Why Does Converting a `uint8` to an `int8` in Go Lead to Unexpected Behavior?

Confusion Surrounding Conversion from uint8 to int8

Converting a uint8 to an int8 in Go can lead to unexpected behavior, as highlighted in the following example:

<code class="go">package main

import (
    "fmt"
)

func main() {
    a := int8(0xfc) // Compile error: constant 252 overflows int8
}</code>

The code fails with a compile-time error because the constant 0xfc exceeds the range of int8 (-128 to 127). However, the following code compiles without errors:

<code class="go">package main

import (
    "fmt"
)

func main() {
    a := 0xfc
    b := int8(a) // Ok
}</code>

The difference lies in the order of operations. In the first example, the constant is converted to int8 before being assigned to a. In the second example, a is first assigned the value 0xfc, and then converted to int8.

According to the Go language specification, constant expressions must be representable by the constant type. In this case, 0xfc cannot be represented by int8, leading to the compile error.

However, when the conversion is performed after assignment, the value of a (which is no longer a constant) is within the range of int8, allowing the conversion to succeed.

For practical use cases, consider the following example:

<code class="go">type Char byte

var c Char = '世' // Error: '世' has value 0x4e16, too large.</code>

This code raises an error because the value of the constant '世' (0x4e16) exceeds the range of byte (0 to 255). To fix this, one can use the following technique:

<code class="go">var c Char = Char('世') // Ok</code>

In this case, '世' is first assigned to a byte type and then converted to Char. This ensures that the value is within the range of the Char type. Similar techniques can be applied when converting bytes to int32 with considerations for the sign of the resulting value.

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