Do Doubles Accurately Represent All Integer Values?

Are All Integer Values Accurately Represented as Doubles?

Converting an integer to a double raises the question of whether it will maintain perfect representation. This article addresses this query by considering the IEEE 754 standard for floating-point arithmetic.

Double Representation

Doubles use the form mantissa * 2^exponent, employing some bits for the non-integer portion. By setting a large enough exponent, integers can be represented solely using the fractional part.

32-bit Integers

IEEE 754 double-precision can accommodate integers up to 53 bits (52 significand bits plus the implicit leading bit). Therefore, all 32-bit integers can be perfectly represented as doubles.

64-bit Integers

However, this guarantee does not extend to 64-bit integers, as their size exceeds the capacity of double representation.

Practical Example

To illustrate this concept, consider the code snippet:

<code class="cpp">int a = 3;
int b = 4;
double d_a = a;
double d_b = b;
double int_sum = a + b;
double d_sum = d_a + d_b;

if (double(int_sum) == d_sum)
{
    std::cout <p>This code will consistently output "Same" for 32-bit integers, demonstrating that they are preserved as doubles.</p>
<p><strong>Fractional Differences</strong></p>
<p>Another aspect to consider is the possibility of a double conversion introducing a fractional difference that is later lost during the conversion back to integer. However, due to the exponential nature of double representation, this is not possible. The smallest difference between two doubles is always a power of two, ensuring that rounding never causes a change in the integer value.</p>
<p><strong>Conclusion</strong></p>
<p>While 32-bit integers can be perfectly represented as doubles, this is not true for 64-bit integers. Nonetheless, double precision offers ample capacity for most real-world integer applications.</p></code>

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