英文:
Does Go fix or share C's hazardous implicit -> unsigned conversion?
问题
在《21世纪的C语言》一书中，Ben Klemens描述了C语言如何自动将有符号数转换为无符号数，例如在比较表达式中。Go语言是否也有这种危险的行为，或者Go语言对此有不同的处理方式？
英文:
In "21st Century C", Ben Klemens describes how C automatically converts signed numbers to unsigned, in comparison expressions for example. Does Go share this same hazardous behavior, or does Go approach it differently?
答案1
得分: 2
在Go语言中，没有隐式转换。如果要比较或进行算术运算的两个值的类型不同，必须进行手动和显式的类型转换。
a := 3 // 数值常量3默认为int类型
b := uint(2)
c := a < b // 编译器错误
d := a < int(b) // 正确
英文:
There are no implicit conversions in Go. In order to compare or do arithmetic with two values of different types, you must do manual and explicit type conversion.
a := 3 // numerical constant 3 defaults to int
b := uint(2)
c := a < b // compiler error
d := a < int(b) // OK
答案2
得分: 2
Go在所有操作中，包括比较，只使用显式类型转换。
var a uint64
var b int64
a = 1
b = 1
if a == b {
fmt.Println("Equal")
}
这段代码会导致错误：
tmp/sandbox428200464/main.go:13: invalid operation: a == b (mismatched types uint64 and int64)
要进行比较，你必须显式地将变量转换为相同的类型：
if int64(a) == b {...}
所以，可以说，这个问题肯定是修复了。
英文:
Go uses only explicit type conversions for all the operations including comparison.
var a uint64
var b int64
a = 1
b = 1
if a == b {
fmt.Println("Equal")
}
This snippet is going to cause an error:
> tmp/sandbox428200464/main.go:13: invalid operation: a == b (mismatched types uint64 and int64)
To do the comparison you must cast the variables to the same type explicitly:
if int64(a) == b {...}
So, definitely yes, it is fixed if one can say so.
Playgound for the snippet.
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