go - 转换负数

Question

当将负数转换为无符号整数并稍后添加该值时，它会导致减去。

a := (uint8)(10)
b := (int8)(-8)
fmt.Println(a + (uint8)(b)) // result: 2

这是一种惯用的方法还是应该更明确地完成？

Answer 1

由于类型是无符号的，所以它是一个 overflow :
uint8(b) 是 248，所以 a + uint8(b) 是 10+248=258=> 255 0 1 2 => 2

my question is more about how to subtract from unsigned integers when the value (sometimes you want to add and sometimes subtract) is coming from an argument (that must be a signed type) which makes it so that you have to do type conversion before subtracting/adding.

同时使用 int8:

    a := int8(10)
    b := int8(-8)
    fmt.Println(a + b) // 2
    fmt.Println(a - b) // 18

您可以避免溢出，例如 this :

    a := uint8(10)
    b := int8(-8)
    c := uint8(b)
    d := uint16(a) + uint16(c)
    fmt.Println(d) // 258

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你应该删除这里多余的括号:

a := (uint8)(10)
b := (int8)(-8)
fmt.Println(a + (uint8)(b))

使用这个:

a := uint8(10)
b := int8(-8)
fmt.Println(a + uint8(b))

---

参见:
confusion about convert `uint8` to `int8`