Printed Double Value Differs from Originally Set Value

How do I print a double value with full precision using cout?

You can set the precision directly on std::cout and use the std::fixed format specifier.

double d = 3.14159265358979;
cout.precision(17);
cout << "Pi: " << fixed << d << endl;

You can #include <limits> to get the maximum precision of a float or double.

#include <limits>

typedef std::numeric_limits< double > dbl;

double d = 3.14159265358979;
cout.precision(dbl::max_digits10);
cout << "Pi: " << d << endl;

How to get the difference between two dictionaries in Python?

Try the following snippet, using a dictionary comprehension:

value = { k : second_dict[k] for k in set(second_dict) - set(first_dict) }

In the above code we find the difference of the keys and then rebuild a dict taking the corresponding values.

How do I display a decimal value to 2 decimal places?

decimalVar.ToString("#.##"); // returns ".5" when decimalVar == 0.5m

or

decimalVar.ToString("0.##"); // returns "0.5"  when decimalVar == 0.5m

or

decimalVar.ToString("0.00"); // returns "0.50"  when decimalVar == 0.5m

How to make lists contain only distinct element in Python?

The simplest is to convert to a set then back to a list:

my_list = list(set(my_list))

One disadvantage with this is that it won't preserve the order. You may also want to consider if a set would be a better data structure to use in the first place, instead of a list.

Printing double values accordingly in java

try to use this format: "#0.##"

Java:Why should we use BigDecimal instead of Double in the real world?

It's called loss of precision and is very noticeable when working with either very big numbers or very small numbers. The binary representation of decimal numbers with a radix is in many cases an approximation and not an absolute value. To understand why you need to read up on floating number representation in binary. Here is a link: http://en.wikipedia.org/wiki/IEEE_754-2008. Here is a quick demonstration:

in bc (An arbitrary precision calculator language) with precision=10:

(1/3+1/12+1/8+1/15) = 0.6083333332

(1/3+1/12+1/8) = 0.541666666666666

(1/3+1/12) = 0.416666666666666

Java double:

0.6083333333333333

0.5416666666666666

0.41666666666666663

Java float:

0.60833335

0.5416667

0.4166667

If you are a bank and are responsible for thousands of transactions every day, even though they are not to and from one and same account (or maybe they are) you have to have reliable numbers. Binary floats are not reliable - not unless you understand how they work and their limitations.

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