string to float conversion? - c++

I'm wondering what sort of algorithm could be used to take something like "4.72" into a float data type, equal to
float x = 4.72;

scanf, operator>> for istreams, and strtof would be the obvious choices.
There is also atof, but, like atoi, it lacks a way to tell you there was an error in the input, so it's generally best to avoid both.

For C++ you can use boost::lexical_cast:
std::string str( "4.72" );
float x = boost::lexical_cast< float >( str );
For C you can use sscanf:
char str[]= "4.72";
float x;
sscanf( str, "%f", &x );

For C++ This is the algorithm I use:
bool FromString(const string& str, double& number) {
std::istringstream i(str);
if (!(i >> number)) {
// Number conversion failed
return false;
}
return true;
}
I used atof() in the past for the conversion, but I found this problematic because if no valid conversion can be made, it will return (0.0). So, you would not know if it failed and returned zero, or if the string actually had "0" in it.

For C strtod() and C99 friends strtof() and strtold() (description on same link) already have that algorithm implemented.
If you are having problems writing your own, post your code and specific questions about it.

As you've asked for an algorithm, not a method, here is my explanation for a simple algorithm (and an implementation in C):
Initialize 4 integer variables, one for the value before dot, one for the after part, one for the power of the mantissa, one for the sign. Let's say, f, m, d, sign = 1.
First look for + or - sign at the beginning. If there are no sign characters or + sign then continue. If the first character is -, then sign = -1.
Then, read an integer value into f until a . or NULL character.
If you end up with a dot character, then start reading the mantissa part as in the previous step into m. But this time also multiply d by 10 with each digit.
In the end, return sign*(f + (float) m/d). The casting makes sure the division is done in floating-points and the type of the expression is float.
I guess, reading the code might be easier. So here is the code:
float atof(char *s)
{
int f, m, sign, d=1;
f = m = 0;
sign = (s[0] == '-') ? -1 : 1;
if (s[0] == '-' || s[0] == '+') s++;
for (; *s != '.' && *s; s++) {
f = (*s-'0') + f*10;
}
if (*s == '.')
for (++s; *s; s++) {
m = (*s-'0') + m*10;
d *= 10;
}
return sign*(f + (float)m/d);
}

You can use boost:lexical_cast
http://www.boost.org/doc/libs/1_44_0/libs/conversion/lexical_cast.htm

I assume you want an actual algorithm, not a library function that already does it. I don't have time to write and test actual code, but here is what I would do:
Initialize a float that will be used as an accumulator to 0.
Figure out where the decimal place is in the string, this will let you know what "column" each of the digits is (i.e. 100s, 10s, 1s, 1/10ths, etc).
Start at the beginning of the string.
Take that digit, convert to an int (trivially done by subtracting 0x30 from the ASCII value)
Multiply the value by the place column (for the first digit in your example, that would be 4 * 1 == 4, for the next digit 7 * 0.1 == 0.7).
Add the result to the accumulator
Repeat from step 4 for each remaining digit.
The accumulator now contains your result.
Due to the roundoff of converting between base 10 and base 2 in every iteration of this loop, the result you get from this algorithm may not be the closest possible binary representation to the original value. I don't really know of a good way to improve it though... perhaps someone else can chime in with that.

The atof() function can be helpful.
http://www.cplusplus.com/reference/clibrary/cstdlib/atof/

From cplusplus.com: "stringstream provides an interface to manipulate strings as if they were input/output streams."
You can initialize a stringstream with your string then read a float from the stringstream using operator>> just like you would with cin.
Here is an example:
#include<iostream>
#include<string>
#include<sstream>
using namespace std;
int main() {
string s = "4.72";
stringstream sstrm(s);
float x;
sstrm >> x;
cout << x << endl;
}

Related

How to make output without 00000? [duplicate]

Is there any way to set the precision of the result when converting a double to string using std::to_string()?
No.
Returns: Each function returns a string object holding the character representation of the value of
its argument that would be generated by calling sprintf(buf, fmt, val) with a format specifier of
"%d", "%u", "%ld", "%lu", "%lld", "%llu", "%f", "%f", or "%Lf", respectively, where buf designates
an internal character buffer of sufficient size.
I was just looking for a solution to this as I was overloading the std::cout << operator and therefore it would have been tricky to do the std::stringstream workaround. I solved my problem by using the std::substr() function to locate the decimal and take a chosen number of digits past it.
std::string trimmedString = std::to_string(doubleVal).substr(0, std::to_string(doubleVal).find(".") + precisionVal + 1);
This will give you "precisionVal" 0's after your number.
Ex:
double doubleVal = 3;
int preisionVal = 2
3.000000 becomes 3.00
I believe that using std::stringstream with setprecision would be the most flexible/portable choice, but if you know your data, as an workaround, you could try to substring the to_string result. For example:
std::string seriesSum(int n)
{
double sum = 0, div = 1;
for(int i = 0; i < n; i++) {
sum += 1.0 / div;
div += 3;
}
return std::to_string(round(sum * 100)/100).substr(0,4);
}
In the code above I'm printing with two decimal places 0.00 by taking the first 4 digits of the string, but it only works because I know the integer part is never going above one digit. You could also use string.find() to search for the decimal separator and use it's position to calculate the size of the substring, making it a bit more dynamic.

Scanf-reading variables which have to meet ceratin specifications

I have to solve an exercise which at first glance seemed really simple to me. However, I am not so sure how to solve it. Given the fact that variables a and b have certain admissable values, should I verify whether they are in the correct range after reading them? And regarding variable b, in the examples it is given in scientific notation, so should I read it using the e specifier? And last but not least, the assignment supression of ',' is done using "%*[,]"?
Write a set of statements to read from the standard input three variables a, b and c, where the variables have the following types:
a – unsigned short integer, b – float, c – string of at most 31 characters
The input values are separated by one comma. The first variable is given as an octal number. The admissible values for a and b are as follows:
-4331 < a <= 28731
-1035 < b < 9749273
Suitable error messages should be given if the values are not within the allowed range or in the wrong format. Sample input:
12745, -1.e-3, Is this a good one?
0xEE04F, 21e75, Maybe this one...
70073, 21e+6, Or this one. Could this string be too long to put into allocated area?
OP: should I verify whether they are in the correct range after reading them?
A: Yes. The scanning will check gross range error. Then code checks local range.
OP: read it using the e specifier?
A: a, e, f, g are all equivalent in scanf() family. So use what seems appropriate: e is good.
OP: the assignment suppression of ',' is done using "%*[,]"
A: Almost: One could use "%*1[,]", but "," is sufficient.
Use fgets() and sscanf() and then check range.
char buf[1000];
if (fgets(buf, sizeof buf, stdin) == NULL) {
handle_EOF_or_IOError();
}
short Number;
float x;
char s[31+1];
if (3 != sscanf(buf, "%ho,%e, %31[^\n]", &Number, &x, s)) {
handle_ParseError();
}
short NumberTooSmall = -4331;
short NumberMax = 28731
float xTooSmall = -1035.0f;
float xTooBig = 9749273.0f;
if ((Number <= NumberTooSmall) || (Number > NumberMax) ||
(x <= xTooSmall) || (x >= xTooBig)) {
handle_RangeError();
}
Note: there is a subtle issue on when the parsed string begin in "12745, -1.e-3, Is this a good one?", etc. OP says that , is the seperator. This implies the string starts with " Is this ...". I think the OP wants to skip leading spaces and get "Is this ...".
Assuming IEEE binary32 for float, 9749273.0 needs 24 bits to represent exactly. Since this format has 23+1 bits of precision for normal numbers, all is well. Care should be observed with range testing of floating point values.

Get printf to ignore the negative sign on values of zero

I'm trying to write a (mostly)* C program that sorts numerical results and eliminates duplicates. The results are stored as STRUCTS that contain a string, an integer, and 4 doubles. The doubles are what is relevant for determining if two results are duplicates.
To do this, I sprintf a string using the 4 doubles to some precision i.e.
#define PRECISION 5
sprintf(hashString, "%.*lf %.*lf %.*lf %.*lf", PRECISION, result.v1, PRECISION, result.v2, PRECISION, result.v3, PRECISION, result.v4);
I then use this as a hashkey for a tr1::unordered_map<string, ResultType>. Then the program checks to see if the hashtable already contains an entry for that key, if so, the result is a duplicate and can be discarded. Otherwise, it gets added to the hashtable.
The problem is that sometimes one of my values will be rounded to zero from, for example, -10E-9, by sprintf; As a result, the string will contain "-0.00000" rather than "0.00000". These two values will obviously generate different hashkeys, despite representing the same result.
Is there something built into sprintf or even the C language that will allow me to deal with this? I've come up with a bit of a work around (see post below) -- but if there's something built in, I would much rather use that.
*the program is written in C because that's the language I'm most comfortable in, but I'll end up compiling with g++ in order to use the unordered_map.
I've come up with the following workaround. But A) I'm hoping there's a builtin solution and B) I don't have a very deep understanding of atof or floating point math, so I'm not sure if the condition if(doubleRepresentation == 0.0) will always trip when it should.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define PRECISION 5
#define ACCURACY 10E-6
double getRidOfNegZeros (double number)
{
char someNumAsStr[PRECISION + 3]; // +3 accounts for a possible minus sign, the leading 0 or 1, and the decimal place.
sprintf(someNumAsStr, "%.*lf", PRECISION, number);
double doubleRepresentation = atof(someNumAsStr);
if((doubleRepresentation < ACCURACY) && (doubleRepresentation > -ACCURACY))
{
doubleRepresentation = 0.0;
}
return doubleRepresentation;
}
int main()
{
printf("Enter a number: \n");
double somenum;
scanf("%lf",&somenum);
printf("The new representation of double \"%.*lf\" is \"%.*lf\"\n", PRECISION, somenum, PRECISION, getRidOfNegZeros(somenum));
return 0;
}
Rather than sprintf()ing the doubles to a big string and using that as the key in a map, why not just put your structs into the map? You can do this easily enough if you just write a less-than operator for your structs which considers the floating-point values you want to use as the key. Something like this:
bool operator <(const MyStruct &lhs, const MyStruct &rhs)
{
return lhs.v1 < rhs.v1 ||
(lhs.v1 == rhs.v1 && lhs.v2 < rhs.v2); // ...
}
Then you can replace your tr1::unordered_map<string, ResultType> with std::map<ResultType>, and avoid the whole string printing business all together. If you want you can add some epsilon to the comparison function so that numbers that are nearly the same are stably sorted.
If you know that you only care about differences of 0.00001 (based on your definition of PRECISION), you can round the values to integers first. Something like this may work:
#include <math.h>
#include <stdio.h>
#define SCALE 1e5 // instead of PRECISION 5
sprintf(hashString, "%d %d %d %d",
(int)round(result.v1 * SCALE),
(int)round(result.v2 * SCALE),
(int)round(result.v3 * SCALE),
(int)round(result.v4 * SCALE));
This also requires a bound on the magnitude of the floating-point values. You don't want to overflow your integer values.
You can also bypass the string formatting and simply do the rounding calculations as part of a structure-level hash, as others have suggested.
Perhaps implement a utility function to round/snap values to positive zero. Use precision digit count similar to printf style format syntax.
// Prevent display of -0 values by snapping to positive zero
// \a_number original number
// \a_precisionCount number of digits of decimal precision eg. 2 for #.##, 0 for whole integer. Default 0 (whole integer number.)
// \returns number rounded to positive zero if result would have produced -0.00 for precision.
template <class Real>
Real PosZero(const Real& a_number, const int a_precisionCount = 0)
{
Real precisionValue = Real(0.5) * pow(Real(0.10), Real(a_precisionCount));
if( (a_number > -abs(precisionValue)) && (a_number < abs(precisionValue)) )
{
return +0.0;
}
return a_number;
}
Test:
f32 value = -0.049f;
int precision = 4; // Test precision from param
printf("%.0f, %.2f, %.*f", PosZero(value), PosZero(value,2), precision, PosZero(value,precision));
Test output:
"0, -0.05, -0.0490"
This is intended to be a general solution for people wanting to avoid negative zeros in formatted strings. Not specific to the original poster's use of creating a key or hash.
#include <string>
#define PRECISION 5
#define LIMIT 5e-6
std::string string_rep (double x) {
char buf[32];
double xtrunc = ((x > -LIMIT) && (x < LIMIT)) ? 0.0 : x;
std::sprintf (buf, "%.*f", PRECISION, xtrunc);
return std::string(buf);
}
std::string make_key (double x, double y, double z, double w) {
std::string strx = string_rep (x);
std::string stry = string_rep (y);
std::string strz = string_rep (z);
std::string strw = string_rep (w);
return strx + " " + stry + " " + strz + " " + strw;
}
If you're only using this for the purposes of hashing the double values, then don't bother converting them to a string—just hash the double values directly. Any hash library worth its salt will have the ability to hash arbitrary binary blobs of data.
If for some strange reason your hash library only supports null-terminated C strings, then print out the raw bytes of the double value:
// Alias the double value as a byte array
unsigned char *d = (unsigned char *)&result.v1;
// Prefer snprintf to sprintf!
spnrintf(hashString, hashStringLength, "%02x%02x%02x%02x%02x%02x%02x%02x",
d[0], d[1], d[2], d[3], d[4], d[5], d[6], d[7]);
// ...and so on for each double value
This ensures that unequal values will definitely be given unequal strings.

How to convert string (22.123) format number into float variable format without using any API in c++

How to convert string (22.123) format number into float variable format without using any API in c++. This is just to understand more about the inside coding.. thnx
something like:
double string_to_double(std::string s)
{
int p = 0;
int p_dec = s.length();
double val = 0;
for (int i=0; i<s.length(); ++i)
{
double digit = (double)(s[i] - '0');
if (s[i] == '.') { p_dec = p; }
else { val += digit*powf(10,p--); }
}
val /= powf(10, p_dec);
}
Basic algorithm, assuming no input in the form 1.2e-4:
(1) Read an integer before the dot. If the number of digits is > 16 (normal precision of double), convert that integer into floating point directly and return.
(2) Read an at most 16 digits dot as an integer. Compute (that integer) ÷ 10digits read. Sum up this with the integer in step (1) and return.
This only involve 2 floating point operation: one + and one ÷, and a bunch of integer arithmetics. The advantage over multiplications and divisions by powers of 10 is that the error won't accumulate unnecessarily.
(To read 16-digit integers you need a 64-bit int.)
In reality, you should use sscanf(str, "%lf", ...), std::istringstream, or boost::lexical_cast<double>.
go over the number digit by digit by using a bunch of multiplications and divisions by powers of 10 and construct the string character by character.
If you just want an idea of how to do it, the other answer, if you want an accurate result, the problem is not so simple and you should refer to the literature on the subject. An example: ftp://ftp.ccs.neu.edu/pub/people/will/howtoread.ps
I'm pretty sure that the Plauger Standard C Library book has a disc with the source of strtod.
http://www.amazon.co.uk/Standard-C-Library-P-J-Plauger/dp/0131315099
and there are online versions too:
http://www.google.co.uk/search?hl=en&client=firefox-a&hs=IvI&rls=org.mozilla%3Aen-GB%3Aofficial&q=strtod+source+code

C/C++ counting the number of decimals?

Lets say that input from the user is a decimal number, ex. 5.2155 (having 4 decimal digits). It can be stored freely (int,double) etc.
Is there any clever (or very simple) way to find out how many decimals the number has? (kinda like the question how do you find that a number is even or odd by masking last bit).
Two ways I know of, neither very clever unfortunately but this is more a limitation of the environment rather than me :-)
The first is to sprintf the number to a big buffer with a "%.50f" format string, strip off the trailing zeros then count the characters after the decimal point. This will be limited by the printf family itself. Or you could use the string as input by the user (rather than sprintfing a floating point value), so as to avoid floating point problems altogether.
The second is to subtract the integer portion then iteratively multiply by 10 and again subtract the integer portion until you get zero. This is limited by the limits of computer representation of floating point numbers - at each stage you may get the problem of a number that cannot be represented exactly (so .2155 may actually be .215499999998). Something like the following (untested, except in my head, which is about on par with a COMX-35):
count = 0
num = abs(num)
num = num - int(num)
while num != 0:
num = num * 10
count = count + 1
num = num - int(num)
If you know the sort of numbers you'll get (e.g., they'll all be 0 to 4 digits after the decimal point), you can use standard floating point "tricks" to do it properly. For example, instead of:
while num != 0:
use
while abs(num) >= 0.0000001:
Once the number is converted from the user representation (string, OCR-ed gif file, whatever) into a floating point number, you are not dealing with the same number necessarily. So the strict, not very useful answer is "No".
If (case A) you can avoid converting the number from the string representation, the problem becomes much easier, you only need to count the digits after the decimal point and subtract the number of trailing zeros.
If you cannot do it (case B), then you need to make an assumption about the maximum number of decimals, convert the number back into string representation and round it to this maximum number using the round-to-even method. For example, if the user supplies 1.1 which gets represented as 1.09999999999999 (hypothetically), converting it back to string yields, guess what, "1.09999999999999". Rounding this number to, say, four decimal points gives you "1.1000". Now it's back to case A.
Off the top of my head:
start with the fractional portion: .2155
repeatedly multiply by 10 and throw away the integer portion of the number until you get zero. The number of steps will be the number of decimals. e.g:
.2155 * 10 = 2.155
.155 * 10 = 1.55
.55 * 10 = 5.5
.5 * 10 = 5.0
4 steps = 4 decimal digits
Something like this might work as well:
float i = 5.2154;
std::string s;
std::string t;
std::stringstream out;
out << i;
s = out.str();
t = s.substr(s.find(".")+1);
cout<<"number of decimal places: " << t.length();
What do you mean "stored freely (int"? Once stored in an int, it has zero decimals left, clearly. A double is stored in a binary form, so no obvious or simple relation to "decimals" either. Why don't you keep the input as a string, just long enough to count those decimals, before sending it on to its final numeric-variable destination?
using the Scientific Notation format (to avoid rounding errors):
#include <stdio.h>
#include <string.h>
/* Counting the number of decimals
*
* 1. Use Scientific Notation format
* 2. Convert it to a string
* 3. Tokenize it on the exp sign, discard the base part
* 4. convert the second token back to number
*/
int main(){
int counts;
char *sign;
char str[15];
char *base;
char *exp10;
float real = 0.00001;
sprintf (str, "%E", real);
sign= ( strpbrk ( str, "+"))? "+" : "-";
base = strtok (str, sign);
exp10 = strtok (NULL, sign);
counts=atoi(exp10);
printf("[%d]\n", counts);
return 0;
}
[5]
If the decimal part of your number is stored in a separate int, you can just count the its decimal digits.
This is a improvement on andrei alexandrescu's improvement. His version was already faster than the naive way (dividing by 10 at every digit). The version below is constant time and faster at least on x86-64 and ARM for all sizes, but occupies twice as much binary code, so it is not as cache-friendly.
Benchmarks for this version vs alexandrescu's version on my PR on facebook folly.
Works on unsigned, not signed.
inline uint32_t digits10(uint64_t v) {
return 1
+ (std::uint32_t)(v>=10)
+ (std::uint32_t)(v>=100)
+ (std::uint32_t)(v>=1000)
+ (std::uint32_t)(v>=10000)
+ (std::uint32_t)(v>=100000)
+ (std::uint32_t)(v>=1000000)
+ (std::uint32_t)(v>=10000000)
+ (std::uint32_t)(v>=100000000)
+ (std::uint32_t)(v>=1000000000)
+ (std::uint32_t)(v>=10000000000ull)
+ (std::uint32_t)(v>=100000000000ull)
+ (std::uint32_t)(v>=1000000000000ull)
+ (std::uint32_t)(v>=10000000000000ull)
+ (std::uint32_t)(v>=100000000000000ull)
+ (std::uint32_t)(v>=1000000000000000ull)
+ (std::uint32_t)(v>=10000000000000000ull)
+ (std::uint32_t)(v>=100000000000000000ull)
+ (std::uint32_t)(v>=1000000000000000000ull)
+ (std::uint32_t)(v>=10000000000000000000ull);
}
Years after the fight but as I have made my own solution in three lines :
string number = "543.014";
size_t dotFound;
stoi(number, &dotFound));
string(number).substr(dotFound).size()
Of course you have to test before if it is really a float
(With stof(number) == stoi(number) for example)
int main()
{
char s[100];
fgets(s,100,stdin);
unsigned i=0,sw=0,k=0,l=0,ok=0;
unsigned length=strlen(s);
for(i=0;i<length;i++)
{
if(isprint(s[i]))
{
if(sw==1)
{
k++;
if(s[i]=='0')
{
ok=0;
}
if(ok==0)
{
if(s[i]=='0')
l++;
else
{
ok=1;
l=0;
}
}
}
if(s[i]=='.')
{
sw=1;
}
}
}
printf("%d",k-l);
return 0;
}
This is a robust C++ 11 implementation suitable for float and double types:
template <typename T>
std::enable_if_t<(std::is_floating_point<T>::value), std::size_t>
decimal_places(T v)
{
std::size_t count = 0;
v = std::abs(v);
auto c = v - std::floor(v);
T factor = 10;
T eps = std::numeric_limits<T>::epsilon() * c;
while ((c > eps && c < (1 - eps)) && count < std::numeric_limits<T>::max_digits10)
{
c = v * factor;
c = c - std::floor(c);
factor *= 10;
eps = std::numeric_limits<T>::epsilon() * v * factor;
count++;
}
return count;
}
It throws the value away each iteration and instead keeps track of a power of 10 multiplier to avoid rounding issues building up. It uses machine epsilon to correctly handle decimal numbers that cannot be represented exactly in binary such as the value of 5.2155 as stipulated in the question.
Based on what others wrote, this has worked well for me. This solution does handle the case where a number can't be represented exactly in binary.
As suggested by others, the condition for the while loop indicates the precise behavior. My update uses the machine epsilon value to test whether the remainder on any loop is representable by the numeric format. The test should not compare to 0 or a hardcoded value like 0.000001.
template<class T, std::enable_if_t<std::is_floating_point_v<T>, T>* = nullptr>
unsigned int NumDecimalPlaces(T val)
{
unsigned int decimalPlaces = 0;
val = std::abs(val);
val = val - std::round(val);
while (
val - std::numeric_limits<T>::epsilon() > std::numeric_limits<T>::epsilon() &&
decimalPlaces <= std::numeric_limits<T>::digits10)
{
std::cout << val << ", ";
val = val * 10;
++decimalPlaces;
val = val - std::round(val);
}
return val;
}
As an example, if the input value is 2.1, the correct solution is 1. However, some other answers posted here would output 16 if using double precision because 2.1 can't be precisely represented in double precision.
I would suggest reading the value as a string, searching for the decimal point, and parsing the text before and after it as integers. No floating point or rounding errors.
char* fractpart(double f)
{
int intary={1,2,3,4,5,6,7,8,9,0};
char charary={'1','2','3','4','5','6','7','8','9','0'};
int count=0,x,y;
f=f-(int)f;
while(f<=1)
{
f=f*10;
for(y=0;y<10;y++)
{
if((int)f==intary[y])
{
chrstr[count]=charary[y];
break;
}
}
f=f-(int)f;
if(f<=0.01 || count==4)
break;
if(f<0)
f=-f;
count++;
}
return(chrstr);
}
Here is the complete program
#include <iostream.h>
#include <conio.h>
#include <string.h>
#include <math.h>
char charary[10]={'1','2','3','4','5','6','7','8','9','0'};
int intary[10]={1,2,3,4,5,6,7,8,9,0};
char* intpart(double);
char* fractpart(double);
int main()
{
clrscr();
int count = 0;
double d = 0;
char intstr[10], fractstr[10];
cout<<"Enter a number";
cin>>d;
strcpy(intstr,intpart(d));
strcpy(fractstr,fractpart(d));
cout<<intstr<<'.'<<fractstr;
getche();
return(0);
}
char* intpart(double f)
{
char retstr[10];
int x,y,z,count1=0;
x=(int)f;
while(x>=1)
{
z=x%10;
for(y=0;y<10;y++)
{
if(z==intary[y])
{
chrstr[count1]=charary[y];
break;
}
}
x=x/10;
count1++;
}
for(x=0,y=strlen(chrstr)-1;y>=0;y--,x++)
retstr[x]=chrstr[y];
retstr[x]='\0';
return(retstr);
}
char* fractpart(double f)
{
int count=0,x,y;
f=f-(int)f;
while(f<=1)
{
f=f*10;
for(y=0;y<10;y++)
{
if((int)f==intary[y])
{
chrstr[count]=charary[y];
break;
}
}
f=f-(int)f;
if(f<=0.01 || count==4)
break;
if(f<0)
f=-f;
count++;
}
return(chrstr);
}
One way would be to read the number in as a string. Find the length of the substring after the decimal point and that's how many decimals the person entered. To convert this string into a float by using
atof(string.c_str());
On a different note; it's always a good idea when dealing with floating point operations to store them in a special object which has finite precision. For example, you could store the float points in a special type of object called "Decimal" where the whole number part and the decimal part of the number are both ints. This way you have a finite precision. The downside to this is that you have to write out methods for arithmetic operations (+, -, *, /, etc.), but you can easily overwrite operators in C++. I know this deviates from your original question, but it's always better to store your decimals in a finite form. In this way you can also answer your question of how many decimals the number has.