Bash脚本。通过所有ASCII字符迭代 [英] Bash script. Iterate through all ASCII chars

问题描述

我知道如何通过人物迭代

I know how to iterate through characters

for c in {a..z}

但我无法弄清楚，如何通过所有ASCII字符迭代。
我敢肯定，它应该是很容易的。

But I can't figure out, how to iterate through all ASCII chars. I'm sure, it should be really easy.

推荐答案

你可以做的就是迭代从0到127，然后将十进制值转换为字符的ASCII值（或后）。

What you can do is to iterate from 0 to 127 and then convert the decimal value to its ASCII value(or back).

您可以使用这些功能来做到这一点：

You can use these functions to do it:

# POSIX
# chr() - converts decimal value to its ASCII character representation
# ord() - converts ASCII character to its decimal value

chr() {
  [ ${1} -lt 256 ] || return 1
  printf \\$(printf '%03o' $1)
}

# Another version doing the octal conversion with arithmetic
# faster as it avoids a subshell
chr () {
  [ ${1} -lt 256 ] || return 1
  printf \\$(($1/64*100+$1%64/8*10+$1%8))
}

# Another version using a temporary variable to avoid subshell.
# This one requires bash 3.1.
chr() {
  local tmp
  [ ${1} -lt 256 ] || return 1
  printf -v tmp '%03o' "$1"
  printf \\"$tmp"
}

ord() {
  LC_CTYPE=C printf '%d' "'$1"
}

# hex() - converts ASCII character to a hexadecimal value
# unhex() - converts a hexadecimal value to an ASCII character

hex() {
   LC_CTYPE=C printf '%x' "'$1"
}

unhex() {
   printf \\x"$1"
}

# examples:

chr $(ord A)    # -> A
ord $(chr 65)   # -> 65

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