Convert the specified double-precision floating point number to a 64-bit signed integer in C#

To convert a double-precision floating point number to its 64-bit signed integer representation, C# provides the BitConverter.DoubleToInt64Bits() method. This method converts the binary representation of a double value into a long integer without changing the underlying bit pattern.

This is useful when you need to examine the internal binary structure of floating-point numbers or perform low-level operations on their bit representations.

Syntax

Following is the syntax for converting a double to its 64-bit signed integer representation −

long result = BitConverter.DoubleToInt64Bits(doubleValue);

Parameters

The DoubleToInt64Bits() method takes one parameter −

• value: A double-precision floating point number to convert.

Return Value

Returns a 64-bit signed integer whose bit pattern is equivalent to the input double value.

Using DoubleToInt64Bits() with Positive Values

Example

using System;

public class Demo {
   public static void Main() {
      double d = 5.646587687;
      Console.WriteLine("Original double value: " + d);
      
      long res = BitConverter.DoubleToInt64Bits(d);
      Console.WriteLine("64-bit signed integer: " + res);
      
      // Convert back to verify
      double backToDouble = BitConverter.Int64BitsToDouble(res);
      Console.WriteLine("Converted back to double: " + backToDouble);
   }
}

The output of the above code is −

Original double value: 5.646587687
64-bit signed integer: 4618043510978159912
Converted back to double: 5.646587687

Using DoubleToInt64Bits() with Small Values

Example

using System;

public class Demo {
   public static void Main() {
      double d = 0.001;
      Console.WriteLine("Original double value: " + d);
      
      long res = BitConverter.DoubleToInt64Bits(d);
      Console.WriteLine("64-bit signed integer: " + res);
      
      // Show hexadecimal representation for better understanding
      Console.WriteLine("Hexadecimal representation: 0x" + res.ToString("X"));
   }
}

The output of the above code is −

Original double value: 0.001
64-bit signed integer: 4562254508917369340
Hexadecimal representation: 0x3F50624DD2F1A9FC

Using DoubleToInt64Bits() with Special Values

Example

using System;

public class Demo {
   public static void Main() {
      double[] specialValues = { 0.0, -0.0, double.PositiveInfinity, 
                                double.NegativeInfinity, double.NaN };
      
      foreach (double value in specialValues) {
         long bits = BitConverter.DoubleToInt64Bits(value);
         Console.WriteLine($"Value: {value,18} -> Bits: {bits,20}");
      }
   }
}

The output of the above code is −

Value:                  0 -> Bits:                    0
Value:                  0 -> Bits:          -9223372036854775808
Value:           Infinity -> Bits:           9218868437227405312
Value:          -Infinity -> Bits:          -4503599627370496
Value:                NaN -> Bits:          -2251799813685248

How It Works

The DoubleToInt64Bits() method performs a direct bit-level conversion without any mathematical transformation. The IEEE 754 double-precision format uses 64 bits arranged as follows −

• 1 bit for the sign
• 11 bits for the exponent
• 52 bits for the mantissa (fractional part)

This method is particularly useful for debugging floating-point issues, implementing custom serialization, or working with bit manipulation algorithms.

Conclusion

The BitConverter.DoubleToInt64Bits() method converts a double-precision floating point number to its 64-bit signed integer bit representation. This is useful for examining the internal structure of floating-point numbers and performing low-level bit operations without losing precision.