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Why Does My Go Constant Overflow When Using Bitwise Operations?

Patricia Arquette
Patricia ArquetteOriginal
2024-11-01 08:55:30107browse

Why Does My Go Constant Overflow When Using Bitwise Operations?

Trouble with Go Constant Overflow: Resolving Int Overflows

In the realm of Go programming, a common question arises: "Why does my constant overflow when userid := 12345 and did := (userid &^(0xFFFF << 48))?"

When compiling such code, the dreaded error emerges: "constant -18446462598732840961 overflows int." But what lies beneath this misleading message?

The Essence of the Issue

The culprit behind this woe is the untyped nature of the constant ^(0xFFFF << 48). Go assigns untyped constants colossal values, exceeding the bounds of conventional integers.

0xffff << 48 yields 0xffff000000000000. Reversing its sign using the bitwise negation operator (^) results in -0xffff000000000001.

Now, the variable userid, initialized with the value 12345, inherits the type int. When the bitwise AND operation (&) is performed, the compiler assumes that the untyped constant should be downgraded to the type int. This assumption leads to a clash, as the magnitude of the constant surpasses the maximum value admissible for int.

The Solution: Taming Untyped Giants

To resolve this issue, embrace the power of type casting. If you intended to assign the constant 0x0000ffffffffffff, consider utilizing 1<<48 - 1. This expression, when employed in 64-bit environments, ensures compatibility with larger integers.

For true portability, eschew int in favor of int64, widening the scope of your code to accommodate both 32-bit and 64-bit architectures.

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