < fixed<<setprecision(10)<<f1<<endl;
we get random numbers for the remaining 7 digits? But it is not the same case in double.
Two things to be aware of:
floats are stored in binary.
float has a maximum of 24 significant bits. This is equivalent to 7.22 significant digits.
So, to your computer, there's no such number as 3.14. The closest you can get using float is 3.1400001049041748046875.
double has 53 significant bits (~15.95 significant digits), so you get a more accurate approximation, 3.140000000000000124344978758017532527446746826171875. The "noise" digits don't show up with setprecision(10), but would with setprecision(17) or higher.
They're not really "random" -- they're the (best available) decimal representation of that binary fraction (will be exact only for fractions whose denominator is a power of two, e.g., 3.125 would display exactly).
Of course that changes depending on the number of bits available to represent the binary fraction that best approaches the decimal one you originally entered as a literal, i.e., single vs double precision floats.
Not really a C++ specific issue (applies to all languages using binary floats, typically to exploit the machine's underlying HW, i.e., most languages). For a very bare-bone tutorial, I recommend reading this.