英文:
No panic when converting int to uint?
问题
我对以下类型转换感到困惑。我本来期望两个uint转换都会引发恐慌。
a := -1
_ = uint(a) // 为什么没有恐慌？
_ = uint(-1) // 引发恐慌：常量-1溢出uint
为什么第2行没有引发恐慌？
链接：https://play.golang.org/p/jcfDL8km2C
英文:
I'm confused about the following type conversion. I would expect both uint conversions to panic.
a := -1
_ = uint(a) // why no panic?
_ = uint(-1) // panics: constant -1 overflows uint
Why doesn't it panic in line 2?
答案1
得分: 5
如在问题 6923中提到：
> T(c)，其中T是一个类型，c是一个常量，意味着将c视为具有类型T而不是默认类型之一。如果c无法表示为T，则会报错，但对于float和complex常量，只要值不太大，我们会将其静默舍入为T。
在这里：
const x uint = -1
var x uint = -1
> 这不起作用，因为-1不能（隐式地）**转换**为uint。
_ = uint(a) // 为什么没有恐慌？
因为a不是无类型常量，而是一个有类型的变量（int）。请参见Playground和“Golang常量溢出uint64有什么问题”：
package main
import "fmt"
func main() {
a := -1
_ = uint(a) // 为什么没有恐慌？
var b uint
b = uint(a)
fmt.Println(b)
// _ = uint(-1) // 恐慌：main.go:7: constant -1 overflows uint
}
结果：在32位系统上为4294967295，在64位系统上为18446744073709551615，由starriet在评论中注释。
这是非常量数值转换的特定规则：
> 在整数类型之间进行转换时，如果值是有符号整数，则将其符号扩展为隐式无限精度；否则将其零扩展。然后将其截断以适应结果类型的大小。
英文:
As mentioned in issue 6923:
> T(c) where T is a type and c is a constant means to treat c as having type T rather than one of the default types.
It gives an error if c can not be represented in T, except that for float and complex constants we quietly round to T as long as the value is not too large.
Here:
const x uint = -1
var x uint = -1
> This doesn't work because -1 cannot be (implicitly) converted to a uint.
_ = uint(a) // why no panic?
Because a is not an untyped constant, but a typed variable (int). See Playground and "what's wrong with Golang constant overflows uint64":
package main
import "fmt"
func main() {
a := -1
_ = uint(a) // why no panic?
var b uint
b = uint(a)
fmt.Println(b)
// _ = uint(-1) // panics: main.go:7: constant -1 overflows uint
}
Result: 4294967295 (on 32-bits system) or 18446744073709551615 (on 64-bits system), as commented by starriet
That are specific rules for the conversion of non-constant numeric values:
> When converting between integer types, if the value is a signed integer, it is sign extended to implicit infinite precision; otherwise it is zero extended.
It is then truncated to fit in the result type's size.
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