Convert string array to byte array in go

huangapple go评论151阅读模式
英文:

Convert string array to byte array in go

问题

给定一个只包含单个字符的字符串数组,例如:

ex := [...]string{"a","o",".",".","2",".",".","9"}

是否有一种方法可以获得一个字节数组,其内容与字符串数组相同,但是使用字节而不是字符串?

英文:

Given an string array which only contains single characters such as:

ex := [...]string{"a","o",".",".","2",".",".","9"}

is there a way to get a byte array with same content but with bytes instead of strings?

答案1

得分: 3

使用转换将每个string转换为[]byte

ex := [...]string{"a", "o", ".", ".", "2", ".", ".", "9"}
var ey [len(ex)][]byte
for i := range ex {
    ey[i] = []byte(ex[i])
}

如果你的目的是获取连接字符串的字节数组,可以使用这段代码。这段代码仅在字符串为单个ASCII字符时有效。

ex := [...]string{"a", "o", ".", ".", "2", ".", ".", "9"}
var ey [len(ex)]byte
for i := range ex {
    ey[i] = ex[i][0]
}

如果你想要获取连接字符串的字节切片,可以使用以下表达式:[]byte(strings.Join(ex[:], ""))

我不知道你这样做的上下文,但我猜想使用切片数组更合适:

ex := []string{"a", "o", ".", ".", "2", ".", ".", "9"}
ey := make([][]byte, len(ex))
for i := range ex {
    ey[i] = []byte(ex[i])
}

..

s := []byte(strings.Join(ex, ""))
英文:

Use a conversion to convert each string to a []byte.

ex := [...]string{"a", "o", ".", ".", "2", ".", ".", "9"}
var ey [len(ex)][]byte
for i := range ex {
	ey[i] = []byte(ex[i])
}

Use this code if your intent is to get a byte array of the joined strings. This code only works when the strings are single ASCII characters.

ex := [...]string{"a", "o", ".", ".", "2", ".", ".", "9"}
var ey [len(ex)]byte
for i := range ex {
	ey[i] = ex[i][0]
}

Use this expression of you want to get a slice of bytes of the joined strings: []byte(strings.Join(ex[:], ""))

I don't know the your context for doing this, but my guess is that it's more appropriate to use a slice than an array:

ex := []string{"a", "o", ".", ".", "2", ".", ".", "9"}
ey := make([][]byte, len(ex))
for i := range ex {
	ey[i] = []byte(ex[i])
}

..

s := []byte(strings.Join(ex, ""))

答案2

得分: 0

这似乎可以实现:

package main

import (
   "fmt"
   "strings"
)

func main() {
   a := [...]string{"a","o",".",".","2",".",".","9"}
   var b [len(a)]byte
   c := strings.Join(a[:], "")
   copy(b[:], c)
   // [8]uint8{0x61, 0x6f, 0x2e, 0x2e, 0x32, 0x2e, 0x2e, 0x39}
   fmt.Printf("%#v\n", b)
}

https://godocs.io/strings#Join

英文:

This seems to do it:

package main

import (
   "fmt"
   "strings"
)

func main() {
   a := [...]string{"a","o",".",".","2",".",".","9"}
   var b [len(a)]byte
   c := strings.Join(a[:], "")
   copy(b[:], c)
   // [8]uint8{0x61, 0x6f, 0x2e, 0x2e, 0x32, 0x2e, 0x2e, 0x39}
   fmt.Printf("%#v\n", b)
}

https://godocs.io/strings#Join

答案3

得分: 0

根据这是否是代码生成流程的一部分,你可以有几种方法来实现。

直接方式:

bs := [...]byte{'a', 'o', '.', '.', '2', '.', '.', '9'}

或者间接方式:

ex := [...]string{"a", "o", ".", ".", "2", ".", ".", "9"}

bs := [...]byte{
    ex[0][0],
    ex[1][0],
    ex[2][0],
    ex[3][0],
    ex[4][0],
    ex[5][0],
    ex[6][0],
    ex[7][0],
} // 类型为 [8]int8,即 [8]byte

根据你的使用情况,这些方法可能过于死板。对于动态初始化方法,请参考这里的其他答案。

https://play.golang.org/p/iMEjFpCKAaW

英文:

Depending on if this is part of a code-generation pipeline, you can do this a couple of ways.

Directly:

bs := [...]byte{'a', 'o', '.', '.', '2', '.', '.', '9'}

or indirectly:

ex := [...]string{"a", "o", ".", ".", "2", ".", ".", "9"}

bs := [...]byte{
	ex[0][0],
	ex[1][0],
	ex[2][0],
	ex[3][0],
	ex[4][0],
	ex[5][0],
	ex[6][0],
	ex[7][0],
} // type [8]int8  i.e. [8]byte

https://play.golang.org/p/iMEjFpCKAaW

Depending on your use case, these ways may be too rigid. For dynamic initialization methods see the other answers here.

huangapple
  • 本文由 发表于 2021年10月28日 06:39:55
  • 转载请务必保留本文链接:https://go.coder-hub.com/69746257.html
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