Precision is the number of digits in a number. Scale is the number of
digits to the right of the decimal point in a number. For example, the
number 123.45 has a precision of 5 and a scale of 2.
I need to convert a double with a maximum scale of 7(i.e. it may have 7 digits after the decimal point) to a __int128. However, given a number, I don't know in advance, the actual scale the number has.
#include <iostream>
#include "json.hpp"
using json = nlohmann::json;
#include <string>
static std::ostream& operator<<(std::ostream& o, const __int128& x) {
if (x == std::numeric_limits<__int128>::min()) return o << "-170141183460469231731687303715884105728";
if (x < 0) return o << "-" << -x;
if (x < 10) return o << (char)(x + '0');
return o << x / 10 << (char)(x % 10 + '0');
}
int main()
{
std::string str = R"({"time": [0.143]})";
std::cout << "input: " << str << std::endl;
json j = json::parse(str);
std::cout << "output: " << j.dump(4) << std::endl;
double d = j["time"][0].get<double>();
__int128_t d_128_bad = d * 10000000;
__int128_t d_128_good = __int128(d * 1000) * 10000;
std::cout << std::setprecision(16) << std::defaultfloat << d << std::endl;
std::cout << "d_128_bad: " << d_128_bad << std::endl;
std::cout << "d_128_good: " << d_128_good << std::endl;
}
Output:
input: {"time": [0.143]}
output: {
"time": [
0.143
]
}
0.143
d_128_bad: 1429999
d_128_good: 1430000
As you can see, the converted double is not the expected 1430000 instead it is 1429999. I know the reason is that a float point number can not be represented exactly. The problem can be solved if I know the number of digit after the decimal point.
For example,
I can instead use __int128_t(d * 1000) * 10000. However, I don't know the scale of a given number which might have a maximum of scale 7.
Question> Is there a possible solution for this? Also, I need to do this conversion very fast.
I'm not familiar with this library, but it does appear to have a mechanism to get a json object's string representation (dump()). I would suggest you parse that into your value rather than going through the double intermediate representation, as in that case you will know the scale of the value as it was written.
Related
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
int(main){
std::vector<int> vObj;
float n = 0.59392;
int nCopy = n;
int temNum = 0;;
while (fmod(nCopy, 1) != 0) {
temNum = (nCopy * 10); cout << endl << nCopy << endl;
nCopy *= 10;
vObj.push_back(temNum);
cout << "\n\n Cycle\n\n";
cout << "Temp Num: " << temNum << "\n\nN: " << nCopy << endl;
}
return 0;
}
For example, I input 0.59392 but eventually when the code reaches the bottom, where it should be going
5939.2 and then go to
59392 and stop but for some reason
it keeps going.
yeah , so you have 3 major problems in your code , first of all : it's int main() not int(main) . second : the variable named **nCopy ** is not supposed to be a integer data type , third one : you have to know what the actual representation of the float number , but first this is my solution for your problem , it's not the best one , but it works for this case :
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
int main() {
std::vector<int> vObj;
double n = 0.59392;
double nCopy = n;
int temNum = 0;;
while (fmod(nCopy, 1) != 0) {
temNum = (nCopy * 10); cout << endl << nCopy << endl;
nCopy *= 10;
vObj.push_back(temNum);
cout << "\n\n Cycle\n\n";
cout << "Temp Num: " << temNum << "\n\nN: " << nCopy << endl;
}
return 0;
}
so the explanation is as follow , the double data types gives higher precision than float , that's why I used double instead of float , but it will lack accuracy when the number becomes big .
second of : you have to how is float or double is represented , as the value 0.59392 is actually stored in the memory as value 0.593900024890899658203125 when using float according to IEEE 754 standard , so there are other types of decimals to solve this problem where the difference between them is as follow
Decimal representation gives lower accuracy but higher range with big numbers and high accuracy when talking about small numbers, most 2 used standards are binary integer decimal (BID) and densely packed decimal (DPD)
float and doubles gives higher accuracy than Decimal when talking about big numbers but lower range ,they follow IEEE 754 standard
Fixed-Point types have the lowest range but they are the most accurate one and they are the fastest ones
but unfortunately , C++ only supports float and double types of numbers , but I believe there is external libraries out there to define a decimal data type.
I have always found iomanip confusing and counter intuitive. I need help.
A quick internet search finds (https://www.vedantu.com/maths/precision) "We thus consider precision as the maximum number of significant digits after the decimal point in a decimal number" (the emphasis is mine). That matches my understanding too. However I wrote a test program and:
stm << std::setprecision(3) << 5.12345678;
std::cout << "5.12345678: " << stm.str() << std::endl;
stm.str("");
stm << std::setprecision(3) << 25.12345678;
std::cout << "25.12345678: " << stm.str() << std::endl;
stm.str("");
stm << std::setprecision(3) << 5.1;
std::cout << "5.1: " << stm.str() << std::endl;
stm.str("");
outputs:
5.12345678: 5.12
25.12345678: 25.1
5.1: 5.1
If the precision is 3 then the output should be:
5.12345678: 5.123
25.12345678: 25.123
5.1: 5.1
Clearly the C++ standard has a different interpretation of the meaning of "precision" as relates to floating point numbers.
If I do:
stm.setf(std::ios::fixed, std::ios::floatfield);
then the first two values are formatted correctly, but the last comes out as 5.100.
How do I set the precision without padding?
You can try using this workaround:
decltype(std::setprecision(1)) setp(double number, int p) {
int e = static_cast<int>(std::abs(number));
e = e != 0? static_cast<int>(std::log10(e)) + 1 + p : p;
while(number != 0.0 && static_cast<int>(number*=10) == 0 && e > 1)
e--; // for numbers like 0.001: those zeros are not treated as digits by setprecision.
return std::setprecision(e);
}
And then:
auto v = 5.12345678;
stm << setp(v, 3) << v;
Another more verbose and elegant solution is to create a struct like this:
struct __setp {
double number;
bool fixed = false;
int prec;
};
std::ostream& operator<<(std::ostream& os, const __setp& obj)
{
if(obj.fixed)
os << std::fixed;
else os << std::defaultfloat;
os.precision(obj.prec);
os << obj.number; // comment this if you do not want to print immediately the number
return os;
}
__setp setp(double number, int p) {
__setp setter;
setter.number = number;
int e = static_cast<int>(std::abs(number));
e = e != 0? static_cast<int>(std::log10(e)) + 1 + p : p;
while(number != 0.0 && static_cast<int>(number*=10) == 0)
e--; // for numbers like 0.001: those zeros are not treated as digits by setprecision.
if(e <= 0) {
setter.fixed = true;
setter.prec = 1;
} else
setter.prec = e;
return setter;
}
Using it like this:
auto v = 5.12345678;
stm << setp(v, 3);
I don't think you can do it nicely. There are two candidate formats: defaultfloat and fixed. For the former, "precision" is the maximum number of digits, where both sides of the decimal separator count. For the latter "precision" is the exact number of digits after the decimal separator.
So your solution, I think, is to use fixed format and then manually clear trailing zeros:
#include <iostream>
#include <iomanip>
#include <sstream>
void print(const double number)
{
std::ostringstream stream;
stream << std::fixed << std::setprecision(3) << number;
auto string=stream.str();
while(string.back()=='0')
string.pop_back();
if(string.back()=='.') // in case number is integral; beware of localization issues
string.pop_back();
std::cout << string << "\n";
}
int main()
{
print(5.12345678);
print(25.12345678);
print(5.1);
}
The fixed format gives almost what you want except that it preserves trailing zeros. There is no built-in way to avoid that but you can easily remove those zeros manually. For example, in C++20 you can do the following using std::format:
std::string format_fixed(double d) {
auto s = fmt::format("{:.3f}", d);
auto end = s.find_last_not_of('0');
return end != std::string::npos ? std::string(s.c_str(), end + 1) : s;
}
std::cout << "5.12345678: " << format_fixed(5.12345678) << "\n";
std::cout << "25.12345678: " << format_fixed(25.12345678) << "\n";
std::cout << "5.1: " << format_fixed(5.1) << "\n";
Output:
5.12345678: 5.123
25.12345678: 25.123
5.1: 5.1
The same example with the {fmt} library, std::format is based on: godbolt.
Disclaimer: I'm the author of {fmt} and C++20 std::format.
I need truncation of 2 decimal digits after decimal comma.
I am using following code in C++:
auto doubleTmp = value * 100.00;
int64_t tmp = static_cast<int64_t>(doubleTmp);
double res = ( static_cast<double>(tmp) ) /100.00;
but for example when I set value = 70.82 doubleTmp is 70.8199999 and result is 70.81. What will better way for this and why?
The problem is that neither the input value nor the result res is representable in a computer memory accurately for 70.82. As #MatthieuBrucher suggested, you should use std::lround; consider the following code:
auto value = 70.82;
std::cout << std::fixed << std::setprecision(20) << value << std::endl;
auto tmp = std::lround(value * 100.0);
std::cout << tmp << std::endl;
double res = static_cast<double>(tmp) / 100.00;
std::cout << std::fixed << std::setprecision(20) << res << std::endl;
Which gives the following output:
70.81999999999999317879
7082
70.81999999999999317879
However, you can store the result as a pair of integral numbers, where the first one will represent the integral part and the second one the fractional part:
auto res_integral = tmp / 100;
auto res_fractional = tmp % 100;
std::cout << res_integral << "." << res_fractional << std::endl;
Or, simply store it as tmp with the knowledge that you are storing 100*x instead of x.
Let's say that you have a function:
string function(){
double f = 2.48452
double g = 2
double h = 5482.48552
double i = -78.00
double j = 2.10
return x; // ***
}
* for x we insert:
if we will insert f, function returns: 2.48
if we will insert g, function returns: 2
if we will insert h, function returns: 5482.49
if we will insert i, function returns:-78
if we will insert j, function returns: 2.1
They are only example, who shows how the funcion() works. To precise:
The function for double k return rounded it to: k.XX,
but for:
k=2.20
it return 2.2 as string.
How it implements?
1) Just because you see two digits, it doesn't mean the underlying value was necessarily rounded to two digits.
The precision of the VALUE and the number of digits displayed in the FORMATTED OUTPUT are two completely different things.
2) If you're using cout, you can control formatting with "setprecision()":
http://www.cplusplus.com/reference/iomanip/setprecision/
EXAMPLE (from the above link):
// setprecision example
#include <iostream> // std::cout, std::fixed
#include <iomanip> // std::setprecision
int main () {
double f =3.14159;
std::cout << std::setprecision(5) << f << '\n';
std::cout << std::setprecision(9) << f << '\n';
std::cout << std::fixed;
std::cout << std::setprecision(5) << f << '\n';
std::cout << std::setprecision(9) << f << '\n';
return 0;
}
sample output:
3.1416
3.14159
3.14159
3.141590000
Mathematically, 2.2 is exactly the same as 2.20, 2.200, 2.2000, and so on. If you want to see more insignificant zeros, use [setprecision][1]:
cout << fixed << setprecision(2);
cout << 2.2 << endl; // Prints 2.20
To show up to 2 decimal places, but not showing trailing zeros you can do something such as:
std::string function(double value)
{
// get fractional part
double fracpart = value - static_cast<long>(value);
// compute fractional part rounded to 2 decimal places as an int
int decimal = static_cast<int>(100*fabs(fracpart) + 0.5);
if (decimal >= 100) decimal -= 100;
// adjust precision based on the number of trailing zeros
int precision = 2; // default 2 digits precision
if (0 == decimal) precision = 0; // 2 trailing zeros, don't show decimals
else if (0 == (decimal % 10)) precision = 1; // 1 trailing zero, keep 1 decimal place
// convert value to string
std::stringstream str;
str << std::fixed << std::setprecision(precision) << value;
return str.str();
}
I would like to output a floating-point number as a percentage, with up to three decimal places.
I know that iostreams have three different ways of presenting floats:
"default", which displays using either the rules of fixed or scientific, depending on the number of significant digits desired as defined by setprecision;
fixed, which displays a fixed number of decimal places defined by setprecision; and
scientific, which displays a fixed number of decimal places but using scientific notation, i.e. mantissa + exponent of the radix.
These three modes can be seen in effect with this code:
#include <iostream>
#include <iomanip>
int main() {
double d = 0.00000095;
double e = 0.95;
std::cout << std::setprecision(3);
std::cout.unsetf(std::ios::floatfield);
std::cout << "d = " << (100. * d) << "%\n";
std::cout << "e = " << (100. * e) << "%\n";
std::cout << std::fixed;
std::cout << "d = " << (100. * d) << "%\n";
std::cout << "e = " << (100. * e) << "%\n";
std::cout << std::scientific;
std::cout << "d = " << (100. * d) << "%\n";
std::cout << "e = " << (100. * e) << "%\n";
}
// output:
// d = 9.5e-05%
// e = 95%
// d = 0.000%
// e = 95.000%
// d = 9.500e-05%
// e = 9.500e+01%
None of these options satisfies me.
I would like to avoid any scientific notation here as it makes the percentages really hard to read. I want to keep at most three decimal places, and it's ok if very small values show up as zero. However, I would also like to avoid trailing zeros in fractional places for cases like 0.95 above: I want that to display as in the second line, as "95%".
In .NET, I can achieve this with a custom format string like "0.###%", which gives me a number formatted as a percentage with at least one digit left of the decimal separator, and up to three digits right of the decimal separator, trailing zeros skipped: http://ideone.com/uV3nDi
Can I achieve this with iostreams, without writing my own formatting logic (e.g. special casing small numbers)?
I'm reasonably certain nothing built into iostreams supports this directly.
I think the cleanest way to handle it is to round the number before passing it to an iostream to be printed out:
#include <iostream>
#include <vector>
#include <cmath>
double rounded(double in, int places) {
double factor = std::pow(10, places);
return std::round(in * factor) / factor;
}
int main() {
std::vector<double> values{ 0.000000095123, 0.0095123, 0.95, 0.95123 };
for (auto i : values)
std::cout << "value = " << 100. * rounded(i, 5) << "%\n";
}
Due to the way it does rounding, this has a limitation on the magnitude of numbers it can work with. For percentages this probably isn't an issue, but if you were working with a number close to the largest that can be represented in the type in question (double in this case) the multiplication by pow(10, places) could/would overflow and produce bad results.
Though I can't be absolutely certain, it doesn't seem like this would be likely to cause an issue for the problem you seem to be trying to solve.
This solution is terrible.
I am serious. I don't like it. It's probably slow and the function has a stupid name. Maybe you can use it for test verification, though, because it's so dumb I guess you can easily see it pretty much has to work.
It also assumes decimal separator to be '.', which doesn't have to be the case. The proper point could be obtained by:
char point = std::use_facet< std::numpunct<char> >(std::cout.getloc()).decimal_point();
But that's still not solving the problem, because the characters used for digits could be different and in general this isn't something that should be written in such a way.
Here it is.
template<typename Floating>
std::string formatFloatingUpToN(unsigned n, Floating f) {
std::stringstream out;
out << std::setprecision(n) << std::fixed;
out << f;
std::string ret = out.str();
// if this clause holds, it's all zeroes
if (std::abs(f) < std::pow(0.1, n))
return ret;
while (true) {
if (ret.back() == '0') {
ret.pop_back();
continue;
} else if (ret.back() == '.') {
ret.pop_back();
break;
} else
break;
}
return ret;
}
And here it is in action.