本文介绍了关于将uint8转换为int8的困惑的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想将uint8
转换为int
,所以我写了一个const 0xfc
,并尝试使用int8(0xfc)
对其进行转换.但是,代码会引发错误:
I want to convert uint8
to int
, so I write a const 0xfc
, and try to use int8(0xfc)
to convert it. However the code raises an error:
package main
import (
"fmt"
)
func main() {
a := int8(0xfc) // compile error: constant 252 overflows int8
b := a
fmt.Println(b)
}
但是如果我在分配后推迟类型转换,则代码可以解决.
But if I defer the type conversion after assignment, the code can work around.
package main
import (
"fmt"
)
func main() {
a := 0xfc
b := int8(a) // ok
fmt.Println(b)
}
我的问题:
- 这两个代码之间有什么区别吗?
- 为什么第一个会引发编译错误?
推荐答案
uint(-1) // -1 cannot be represented as a uint
int(3.14) // 3.14 cannot be represented as an int
int64(Huge) // 1267650600228229401496703205376 cannot be represented as an int64
Four * 300 // operand 300 cannot be represented as an int8 (type of Four)
Four * 100 // product 400 cannot be represented as an int8 (type of Four)
- see:https://blog.golang.org/constants
type Char byte
var c Char = '世' // Error: '世' has value 0x4e16, too large.
如果编译器抱怨您使用了常量,则可能是真正的错误.
If the compiler complains about your use of a constant, it's likely a real bug like this.
是的,这是要走的路:
var b byte = 0xff
i32 := int32(int8(b))
fmt.Println(i32) // -1
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