英文:
Golang hex encoding
问题
我正在尝试重新创建一个我在Python中找到的PKCS #5填充算法。
我遇到困难的主要部分是这一行：
return data + (chr(pad_count) * pad_count).encode('utf-8')
它实际上是将pad_count（一个介于1和16之间的整数）重复作为一个字符，重复1到16次。我在Go语言中很难得到类似的结果。
例如，pad_count为11将返回字符串：
\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b
我目前得到的最接近的结果是：
b := make([]byte, 2)
binary.LittleEndian.PutUint16(b, uint16(padCount))
fmt.Println("Pad: ", padCount, "Hex: ", hex.EncodeToString(b))
它将返回：
Pad: 11 Hex: 0b00
这已经非常接近了，显然我可以截取子字符串，并自己添加\x，但是否有更好的方法来解决这个问题呢？另外，如果我截取子字符串，我觉得不能保证适用于所有的组合情况。
英文:
I'm attempting to recreate a PKCS #5 Padding algorithm I've found written in python.
The main line I'm struggling to recreate is this
return data + (chr(pad_count) * pad_count).encode('utf-8')
which essentially repeats pad_count (an integer, between 1 and 16), as a char, between 1 and 16 times. I'm having trouble getting a similar result in Go.
For example, pad_count of 11 will return the string
\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b\x0b
The closeset I've come is this:
b := make([]byte, 2)
binary.LittleEndian.PutUint16(b, uint16(padCount))
fmt.Println("Pad: ", padCount, "Hex: ", hex.EncodeToString(b))
which will return:
Pad: 11 Hex: 0b00
This is pretty close, and obviously I could take a substring, and add the \x myself, but is there a better way to go about this? Also if I substring, I feel there is no guarantee that would work for all the combinations.
答案1
得分: 7
正如James Henstridge已经提到的，你想要的格式(\x0b...)并不是必需的，而是Python对不可打印字符的表示方式。你可以自己验证一下：
>>> chr(3)
'\x03'
你需要做的是在RFC2898中定义的：
[...] 其中填充字符串
PS由8-(||M|| mod 8)个八位字节组成，每个字节的值为8-(||M|| mod 8)。填充字符串PS将满足以下条件之一：如果 ||M|| mod 8 = 7，则 PS = 01； 如果 ||M|| mod 8 = 6，则 PS = 02 02； ... 如果 ||M|| mod 8 = 0，则 PS = 08 08 08 08 08 08 08 08。
这意味着你不需要使用uint16，而是使用uint8（因为一个八位字节只有8个位），也不需要按照Python的方式格式化字节。所以你需要做的只是使用bytes.Repeat：
bytes.Repeat(paddingChar, paddingCount)
英文:
As James Henstridge already mentioned the formatting you want (\x0b...) is not something that's required but rather python's representation of non-printable characters. See for yourself:
>>> chr(3)
'\x03'
What you have to do is defined in RFC2898:
> [...] where the padding string PS consists of 8-(||M|| mod 8) octets
> each with value 8-(||M|| mod 8). The padding string PS will
> satisfy one of the following statements:
>
> PS = 01, if ||M|| mod 8 = 7 ;
> PS = 02 02, if ||M|| mod 8 = 6 ;
> ...
> PS = 08 08 08 08 08 08 08 08, if ||M|| mod 8 = 0.
This means that you do not need uint16 but uint8 (since an octet has only 8 bits) and you also do not need to format your bytes the way python does. So the only thing you have to do is to use bytes.Repeat:
bytes.Repeat(paddingChar, paddingCount)
答案2
得分: 0
func pad(input []byte, pad_count int) []byte {
out := make([]byte, len(input) + int(pad_count))
copy(out, input)
for i := 0; i < pad_count; i++ {
out[len(input) + i] = byte(pad_count)
}
return out
}
英文:
func pad(input []byte, pad_count int) []byte {
out := make([]byte, len(input) + int(pad_count))
copy(out, input)
for i := 0; i < pad_count; i++ {
out[len(input) + i] = byte(pad_count)
}
return out
}
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