How do you convert a float to a string in C++ while specifying the precision & number of decimal digits?
For example: 3.14159265359 -> "3.14"
A typical way would be to use stringstream:
#include <iomanip>
#include <sstream>
double pi = 3.14159265359;
std::stringstream stream;
stream << std::fixed << std::setprecision(2) << pi;
std::string s = stream.str();
See fixed
Use fixed floating-point notation
Sets the floatfield format flag for the str stream to fixed.
When floatfield is set to fixed, floating-point values are written using fixed-point notation: the value is represented with exactly as many digits in the decimal part as specified by the precision field (precision) and with no exponent part.
and setprecision.
For conversions of technical purpose, like storing data in XML or JSON file, C++17 defines to_chars family of functions.
Assuming a compliant compiler (which we lack at the time of writing),
something like this can be considered:
#include <array>
#include <charconv>
double pi = 3.14159265359;
std::array<char, 128> buffer;
auto [ptr, ec] = std::to_chars(buffer.data(), buffer.data() + buffer.size(), pi,
std::chars_format::fixed, 2);
if (ec == std::errc{}) {
std::string s(buffer.data(), ptr);
// ....
}
else {
// error handling
}
The customary method for doing this sort of thing is to "print to string". In C++ that means using std::stringstream something like:
std::stringstream ss;
ss << std::fixed << std::setprecision(2) << number;
std::string mystring = ss.str();
You can use C++20 std::format:
#include <format>
int main() {
std::string s = std::format("{:.2f}", 3.14159265359); // s == "3.14"
}
or the fmt::format function from the {fmt} library, std::format is based on (godbolt):
#include <fmt/core.h>
int main() {
std::string s = fmt::format("{:.2f}", 3.14159265359); // s == "3.14"
}
where 2 is a precision.
It is not only shorter than using iostreams or sprintf but also significantly faster and is not affected by the locale.
Another option is snprintf:
double pi = 3.1415926;
std::string s(16, '\0');
auto written = std::snprintf(&s[0], s.size(), "%.2f", pi);
s.resize(written);
Demo. Error handling should be added, i.e. checking for written < 0.
Here a solution using only std. However, note that this only rounds down.
float number = 3.14159;
std::string num_text = std::to_string(number);
std::string rounded = num_text.substr(0, num_text.find(".")+3);
For rounded it yields:
3.14
The code converts the whole float to string, but cuts all characters 2 chars after the "."
Here I am providing a negative example where your want to avoid when converting floating number to strings.
float num=99.463;
float tmp1=round(num*1000);
float tmp2=tmp1/1000;
cout << tmp1 << " " << tmp2 << " " << to_string(tmp2) << endl;
You get
99463 99.463 99.462997
Note: the num variable can be any value close to 99.463, you will get the same print out. The point is to avoid the convenient c++11 "to_string" function. It took me a while to get out this trap. The best way is the stringstream and sprintf methods (C language). C++11 or newer should provided a second parameter as the number of digits after the floating point to show. Right now the default is 6. I am positing this so that others won't wast time on this subject.
I wrote my first version, please let me know if you find any bug that needs to be fixed. You can control the exact behavior with the iomanipulator. My function is for showing the number of digits after the decimal point.
string ftos(float f, int nd) {
ostringstream ostr;
int tens = stoi("1" + string(nd, '0'));
ostr << round(f*tens)/tens;
return ostr.str();
}
Related
Updated:
I have been trying all methods I could find in stackoverflow, and still could not find a solution.
My point is, I have a string "23.46" and would like to transfer it to double or float anyway. This is in order to be used in another library.
But no matter how I trancate,floor,*100,add,round, it always gives me 24.4600000001 or something like this.
I know there are some precision issue while transfer. But I do need a way to give me a number that double d = 24.46 precisely.
==============================================================================
I have many string values and some of them are double with converted precision as below:
char pt[100];
sprintf(pt, "%.2lf", i);
return string(pt);
Now on the other side of the code, I need to convert the strings back to double, but I tried strtod and atof with precision loss.
My questions are:
what is the good way to check if a string could be a double?
how to convert string back to double with given precision? I only need it to %.2lf be like:
0.21, 35.45, ...
Thanks so much!
Given that you say that using std::strtod is not giving you a solution to the problem, you can use stringstreams to parse strings as doubles, you can also use its flags to assert if the contents of the string are convertible.
Here is an example with some conversions back and forth, and with checks to see if the whole string (not just some digits in it), is parseable as double:
Live demo
#include <iostream>
#include <sstream>
#include <iomanip>
int main()
{
std::string str = "23.4k7"; //string with a character in the middle
std::istringstream ss(str); //put string in a stream
double num;
ss >> num; //convert string to double
if(!ss.fail() && ss.eof()) { //check if the whole string is parseable
std::cout << "is parseable" << std::endl;
}
else {
std::cout << "is not parseable";
return EXIT_FAILURE;
}
std::stringstream to_num;
to_num << std::fixed << std::setprecision(2) << num; //double to string 2 decimal places
std::cout << to_num.str();
to_num >> num; //final conversion to double
}
Since the string has an alphabetic character in it, this will output:
is not parseable
But if you use a valid number it will output the converted value:
string str:
234.2345
Output:
is parseable
234.23
Note that you could use
Live demo
if(ss >> num)
std::cout << "is parseable";
This, however, has a weakness, it will still parse if you have for instance 123.45rt56, 123.45 will be parsed, the rest will be discarded, the way it is in the sample code, if the string has any character, it will return an error. You can choose the more appropriate way for your needs.
How do you convert a float to a string in C++ while specifying the precision & number of decimal digits?
For example: 3.14159265359 -> "3.14"
A typical way would be to use stringstream:
#include <iomanip>
#include <sstream>
double pi = 3.14159265359;
std::stringstream stream;
stream << std::fixed << std::setprecision(2) << pi;
std::string s = stream.str();
See fixed
Use fixed floating-point notation
Sets the floatfield format flag for the str stream to fixed.
When floatfield is set to fixed, floating-point values are written using fixed-point notation: the value is represented with exactly as many digits in the decimal part as specified by the precision field (precision) and with no exponent part.
and setprecision.
For conversions of technical purpose, like storing data in XML or JSON file, C++17 defines to_chars family of functions.
Assuming a compliant compiler (which we lack at the time of writing),
something like this can be considered:
#include <array>
#include <charconv>
double pi = 3.14159265359;
std::array<char, 128> buffer;
auto [ptr, ec] = std::to_chars(buffer.data(), buffer.data() + buffer.size(), pi,
std::chars_format::fixed, 2);
if (ec == std::errc{}) {
std::string s(buffer.data(), ptr);
// ....
}
else {
// error handling
}
The customary method for doing this sort of thing is to "print to string". In C++ that means using std::stringstream something like:
std::stringstream ss;
ss << std::fixed << std::setprecision(2) << number;
std::string mystring = ss.str();
You can use C++20 std::format:
#include <format>
int main() {
std::string s = std::format("{:.2f}", 3.14159265359); // s == "3.14"
}
or the fmt::format function from the {fmt} library, std::format is based on (godbolt):
#include <fmt/core.h>
int main() {
std::string s = fmt::format("{:.2f}", 3.14159265359); // s == "3.14"
}
where 2 is a precision.
It is not only shorter than using iostreams or sprintf but also significantly faster and is not affected by the locale.
Another option is snprintf:
double pi = 3.1415926;
std::string s(16, '\0');
auto written = std::snprintf(&s[0], s.size(), "%.2f", pi);
s.resize(written);
Demo. Error handling should be added, i.e. checking for written < 0.
Here a solution using only std. However, note that this only rounds down.
float number = 3.14159;
std::string num_text = std::to_string(number);
std::string rounded = num_text.substr(0, num_text.find(".")+3);
For rounded it yields:
3.14
The code converts the whole float to string, but cuts all characters 2 chars after the "."
Here I am providing a negative example where your want to avoid when converting floating number to strings.
float num=99.463;
float tmp1=round(num*1000);
float tmp2=tmp1/1000;
cout << tmp1 << " " << tmp2 << " " << to_string(tmp2) << endl;
You get
99463 99.463 99.462997
Note: the num variable can be any value close to 99.463, you will get the same print out. The point is to avoid the convenient c++11 "to_string" function. It took me a while to get out this trap. The best way is the stringstream and sprintf methods (C language). C++11 or newer should provided a second parameter as the number of digits after the floating point to show. Right now the default is 6. I am positing this so that others won't wast time on this subject.
I wrote my first version, please let me know if you find any bug that needs to be fixed. You can control the exact behavior with the iomanipulator. My function is for showing the number of digits after the decimal point.
string ftos(float f, int nd) {
ostringstream ostr;
int tens = stoi("1" + string(nd, '0'));
ostr << round(f*tens)/tens;
return ostr.str();
}
So I have a form with a label which is supposed to display float value, the problem is that I need to have that number rounded to 2 decimal places whatever happens:
label1->Text = System::Convert::ToString( (float)((float)temperature/204.6) );
I tried looking for few hours but as I found there is no method for direct rounding of that beast equation and as far as I know no way to tell ToString() to round the thing to 2 decimals.
Is there any easy way to round the result to 2 decimal places inside ToString method?
This is quite simple to do:
String^ s = String::Format("{0:N2}", temperature/204.6);
Is there any easy way to round the result to 2 decimal places inside ToString method?
No, not with std::tostring() and if you want to preserve trailing zeroes. Use a std::ostringstream with appropriate I/O manipulators instead:
std::ostringstream oss;
oss << std::fixed << std::setprecision(2) << (temperature/204.6);
label1->Text = oss.str();
Can't You multiple result by 100, cast to int, and then divide by 100 with casting to float?
Here is JavaScript equivalent of toFixed:
#include <iostream>
std::string ToFixed(double number, size_t digits)
{
char format[10];
char str[64];
sprintf_s(format, "%%0.%zdf", digits);
return std::string(str, sprintf_s(str, format, number));
}
int main()
{
std::cout << ToFixed((double)12345 / 204.6, 2) << std::endl;
return 0;
}
Prints:
60.34
I am using C++ and I would like to format doubles in the following obvious way. I have tried playing with 'fixed' and 'scientific' using stringstream, but I am unable to achieve this desired output.
double d = -5; // print "-5"
double d = 1000000000; // print "1000000000"
double d = 3.14; // print "3.14"
double d = 0.00000000001; // print "0.00000000001"
// Floating point error is acceptable:
double d = 10000000000000001; // print "10000000000000000"
As requested, here are the things I've tried:
#include <iostream>
#include <string>
#include <sstream>
#include <iomanip>
using namespace std;
string obvious_format_attempt1( double d )
{
stringstream ss;
ss.precision(15);
ss << d;
return ss.str();
}
string obvious_format_attempt2( double d )
{
stringstream ss;
ss.precision(15);
ss << fixed;
ss << d;
return ss.str();
}
int main(int argc, char *argv[])
{
cout << "Attempt #1" << endl;
cout << obvious_format_attempt1(-5) << endl;
cout << obvious_format_attempt1(1000000000) << endl;
cout << obvious_format_attempt1(3.14) << endl;
cout << obvious_format_attempt1(0.00000000001) << endl;
cout << obvious_format_attempt1(10000000000000001) << endl;
cout << endl << "Attempt #2" << endl;
cout << obvious_format_attempt2(-5) << endl;
cout << obvious_format_attempt2(1000000000) << endl;
cout << obvious_format_attempt2(3.14) << endl;
cout << obvious_format_attempt2(0.00000000001) << endl;
cout << obvious_format_attempt2(10000000000000001) << endl;
return 0;
}
That prints the following:
Attempt #1
-5
1000000000
3.14
1e-11
1e+16
Attempt #2
-5.000000000000000
1000000000.000000000000000
3.140000000000000
0.000000000010000
10000000000000000.000000000000000
There is no way for a program to KNOW how to format the numbers in the way that you are describing, unless you write some code to analyze the numbers in some way - and even that can be quite hard.
What is required here is knowing the input format in your source code, and that's lost as soon as the compiler converts the decimal input source code into binary form to store in the executable file.
One alternative that may work is to output to a stringstream, and then from that modify the output to strip trailing zeros. Something like this:
string obvious_format_attempt2( double d )
{
stringstream ss;
ss.precision(15);
ss << fixed;
ss << d;
string res = ss.str();
// Do we have a dot?
if ((string::size_type pos = res.rfind('.')) != string::npos)
{
while(pos > 0 && (res[pos] == '0' || res[pos] == '.')
{
pos--;
}
res = res.substr(pos);
}
return res;
}
I haven't actually tired it, but as a rough sketch, it should work. Caveats are that if you have something like 0.1, it may well print as 0.09999999999999285 or some such, becuase 0.1 can not be represented in exact form as a binary.
Formatting binary floating-point numbers accurately is quite tricky and was traditionally wrong. A pair of papers published in 1990 in the same journal settled that decimal values converted to binary floating-point numbers and back can have their values restored assuming they don't use more decimal digits than a specific constraint (in C++ represented using std::numeric_limits<T>::digits10 for the appropriate type T):
Clinger's "How to read floating-point numbers accurately" describes an algorithm to convert from a decimal representation to a binary floating-point.
Steele/White's "How to print floating-point numbers accurately" describes how to convert from a binary floating-point to a decimal decimal value. Interestingly, the algorithm even converts to the shortest such decimal value.
At the time these papers were published the C formatting directives for binary floating points ("%f", "%e", and "%g") were well established and they didn't get changed to the take the new results into account. The problem with the specification of these formatting directives is that "%f" assumes to count the digits after the decimal point and there is no format specifier asking to format numbers with a certain number of digits but not necessarily starting to count at the decimal point (e.g., to format with a decimal point but potentially having many leading zeros).
The format specifiers are still not improved, e.g., to include another one for non-scientific notation possibly involving many zeros, for that matter. Effectively, the power of the Steele/White's algorithm isn't fully exposed. The C++ formatting, sadly, didn't improve over the situation and just delegates the semantics to the C formatting directives.
The approach of not setting std::ios_base::fixed and using a precision of std::numeric_limits<double>::digits10 is the closest approximation of floating-point formatting the C and C++ standard libraries offer. The exact format requested could be obtained by getting the digits using using formatting with std::ios_base::scientific, parsing the result, and rewriting the digits afterwards. To give this process a nice stream-like interface it could be encapsulated with a std::num_put<char> facet.
An alternative could be the use of Double-Conversion. This implementation uses an improved (faster) algorithm for the conversion. It also exposes interfaces to get the digits in some form although not directly as a character sequence if I recall correctly.
You can't do what you want to do, because decimal numbers are not representable exactly in floating point format. In otherwords, double can't precisely hold 3.14 exactly, it stores everything as fractions of powers of 2, so it stores it as something like 3 + 9175/65536 or thereabouts (do it on your calculator and you'll get 3.1399993896484375. (I realize that 65536 is not the right denominator for IEEE double, but the gist of it is correct).
This is known as the round trip problem. You can't reliable do
double x = 3.14;
cout << magic << x;
and get "3.14"
If you must solve the round-trip problem, then don't use floating point. Use a custom "decimal" class, or use a string to hold the value.
Here's a decimal class you could use:
https://stackoverflow.com/a/15320495/364818
I am using C++ and I would like to format doubles in the following obvious way.
Based on your samples, I assume you want
Fixed rather than scientific notation,
A reasonable (but not excessive) amount of precision (this is for user display, so a small bit of rounding is okay),
Trailing zeros truncated, and
Decimal point truncated as well if the number looks like an integer.
The following function does just that:
#include <cmath>
#include <iomanip>
#include <sstream>
#include <string>
std::string fixed_precision_string (double num) {
// Magic numbers
static const int prec_limit = 14; // Change to 15 if you wish
static const double log10_fuzz = 1e-15; // In case log10 is slightly off
static const char decimal_pt = '.'; // Better: use std::locale
if (num == 0.0) {
return "0";
}
std::string result;
if (num < 0.0) {
result = '-';
num = -num;
}
int ndigs = int(std::log10(num) + log10_fuzz);
std::stringstream ss;
if (ndigs >= prec_limit) {
ss << std::fixed
<< std::setprecision(0)
<< num;
result += ss.str();
}
else {
ss << std::fixed
<< std::setprecision(prec_limit-ndigs)
<< num;
result += ss.str();
auto last_non_zero = result.find_last_not_of('0');
if (result[last_non_zero] == decimal_pt) {
result.erase(last_non_zero);
}
else if (last_non_zero+1 < result.length()) {
result.erase(last_non_zero+1);
}
}
return result;
}
If you are using a computer that uses IEEE floating point, changing prec_limit to 16 is unadvisable. While this will let you properly print 0.9999999999999999 as such, it also prints 5.1 as 5.0999999999999996 and 9.99999998 as 9.9999999800000001. This is from my computer, your results may vary due to a different library.
Changing prec_limit to 15 is okay, but it still leads to numbers that don't print "correctly". The value specified (14) works nicely so long as you aren't trying to print 1.0-1e-15.
You could do even better, but that might require discarding the standard library (see Dietmar Kühl's answer).
I need help on keeping the precision of a double. If I assign a literal to a double, the actual value was truncated.
int main() {
double x = 7.40200133400;
std::cout << x << "\n";
}
For the above code snippet, the output was 7.402
Is there a way to prevent this type of truncation? Or is there a way to calculate exactly how many floating points for a double? For example, number_of_decimal(x) would give 11, since the input is unknown at run-time so I can't use setprecision().
I think I should change my question to:
How to convert a double to a string without truncating the floating points. i.e.
#include <iostream>
#include <string>
#include <sstream>
template<typename T>
std::string type_to_string( T data ) {
std::ostringstream o;
o << data;
return o.str();
}
int main() {
double x = 7.40200;
std::cout << type_to_string( x ) << "\n";
}
The expected output should be 7.40200 but the actual result was 7.402. So how can I work around this problem? Any idea?
Due to the fact the float and double are internally stored in binary, the literal 7.40200133400 actually stands for the number 7.40200133400000037653398976544849574565887451171875
...so how much precision do you really want? :-)
#include <iomanip>
int main()
{
double x = 7.40200133400;
std::cout << std::setprecision(51) << x << "\n";
}
And yes, this program really prints 7.40200133400000037653398976544849574565887451171875!
You must use setiosflags(ios::fixed) and setprecision(x).
For example, cout << setiosflags(ios::fixed) << setprecision(4) << myNumber << endl;
Also, don't forget to #include <iomanip.h>.
std::cout << std::setprecision(8) << x;
Note that setprecision is persistent and all next floats you print will be printed with that precision, until you change it to a different value. If that's a problem and you want to work around that, you can use a proxy stringstream object:
std::stringstream s;
s << std::setprecision(8) << x;
std::cout << s.str();
For more info on iostream formatting, check out the Input/output manipulators section in cppreference.
Solution using Boost.Format:
#include <boost/format.hpp>
#include <iostream>
int main() {
double x = 7.40200133400;
std::cout << boost::format("%1$.16f") % x << "\n";
}
This outputs 7.4020013340000004.
Hope this helps!
The only answer to this that I've come up with is that there is no way to do this (as in calculate the decimal places) correctly! THE primary reason for this being that the representation of a number may not be what you expect, for example, 128.82, seems innocuous enough, however it's actual representation is 128.8199999999... how do you calculate the number of decimal places there??
Responding to your answer-edit: There is no way to do that. As soon as you assign a value to a double, any trailing zeroes are lost (to the compiler/computer, 0.402, 0.4020, and 0.40200 are the SAME NUMBER). The only way to retain trailing zeroes as you indicated is to store the values as strings (or do trickery where you keep track of the number of digits you care about and format it to exactly that length).
Let s make an analogous request: after initialising an integer with 001, you would want to print it with the leading zeroes. That formatting info was simply never stored.
For further understanding the double precision floating point storage, look at the IEEE 754 standard.
Doubles don't have decimal places. They have binary places. And binary places and decimal places are incommensurable (because log2(10) isn't an integer).
What you are asking for doesn't exist.
The second part of the question, about how to preserve trailing zeroes in a floating point value from value specification to output result, has no solution. A floating point value doesn't retain the original value specification. It seems this nonsensical part was added by an SO moderator.
Regarding the first and original part of the question, which I interpret as how to present all significant digits of 7.40200133400, i.e. with output like 7.402001334, you can just remove trailing zeroes from an output result that includes only trustworthy digits in the double value:
#include <assert.h> // assert
#include <limits> // std::(numeric_limits)
#include <string> // std::(string)
#include <sstream> // std::(ostringstream)
namespace my{
// Visual C++2017 doesn't support comma-separated list for `using`:
using std::fixed; using std::numeric_limits; using std::string;
using std::ostringstream;
auto max_fractional_digits_for_positive( double value )
-> int
{
int result = numeric_limits<double>::digits10 - 1;
while( value < 1 ) { ++result; value *= 10; }
return result;
}
auto string_from_positive( double const value )
-> string
{
ostringstream stream;
stream << fixed;
stream.precision( max_fractional_digits_for_positive( value ) );
stream << value;
string result = stream.str();
while( result.back() == '0' )
{
result.resize( result.size() - 1 );
}
return result;
}
auto string_from( double const value )
-> string
{
return (0?""
: value == 0? "0"
: value < 0? "-" + string_from_positive( -value )
: string_from_positive( value )
);
}
}
#include<iostream>
auto main()
-> int
{
using std::cout;
cout << my::string_from( 7.40200133400 ) << "\n";
cout << my::string_from( 0.00000000000740200133400 ) << "\n";
cout << my::string_from( 128.82 ) << "\n";
}
Output:
7.402001334
0.000000000007402001334
128.81999999999999
You might consider adding logic for rounding to avoid long sequences of 9's, like in the last result.