Why does converting a uint8 to int8 in Go cause an error?

Confusion Converting uint8 to int8

In Go, converting an unsigned 8-bit integer (uint8) to a signed 8-bit integer (int8) can result in an error. Let's explore why.

Consider the following code:

<code class="go">package main

import "fmt"

func main() {
    a := int8(0xfc)  // compile error
    fmt.Println(a)
}</code>

This code raises a compile-time error: "constant 252 overflows int8." To understand this issue, we need to refer to Go's constant expression rules.

According to the language specification, constant expressions must always be representable by values of the constant type. In this case, 0xfc is too large to fit in an int8, whose range is -128 to 127.

If we defer the type conversion, as shown below, the code compiles without errors:

<code class="go">package main

import "fmt"

func main() {
    a := 0xfc
    b := int8(a)  // ok
    fmt.Println(b)
}</code>

This works because 0xfc is interpreted as an integer literal before being converted to an int8. As an integer literal, it can hold values outside the range of int8, but the compiler enforces the type check during the actual conversion.

Some additional points regarding integer conversion in Go:

• Not all integer values can fit into all integer types. For example, int64 cannot hold values larger than 9,223,372,036,854,775,807.
• Unsigned integer types (uint8, uint16, etc.) cannot hold negative values.
• To convert a byte (an alias for uint8) to a signed integer type, you can use the following pattern:

<code class="go">var b byte = 0xff
i32 := int32(int8(b))</code>

This ensures that the sign of the original byte is preserved.

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