英文:
Converting Hex to signed Int in Go
问题
我想在Go语言中将一个十六进制字符串转换为有符号整数值。我的输入值是"FF60"，我希望输出结果是"-160"。当我使用下面的函数时，结果是"65376"，它表示的是无符号表示。
value, err := strconv.ParseInt("FF60", 16, 64)
我本来期望使用ParseUInt函数时的结果是65376。
非常感谢任何帮助。
英文:
I want to convert a hex string to a signed integer Value in Go. My input value is "FF60" and I want the output to be "-160". When I use the following function, the result is "65376", which represents the unsigned representation.
value, err := strconv.ParseInt("FF60", 16, 64)
I would have expected the outcome of 65376 when using the ParseUInt function.
Any help would be really appreciated.
答案1
得分: 3
strconv.ParseInt()的第三个参数指定了你想解析的整数的位数。将0xff60解析为64位整数确实是65376。
实际上，你想将其解析为16位整数，所以将16作为位数传递进去。但这样做会导致错误：
strconv.ParseInt: parsing "FF60": value out of range
这是正确的：0xFF60（即65376）超出了int16的有效范围（有效的int16范围是[-32768..32767]）。
因此，你可以使用strconv.ParseUint()将其解析为无符号16位整数，然后将结果转换为有符号16位整数：
value, err := strconv.ParseUint("FF60", 16, 16)
fmt.Println(value, err)
fmt.Println(int16(value))
这将输出（在Go Playground上尝试）：
65376 <nil>
-160
英文:
The third parameter of strconv.ParseInt() tells the bitsize of the integer you want to parse. 0xff60 parsed as a 64-bit integer is indeed 65376.
You actually want to parse it as a 16-bit integer, so pass 16 as the bitsize. Doing so you will get an error though:
strconv.ParseInt: parsing "FF60": value out of range
Which is true: 0xFF60 (which is 65376) is outside of the valid range of int16 (valid int16 range is [-32768..32767]).
So what you may do is parse it as an unsigned 16-bit integer using strconv.ParseUint(), then convert the result to a signed 16-bit integer:
value, err := strconv.ParseUint("FF60", 16, 16)
fmt.Println(value, err)
fmt.Println(int16(value))
This will output (try it on the Go Playground):
65376 <nil>
-160
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