Why Does Converting a uint64 to an int64 Result in -1?

Conversion of uint64 to int64 with Preservation of Information

In the realm of programming, it is imperative to handle data types with precision. Consider the following code:

Upon execution, the value of y becomes -1, a perplexing outcome since it seemingly truncates significant bits from the original number. To resolve this issue, let's delve into the implications of this conversion.

Preserving Bit Values

Contrary to expectations, the conversion via int64(x) preserves all the bits from x, including the leading 64 bits that are beyond the range of an int64. It signifies that y and x share identical bit patterns in their binary representations:

Therefore, the value of y, despite being represented as -1, is mathematically equivalent to its uint64 counterpart, x.

Example for Clarity

To illustrate, let's consider a slightly different number:

In this case, y = -4, confirming that the lost bits in the conversion are properly maintained. In binary:

The conversion simply interprets the leading 63 bits as the magnitude of a negative number and sets the sign bit accordingly. This allows for lossless conversions between uint64 and int64 within a specific range, preserving the statistical properties crucial for random number generators.

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