英文:
Binary string to unicode
问题
我不确定为什么我的二进制字符串转Unicode的函数不起作用...有人能指出问题或帮我修复吗？我将二进制字符串分块的原因是它太大了，ParseInt无法处理。请参考下面的playground链接查看示例。
func binToString(s []byte) string {
var counter int
chunk := make([]byte, 7)
var buf bytes.Buffer
for i := range s {
if i%8 == 0 {
counter = 0
if i, err := strconv.ParseInt(string(chunk), 2, 64); err == nil {
buf.WriteString(string(i))
}
} else {
chunk[counter] = s[i] //我知道我可以在这里使用取模运算，但我正在测试，使用计数器更容易跟踪和测试
counter++
}
}
return buf.String()
}
转换时，它似乎要么漏掉一个字符，要么添加一个字符（或两个字符）。
这是一个playground链接，展示了函数不按预期工作的示例。
英文:
I'm not 100% sure why my binary string to unicode isn't working..can anyone point out the issue or help me patch it? Also the reason why i chunk out the binary is that it is too large for ParseInt to handle. See the playground link below for an example.
func binToString(s []byte) string {
var counter int
chunk := make([]byte, 7)
var buf bytes.Buffer
for i := range s {
if i%8 == 0 {
counter = 0
if i, err := strconv.ParseInt(string(chunk), 2, 64); err == nil {
buf.WriteString(string(i))
}
} else {
chunk0+网站访问量 = s[i] //i know i can use modulus here too but i was testing and an counter was easier to track and test for me
counter++
}
}
return buf.String()
}
It either seems to miss a character or add an character (or two) on conversion.
Here is a playground link showing an example of the function not working as expected.
答案1
得分: 1
你的函数可以以更简单、更高效的方式实现：
func binToString(s []byte) string {
output := make([]byte, len(s)/8)
for i := 0; i < len(output); i++ {
val, err := strconv.ParseInt(string(s[i*8:(i+1)*8]), 2, 64)
if err == nil {
output[i] = byte(val)
}
}
return string(output)
}
链接：https://play.golang.org/p/Fmo7I-rN3c
英文:
Your function could be implemented in a simpler, more efficient manner:
func binToString(s []byte) string {
output := make([]byte, len(s)/8)
for i := 0; i < len(output); i++ {
val, err := strconv.ParseInt(string(s[i*8:(i+1)*8]), 2, 64)
if err == nil {
output[i] = byte(val)
}
}
return string(output)
}
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