Related
#include <iostream>
using namespace std;
int main() {
int steps=1000000000;
float s = 0;
for (int i=1;i<(steps+1);i++){
s += (i/2.0) ;
}
cout << s << endl;
}
Declaring s as float: 9.0072e+15
Declaring s as double: 2.5e+17 (same result as implementing it in Julia)
I understand double has double precision than float, but float should still handle numbers up to 10^38.
I did read similar topics where results where not the same, but in that cases the differences were very small, here the difference is 25x.
I also add that using long double instead gives me the same result as double. If the matter is the precision, I would have expected to have something a bit different.
The problem is the lack of precision: https://en.wikipedia.org/wiki/Floating_point
After 100 million numbers you are adding 1e8 to 1e16 (or at least numbers of that magnitude), but single precision numbers are only accurate to 7 digits - so it is the same as adding 0 to 1e16; that's why your result is considerably lower for float.
Prefer double over float in most cases.
Problem with floating point precision! Infinite real numbers cannot possibly be represented by the finite memory of a computer. Float, in general, are just approximations of the number they are meant to represent.
For more details, please check the following documentation:
https://softwareengineering.stackexchange.com/questions/101163/what-causes-floating-point-rounding-errors
You didn't mention what type of floating point numbers you are using, but I'm going to assume that you use IEEE 754, or similar.
I understand double has double precision
To be more precise with the terminology, double uses twice as many bits. That's not double the number of reprensentable values, it's 4294967296 times as many representable values, despite being named "double precision".
but float should still handle numbers up to 10^38.
Float can handle a few numbers up to that magnitude. But that does't mean that float values in that range are precise. For example, 3,4028235E+38 can be represented as a single precision float. How much would you imagine is the difference between the previous value representable by float? Is it the machine epsilon? Perhaps 0.1? Maybe 1? No. The difference is about 2E+31.
Now, your numbers aren't quite in that range. But, they're outside the continuous range of whole integers that can be precisely represented by float. The highest value in that range happens to be 16777217, or about 1.7E+7, which is way less than 2.5E+17. So, every addition beyond that range adds some error to the result. You perform a billion calculations so those errors add up.
Conclusions:
Understand that single precision is way less precise than double precision.
Avoid long sequences of calculations where precision errors can accumulate.
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?
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).
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.
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.)
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.
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.
There are three floating point types:
float
double
long double
A simple Venn diagram will explain about:
The set of values of the types
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.
Type float, 32 bits long, has a precision of 7 digits. While it may store values with very large or very small range (+/- 3.4 * 10^38 or * 10^-38), it has only 7 significant digits.
Type double, 64 bits long, has a bigger range (*10^+/-308) and 15 digits precision.
Type long double is nominally 80 bits, though a given compiler/OS pairing may store it as 12-16 bytes for alignment purposes. The long double has an exponent that just ridiculously huge and should have 19 digits precision. Microsoft, in their infinite wisdom, limits long double to 8 bytes, the same as plain double.
Generally speaking, just use type double when you need a floating point value/variable. Literal floating point values used in expressions will be treated as doubles by default, and most of the math functions that return floating point values return doubles. You'll save yourself many headaches and typecastings if you just use double.
Floats have less precision than doubles. Although you already know, read What WE Should Know About Floating-Point Arithmetic for better understanding.
When using floating point numbers you cannot trust that your local tests will be exactly the same as the tests that are done on the server side. The environment and the compiler are probably different on you local system and where the final tests are run. I have seen this problem many times before in some TopCoder competitions especially if you try to compare two floating point numbers.
The built-in comparison operations differ as in when you compare 2 numbers with floating point, the difference in data type (i.e. float or double) may result in different outcomes.
If one works with embedded processing, eventually the underlying hardware (e.g. FPGA or some specific processor / microcontroller model) will have float implemented optimally in hardware whereas double will use software routines. So if the precision of a float is enough to handle the needs, the program will execute some times faster with float then double. As noted on other answers, beware of accumulation errors.
Quantitatively, as other answers have pointed out, the difference is that type double has about twice the precision, and three times the range, as type float (depending on how you count).
But perhaps even more important is the qualitative difference. Type float has good precision, which will often be good enough for whatever you're doing. Type double, on the other hand, has excellent precision, which will almost always be good enough for whatever you're doing.
The upshot, which is not nearly as well known as it should be, is that you should almost always use type double. Unless you have some particularly special need, you should almost never use type float.
As everyone knows, "roundoff error" is often a problem when you're doing floating-point work. Roundoff error can be subtle, and difficult to track down, and difficult to fix. Most programmers don't have the time or expertise to track down and fix numerical errors in floating-point algorithms — because unfortunately, the details end up being different for every different algorithm. But type double has enough precision such that, much of the time, you don't have to worry.
You'll get good results anyway. With type float, on the other hand, alarming-looking issues with roundoff crop up all the time.
And the thing that's not necessarily different between type float and double is execution speed. On most of today's general-purpose processors, arithmetic operations on type float and double take more or less exactly the same amount of time. Everything's done in parallel, so you don't pay a speed penalty for the greater range and precision of type double. That's why it's safe to make the recommendation that you should almost never use type float: Using double shouldn't cost you anything in speed, and it shouldn't cost you much in space, and it will almost definitely pay off handsomely in freedom from precision and roundoff error woes.
(With that said, though, one of the "special needs" where you may need type float is when you're doing embedded work on a microcontroller, or writing code that's optimized for a GPU. On those processors, type double can be significantly slower, or practically nonexistent, so in those cases programmers do typically choose type float for speed, and maybe pay for it in precision.)
Unlike an int (whole number), a float have a decimal point, and so can a double.
But the difference between the two is that a double is twice as detailed as a float, meaning that it can have double the amount of numbers after the decimal point.
I have a class that internally represents some quantity in fixed point as 32-bit integer with somewhat arbitrary denominator (it is neither power of 2 nor power of 10).
For communicating with other applications the quantity is converted to plain old double on output and back on input. As code inside the class it looks like:
int32_t quantity;
double GetValue() { return double(quantity) / DENOMINATOR; }
void SetValue(double x) { quantity = x * DENOMINATOR; }
Now I need to ensure that if I output some value as double and read it back, I will always get the same value back. I.e. that
x.SetValue(x.GetValue());
will never change x.quantity (x is arbitrary instance of the class containing the above code).
The double representation has more digits of precision, so it should be possible. But it will almost certainly not be the case with the simplistic code above.
What rounding do I need to use and
How can I find the critical would-be corner cases to test that the rounding is indeed correct?
Any 32 bits will be represented exactly when you convert to a double, but when you divide then multiply by an arbitrary value you will get a similar value but not exactly the same. You should lose at most one bit per operations, which means your double will be almost the same, prior to casting back to an int.
However, since int casts are truncations, you will get the wrong result when very minor errors turn 2.000 into 1.999, thus what you need to do is a simple rounding task prior to casting back.
You can use std::lround() for this if you have C++11, else you can write you own rounding function.
You probably don't care about fairness much here, so the common int(doubleVal+0.5) will work for positives. If as seems likely, you have negatives, try this:
int round(double d) { return d<0?d-0.5:d+0.5; }
The problem you describe is the same problem which exists with converting between binary and decimal representation just with different bases. At least it exists if you want to have the double representation to be a good approximation of the original value (otherwise you could just multiply the 32 bit value you have with your fixed denominator and store the result in a double).
Assuming you want the double representation be a good approximation of your actual value the conversions are nontrivial! The conversion from your internal representation to double can be done using Dragon4 ("How to print floating point numbers accurately", Steele & White) or Grisu ("How to print floating point numbers quickly and accurately", Loitsch; I'm not sure if this algorithm is independent from the base, though). The reverse can be done using Bellerophon ("How to read floating point numbers accurately", Clinger). These algorithms aren't entirely trivial, though...
This question already has answers here:
Comparing float and double
(3 answers)
Closed 7 years ago.
According to this post, when comparing a float and a double, the float should be treated as double.
The following program, does not seem to follow this statement. The behaviour looks quite unpredictable.
Here is my program:
void main(void)
{
double a = 1.1; // 1.5
float b = 1.1; // 1.5
printf("%X %X\n", a, b);
if ( a == b)
cout << "success " <<endl;
else
cout << "fail" <<endl;
}
When I run the following program, I get "fail" displayed.
However, when I change a and b to 1.5, it displays "success".
I have also printed the hex notations of the values. They are different in both the cases. My compiler is Visual Studio 2005
Can you explain this output ? Thanks.
float f = 1.1;
double d = 1.1;
if (f == d)
In this comparison, the value of f is promoted to type double. The problem you're seeing isn't in the comparison, but in the initialization. 1.1 can't be represented exactly as a floating-point value, so the values stored in f and d are the nearest value that can be represented. But float and double are different sizes, so have a different number of significant bits. When the value in f is promoted to double, there's no way to get back the extra bits that were lost when the value was stored, so you end up with all zeros in the extra bits. Those zero bits don't match the bits in d, so the comparison is false. And the reason the comparison succeeds with 1.5 is that 1.5 can be represented exactly as a float and as a double; it has a bunch of zeros in its low bits, so when the promotion adds zeros the result is the same as the double representation.
I found a decent explanation of the problem you are experiencing as well as some solutions.
See How dangerous is it to compare floating point values?
Just a side note, remember that some values can not be represented EXACTLY in IEEE 754 floating point representation. Your same example using a value of say 1.5 would compare as you expect because there is a perfect representation of 1.5 without any loss of data. However, 1.1 in 32-bit and 64-bit are in fact different values because the IEEE 754 standard can not perfectly represent 1.1.
See http://www.binaryconvert.com
double a = 1.1 --> 0x3FF199999999999A
Approximate representation = 1.10000000000000008881784197001
float b = 1.1 --> 0x3f8ccccd
Approximate representation = 1.10000002384185791015625
As you can see, the two values are different.
Also, unless you are working in some limited memory type environment, it's somewhat pointless to use floats. Just use doubles and save yourself the headaches.
If you are not clear on why some values can not be accurately represented, consult a tutorial on how to covert a decimal to floating point.
Here's one: http://class.ece.iastate.edu/arun/CprE281_F05/ieee754/ie5.html
I would regard code which directly performs a comparison between a float and a double without a typecast to be broken; even if the language spec says that the float will be implicitly converted, there are two different ways that the comparison might sensibly be performed, and neither is sufficiently dominant to really justify a "silent" default behavior (i.e. one which compiles without generating a warning). If one wants to perform a conversion by having both operands evaluated as double, I would suggest adding an explicit type cast to make one's intentions clear. In most cases other than tests to see whether a particular double->float conversion will be reversible without loss of precision, however, I suspect that comparison between float values is probably more appropriate.
Fundamentally, when comparing floating-point values X and Y of any sort, one should regard comparisons as indicating that X or Y is larger, or that the numbers are "indistinguishable". A comparison which shows X is larger should be taken to indicate that the number that Y is supposed to represent is probably smaller than X or close to X. A comparison that says the numbers are indistinguishable means exactly that. If one views things in such fashion, comparisons performed by casting to float may not be as "informative" as those done with double, but are less likely to yield results that are just plain wrong. By comparison, consider:
double x, y;
float f = x;
If one compares f and y, it's possible that what one is interested in is how y compares with the value of x rounded to a float, but it's more likely that what one really wants to know is whether, knowing the rounded value of x, whether one can say anything about the relationship between x and y. If x is 0.1 and y is 0.2, f will have enough information to say whether x is larger than y; if y is 0.100000001, it will not. In the latter case, if both operands are cast to double, the comparison will erroneously imply that x was larger; if they are both cast to float, the comparison will report them as indistinguishable. Note that comparison results when casting both operands to double may be erroneous not only when values are within a part per million; they may be off by hundreds of orders of magnitude, such as if x=1e40 and y=1e300. Compare f and y as float and they'll compare indistinguishable; compare them as double and the smaller value will erroneously compare larger.
The reason why the rounding error occurs with 1.1 and not with 1.5 is due to the number of bits required to accurately represent a number like 0.1 in floating point format. In fact an accurate representation is not possible.
See How To Represent 0.1 In Floating Point Arithmetic And Decimal for an example, particularly the answer by #paxdiablo.
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?
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).
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.
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.)
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.
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.
There are three floating point types:
float
double
long double
A simple Venn diagram will explain about:
The set of values of the types
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.
Type float, 32 bits long, has a precision of 7 digits. While it may store values with very large or very small range (+/- 3.4 * 10^38 or * 10^-38), it has only 7 significant digits.
Type double, 64 bits long, has a bigger range (*10^+/-308) and 15 digits precision.
Type long double is nominally 80 bits, though a given compiler/OS pairing may store it as 12-16 bytes for alignment purposes. The long double has an exponent that just ridiculously huge and should have 19 digits precision. Microsoft, in their infinite wisdom, limits long double to 8 bytes, the same as plain double.
Generally speaking, just use type double when you need a floating point value/variable. Literal floating point values used in expressions will be treated as doubles by default, and most of the math functions that return floating point values return doubles. You'll save yourself many headaches and typecastings if you just use double.
Floats have less precision than doubles. Although you already know, read What WE Should Know About Floating-Point Arithmetic for better understanding.
When using floating point numbers you cannot trust that your local tests will be exactly the same as the tests that are done on the server side. The environment and the compiler are probably different on you local system and where the final tests are run. I have seen this problem many times before in some TopCoder competitions especially if you try to compare two floating point numbers.
The built-in comparison operations differ as in when you compare 2 numbers with floating point, the difference in data type (i.e. float or double) may result in different outcomes.
If one works with embedded processing, eventually the underlying hardware (e.g. FPGA or some specific processor / microcontroller model) will have float implemented optimally in hardware whereas double will use software routines. So if the precision of a float is enough to handle the needs, the program will execute some times faster with float then double. As noted on other answers, beware of accumulation errors.
Quantitatively, as other answers have pointed out, the difference is that type double has about twice the precision, and three times the range, as type float (depending on how you count).
But perhaps even more important is the qualitative difference. Type float has good precision, which will often be good enough for whatever you're doing. Type double, on the other hand, has excellent precision, which will almost always be good enough for whatever you're doing.
The upshot, which is not nearly as well known as it should be, is that you should almost always use type double. Unless you have some particularly special need, you should almost never use type float.
As everyone knows, "roundoff error" is often a problem when you're doing floating-point work. Roundoff error can be subtle, and difficult to track down, and difficult to fix. Most programmers don't have the time or expertise to track down and fix numerical errors in floating-point algorithms — because unfortunately, the details end up being different for every different algorithm. But type double has enough precision such that, much of the time, you don't have to worry.
You'll get good results anyway. With type float, on the other hand, alarming-looking issues with roundoff crop up all the time.
And the thing that's not necessarily different between type float and double is execution speed. On most of today's general-purpose processors, arithmetic operations on type float and double take more or less exactly the same amount of time. Everything's done in parallel, so you don't pay a speed penalty for the greater range and precision of type double. That's why it's safe to make the recommendation that you should almost never use type float: Using double shouldn't cost you anything in speed, and it shouldn't cost you much in space, and it will almost definitely pay off handsomely in freedom from precision and roundoff error woes.
(With that said, though, one of the "special needs" where you may need type float is when you're doing embedded work on a microcontroller, or writing code that's optimized for a GPU. On those processors, type double can be significantly slower, or practically nonexistent, so in those cases programmers do typically choose type float for speed, and maybe pay for it in precision.)
Unlike an int (whole number), a float have a decimal point, and so can a double.
But the difference between the two is that a double is twice as detailed as a float, meaning that it can have double the amount of numbers after the decimal point.