转换负数 [英] Converting negative numbers

问题描述

将负数转换为无符号整数，然后再加上该值将导致相减.

When converting a negative number to an unsigned integer and later adding that value it results in subtracting.

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

这是惯用的方法还是应该更明确地进行?

Is this an idiomatic approach or should it be done more explicitly?

推荐答案

由于类型是未签名的，因此它是 溢出 :
uint8(b)是 248 ，所以 a + uint8(b)是 10 + 248 = 258 = 256 + 2 =>2

Since the type is unsigned it is an overflow:
uint8(b) is 248, so a + uint8(b) is 10+248=258=256+2 => 2

我的问题更多是关于如何在以下情况下从无符号整数中减去值(有时要加，有时要减去)即将到来从参数(必须是带符号的类型)中获取，以便您必须先进行类型转换，然后再进行减法/加法运算.

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.

您可以同时使用 int8 :

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

您可以避免溢出，例如此:

You may avoid the overflow, like this:

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

---

您应在此处删除3对多余的括号:

---

You should remove 3-pair of superfluous Parentheses here:

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

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

使用此:

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

---

请参阅:
关于将uint8转换为int8的困惑

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