This question already has answers here:
How do I print a double value with full precision using cout?
(17 answers)
Closed 9 years ago.
I have a long double and I want to print all digits of it (the complete number without scientific notation) using cout .
Here's the code :-
long double d = 3456489465498484.14159265358979;
cout << "Num: " << d << endl;
output:-
Num: 3.45649e+015
While I want the output to be
Num: 3456489465498484.14159265358979;
I tried precision and set precision but the dont seem to work this way. please help
The internal representation of a floating-point value is typically binary, so initializing it requires converting from decimal to binary and displaying it requires converting from binary to decimal. Most decimal fractions do not have an exact binary representation, and most binary fractions do not have an exact decimal representation, so it is not meaningful, in general, to ask for "all" the digits. You've lost some digits on the way in, and will lose some more on the way out. Decide how many you want, and set the precision to match.
Related
This question already has answers here:
How do I print a double value with full precision using cout?
(17 answers)
Closed 3 years ago.
Why doubles round themselves? How can i prevent it?
If i insert 45000.98 i expect 45000.98, but the number is rounded.
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
double a;
cin >> a; //if i insert 45000.98
cout << a; //output is 45001
cout << endl << setprecision(2) << a; //output is 4.5e+04
}
Double type has 11 bits for exponent and 52 bits for the fractional part, more than enough to give you enough precision to represent 45000.98, but setprecision argument, as far as i recall, receives a characters limit, not the number of digits after decimal point. Use setprecision(8) and you should see 45000.98 as you probably expect.
The double did not round itself; the streaming operation rounded the value — first by default, and later according to your instructions. You requested that your value be rounded to 2 digits of precision, and that's what you got (just the first two digits: 4.5e+04). You are getting scientific notation because you have not requested enough digits to reach the decimal point.
If you want to see all 7 digits of 45000.98 then request at least 7 digits of precision. (You may want to stay under 17 digits though, since that's where you start seeing the artifacts of the floating point representation.)
This question already has answers here:
How do I print a double value with full precision using cout?
(17 answers)
Printing the correct number of decimal points with cout
(13 answers)
c++ cout << [double] not printing decimal places
(2 answers)
Closed 3 years ago.
setlocale(LC_ALL, "Portuguese");
double var = 1.0;
cout << var << endl;
system("pause");
return 0;
output:
1
Press any key to continue . . .
float and double only represent the values of numbers. “1” and “1.0” are both numerals for the same number, so 1 is the correct value of the var you set to 1.0. float and double do not represent the original numerals used to set their values, nor do they represent how much accuracy (relative to some ideal mathematical value) is present. The “1” you see as output is a result of default formatting. Other formatting options are available, but you must specify them yourself.
This question already has answers here:
Printing the correct number of decimal points with cout
(13 answers)
Closed 6 years ago.
As like,
AC = sqrt((AB*AB)+(BC*BC));
BD = sqrt((BC*BC)+(CD*CD));
(where all variables are double type)
Here, i want the value of AC and BD fixed with 2 digits after decimal point and then I want to add AC and BD. Now, if i take (3 digits after decimal point) AC=4.564 and BD=4.789 the the result after adding them is 9.351. Again if i take (2digits after decimal point) AC=4.56 and BD=78 then the result after adding them is 9.34. Now if i print both result fixed with 2 digits after decimal point then it shows 9.35 and 9.34 respectively. But I want 9.34 as result.
Two decimals of accuracy:
Multiply by 100
Add 0.5
Round down to closest integer
Divide by 100
But if you happen to end up with an integer value, you won't have "2 digits after decimal point"
If you're trying to output double values with 2 digits after the decimal then use modifiers to a stream like so:
std::cout << std::setprecision (2) << std::fixed << AC;
This will output a fixed point representation of the floating point number with 2 digits of precision.
Also, you should look at the round, ceil and floor functions and use them as follows:
ans = round (AC * 100.0) / 100.0
to do what you want. You'll find that you cannot represent all numbers as just two digits after the decimal so any comparisons you make should test for closeness of the answer to the target by +/- a small epsilon to the target value.
If you're trying to use the values computed for financial computation then this is very dangerous to do with the double type because certain values cannot be represented exactly with a limited number of bits of precision. You will suffer from accumulation of errors when performing financial computation with float/double types.
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 6 years ago.
I want to calculate the sum of three double numbers and I expect to get 1.
double a=0.0132;
double b=0.9581;
double c=0.0287;
cout << "sum= "<< a+b+c <<endl;
if (a+b+c != 1)
cout << "error" << endl;
The sum is equal to 1 but I still get the error! I also tried:
cout<< a+b+c-1
and it gives me -1.11022e-16
I could fix the problem by changing the code to
if (a+b+c-1 > 0.00001)
cout << "error" << endl;
and it works (no error). How can a negative number be greater than a positive number and why the numbers don't add up to 1?
Maybe it is something basic with summation and under/overflow but I really appreciate your help.
Thanks
Rational numbers are infinitely precise. Computers are finite.
Precision loss is a well known problem in computer programming.
The real question is, how can you remedy it?
Consider using an approximation function when comparing floats for equality.
#include <iostream>
#include <cmath>
#include <limits>
using namespace std;
template <typename T>
bool ApproximatelyEqual(const T dX, const T dY)
{
return std::abs(dX - dY) <= std::max(std::abs(dX), std::abs(dY))
* std::numeric_limits<T>::epsilon();
}
int main() {
double a=0.0132;
double b=0.9581;
double c=0.0287;
//Evaluates to true and does not print error.
if (!ApproximatelyEqual(a+b+c,1.0)) cout << "error" << endl;
}
Floating point numbers in C++ have a binary representation. This means that most numbers that can exactly represented by a decimal fraction with only a few digits cannot be exactly represented by floating point numbers. That's where your error comes from.
One example: 0.1 (decimal) is a periodic fraction in binary:
0.000110011001100110011001100...
Therefore it cannot be exactly be represented with any number of bits with binary encoding.
In order to avoid this type of error, you can use BCD (binary coded decimal) numbers which are supported by some special libraries. The drawbacks are slower calculation speed (not directly supported by the CPU) and slightly higher memory usage.
ANother option is to represent the number by a general fraction and store numerator and denomiator as separate integers.
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
Prevent scientific notation in ostream when using << with double
I get 1e-1 as result after a computation how can I convert the result from exponent to dot notation i.e., 0.1 ? Why is it automatically converted to exponential notation!!
You can use the fixed I/O manipulator to force the number to be printed in fixed-point notation:
double d = 42.0;
std::cout << std::fixed << d;
(std::scientific does the opposite: it forces the number to be printed in scientific notation)
Oracle (generally) doesn't do binary numbers (some support was added in 10g).
Numbers are held in an internal format and, unless you use an implicit or explicit TO_CHAR, it is up to the "client" to display them (or any desired "prettifying").
select to_number('1e-1') num,
to_char(to_number('1e-1'),'9.9EEEE') sci_num,
to_char(to_number('1e-1')) std_num
from dual;
NUM SCI_NUM ST
--------------- --------- --
.10 1.0E-01 .1