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What is the difference between float and double?


Question

I've read about the difference between double precision and single precision. However, in most cases, float and double seem to be interchangeable, i.e. using one or the other does not seem to affect the results. Is this really the case? When are floats and doubles interchangeable? What are the differences between them?

2018/03/10
1
427
3/10/2018 11:00:41 AM

Accepted Answer

Huge difference.

As the name implies, a double has 2x the precision of float[1]. In general a double has 15 decimal digits of precision, while float has 7.

Here's how the number of digits are calculated:

double has 52 mantissa bits + 1 hidden bit: log(253)÷log(10) = 15.95 digits

float has 23 mantissa bits + 1 hidden bit: log(224)÷log(10) = 7.22 digits

This precision loss could lead to greater truncation errors being accumulated when repeated calculations are done, e.g.

float a = 1.f / 81;
float b = 0;
for (int i = 0; i < 729; ++ i)
    b += a;
printf("%.7g\n", b); // prints 9.000023

while

double a = 1.0 / 81;
double b = 0;
for (int i = 0; i < 729; ++ i)
    b += a;
printf("%.15g\n", b); // prints 8.99999999999996

Also, the maximum value of float is about 3e38, but double is about 1.7e308, so using float can hit "infinity" (i.e. a special floating-point number) much more easily than double for something simple, e.g. computing the factorial of 60.

During testing, maybe a few test cases contain these huge numbers, which may cause your programs to fail if you use floats.


Of course, sometimes, even double isn't accurate enough, hence we sometimes have long double[1] (the above example gives 9.000000000000000066 on Mac), but all floating point types suffer from round-off errors, so if precision is very important (e.g. money processing) you should use int or a fraction class.


Furthermore, don't use += to sum lots of floating point numbers, as the errors accumulate quickly. If you're using Python, use fsum. Otherwise, try to implement the Kahan summation algorithm.


[1]: The C and C++ standards do not specify the representation of float, double and long double. It is possible that all three are implemented as IEEE double-precision. Nevertheless, for most architectures (gcc, MSVC; x86, x64, ARM) float is indeed a IEEE single-precision floating point number (binary32), and double is a IEEE double-precision floating point number (binary64).

2020/06/20
528
6/20/2020 9:12:55 AM

Here is what the standard C99 (ISO-IEC 9899 6.2.5 §10) or C++2003 (ISO-IEC 14882-2003 3.1.9 §8) standards say:

There are three floating point types: float, double, and long double. The type double provides at least as much precision as float, and the type long double provides at least as much precision as double. The set of values of the type float is a subset of the set of values of the type double; the set of values of the type double is a subset of the set of values of the type long double.

The C++ standard adds:

The value representation of floating-point types is implementation-defined.

I would suggest having a look at the excellent What Every Computer Scientist Should Know About Floating-Point Arithmetic that covers the IEEE floating-point standard in depth. You'll learn about the representation details and you'll realize there is a tradeoff between magnitude and precision. The precision of the floating point representation increases as the magnitude decreases, hence floating point numbers between -1 and 1 are those with the most precision.

2010/03/06

Given a quadratic equation: x2 − 4.0000000 x + 3.9999999 = 0, the exact roots to 10 significant digits are, r1 = 2.000316228 and r2 = 1.999683772.

Using float and double, we can write a test program:

#include <stdio.h>
#include <math.h>

void dbl_solve(double a, double b, double c)
{
    double d = b*b - 4.0*a*c;
    double sd = sqrt(d);
    double r1 = (-b + sd) / (2.0*a);
    double r2 = (-b - sd) / (2.0*a);
    printf("%.5f\t%.5f\n", r1, r2);
}

void flt_solve(float a, float b, float c)
{
    float d = b*b - 4.0f*a*c;
    float sd = sqrtf(d);
    float r1 = (-b + sd) / (2.0f*a);
    float r2 = (-b - sd) / (2.0f*a);
    printf("%.5f\t%.5f\n", r1, r2);
}   

int main(void)
{
    float fa = 1.0f;
    float fb = -4.0000000f;
    float fc = 3.9999999f;
    double da = 1.0;
    double db = -4.0000000;
    double dc = 3.9999999;
    flt_solve(fa, fb, fc);
    dbl_solve(da, db, dc);
    return 0;
}  

Running the program gives me:

2.00000 2.00000
2.00032 1.99968

Note that the numbers aren't large, but still you get cancellation effects using float.

(In fact, the above is not the best way of solving quadratic equations using either single- or double-precision floating-point numbers, but the answer remains unchanged even if one uses a more stable method.)

2010/03/05

  • A double is 64 and single precision (float) is 32 bits.
  • The double has a bigger mantissa (the integer bits of the real number).
  • Any inaccuracies will be smaller in the double.
2010/03/05

The size of the numbers involved in the float-point calculations is not the most relevant thing. It's the calculation that is being performed that is relevant.

In essence, if you're performing a calculation and the result is an irrational number or recurring decimal, then there will be rounding errors when that number is squashed into the finite size data structure you're using. Since double is twice the size of float then the rounding error will be a lot smaller.

The tests may specifically use numbers which would cause this kind of error and therefore tested that you'd used the appropriate type in your code.

2018/03/10

I just ran into a error that took me forever to figure out and potentially can give you a good example of float precision.

#include <iostream>
#include <iomanip>

int main(){
  for(float t=0;t<1;t+=0.01){
     std::cout << std::fixed << std::setprecision(6) << t << std::endl;
  }
}

The output is

0.000000
0.010000
0.020000
0.030000
0.040000
0.050000
0.060000
0.070000
0.080000
0.090000
0.100000
0.110000
0.120000
0.130000
0.140000
0.150000
0.160000
0.170000
0.180000
0.190000
0.200000
0.210000
0.220000
0.230000
0.240000
0.250000
0.260000
0.270000
0.280000
0.290000
0.300000
0.310000
0.320000
0.330000
0.340000
0.350000
0.360000
0.370000
0.380000
0.390000
0.400000
0.410000
0.420000
0.430000
0.440000
0.450000
0.460000
0.470000
0.480000
0.490000
0.500000
0.510000
0.520000
0.530000
0.540000
0.550000
0.560000
0.570000
0.580000
0.590000
0.600000
0.610000
0.620000
0.630000
0.640000
0.650000
0.660000
0.670000
0.680000
0.690000
0.700000
0.710000
0.720000
0.730000
0.740000
0.750000
0.760000
0.770000
0.780000
0.790000
0.800000
0.810000
0.820000
0.830000
0.839999
0.849999
0.859999
0.869999
0.879999
0.889999
0.899999
0.909999
0.919999
0.929999
0.939999
0.949999
0.959999
0.969999
0.979999
0.989999
0.999999

As you can see after 0.83, the precision runs down significantly.

However, if I set up t as double, such an issue won't happen.

It took me five hours to realize this minor error, which ruined my program.

2018/03/10

Source: https://stackoverflow.com/questions/2386772
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