Why Do C# and C Produce Different Output When Formatting Doubles?

Formatting Doubles for Output in C#

In a recent experiment, an attempt was made to emulate the output formatting capabilities of C in C#. However, discrepancies were observed between the expected and actual output.

The C code:

double i = 10 * 0.69;
printf("%f\n", i);
printf("  %.20f\n", i);
printf("+ %.20f\n", 6.9 - i);
printf("= %.20f\n", 6.9);

produced the following output:

6.900000
6.89999999999999946709
+ 0.00000000000000088818
= 6.90000000000000035527

The C# code:

double i = 10 * 0.69;
Console.WriteLine(i);
Console.WriteLine(String.Format("  {0:F20}", i));
Console.WriteLine(String.Format("+ {0:F20}", 6.9 - i));
Console.WriteLine(String.Format("= {0:F20}", 6.9));

yielded:

6.9
6.90000000000000000000
+ 0.00000000000000088818
= 6.90000000000000000000

Despite i appearing as 6.89999999999999946709 in the debugger, the C# formatting rounds the value to 15 significant decimal digits before applying the format.

This unexpected behavior is because .NET internally stores doubles using a binary representation. Before formatting, .NET converts the double to a decimal string, which inherently introduces rounding error.

To avoid this issue, one can either:

• Access the exact decimal value of the double: Construct a string representation of the decimal value by extracting the internal binary bits manually.
• Use a custom formatting library: Leverage Jon Skeet's DoubleConverter class, which provides methods for retrieving the exact decimal value of a double and rounding the output to a specified precision.

The DoubleConverter usage example:

Console.WriteLine(DoubleConverter.ToExactString(i));
Console.WriteLine(DoubleConverter.ToExactString(6.9 - i));
Console.WriteLine(DoubleConverter.ToExactString(6.9));

// 6.89999999999999946709294817992486059665679931640625
// 0.00000000000000088817841970012523233890533447265625
// 6.9000000000000003552713678800500929355621337890625

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