How Does Go Handle Arithmetic Operations on Extremely Large Constants Without Memory Overflow?

Arithmetic Operations on Constants in Go

Question:

In Go, constant operations on very large numbers seem to work seamlessly. However, how does Go manage to store and perform arithmetic on these constants without exceeding memory limits?

Answer:

Storage of Constants:

Contrary to what you might expect, untyped constants with arbitrary precision are not stored at runtime. Instead, they exist only during compilation time. This means Go does not have to represent them in its executable binaries.

In your example, the constant Huge does not appear in the executable. Instead, a function call to fmt.Println() is recorded with a float64 value of 10.

Precision and Arithmetic:

While constants have arbitrary precision at the source code level, this precision is not guaranteed at runtime. Here's how Go manages arithmetic on constants:

• Default Type Inference: Untyped constants like 1e1000 are assigned a default type based on their value (e.g., float64 in your example).
• Limited Precision: At runtime, only fixed-precision types are available (int, float32, float64, etc.).
• Loss of Precision: If the result of a constant arithmetic operation exceeds the precision of the default type, it is converted to the nearest representable value. Thus, the result of Huge / 1e999 becomes 10.0 (float64).
• Accuracy Guarantees: Despite the potential loss of precision, Go's compiler guarantees that the result of a constant expression is always within a certain range. For example, integer constants must have at least 256 bits of precision.

Implementing Arbitrary Precision:

While the Go standard library does not provide arbitrary precision out of the box, there are packages like math/big and go/constant that allow you to represent and manipulate values with "arbitrary" precision. These packages use big integers and fractions to store and operate on numbers beyond the limits of Go's built-in types.

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