Why Does Converting uint8 to int8 with a Constant Fail in Go?

Confusion about Converting uint8 to int8

When converting a uint8 (unsigned 8-bit integer) to an int8 (signed 8-bit integer), the result may not always be as expected. This article explores the differences between the following two approaches and explains why the first one raises a compile error:

<code class="go">a := int8(0xfc) // compile error
// deferring type conversion
a := 0xfc
b := int8(a) </code>

The Difference Between the Codes

The difference between the two codes lies in the order of operations:

• In the first example, the type conversion (int8(0xfc)) occurs before assignment, while in the second example, the assignment (a := 0xfc) occurs before the type conversion (int8(a)).

Compile Error in the First Code

The first code raises a compile error because the constant 0xfc exceeds the range of the int8 type. int8 can represent values between -128 and 127, while 0xfc is 252 in decimal, which exceeds this range. According to the Go language specification, constant expressions must always be accurately representable by values of the constant type.

Legal and Illegal Constant Expressions

Refer to https://golang.org/ref/spec#Constant_expressions for a detailed explanation of legal and illegal constant expressions. The blog post at https://blog.golang.org/constants provides additional insights into the limitations of integer types and the importance of type-checking constants to catch errors.

Solving the Conversion Issue

If your goal is to convert a byte (which is a uint8) to an int32 while considering its sign, the recommended approach is to first convert it to an int8 and then to an int32:

<code class="go">var b byte = 0xff
i32 := int32(int8(b)) // -1</code>

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