从stdin读取时丢失换行符

Question

我一直在为此而努力。 我想在bash中按字符循环多行字符串，但是丢失了所有换行符。当我没有发现任何明显错误时，我做的第一件事就是对它运行shellcheck，这对程序来说似乎很好。

script.sh ：

#!/usr/bin/env bash

transform_single() {
        if [[ $# -ne 1 ]]; then
                echo 'Error: illegal number of args' 1>&2
        fi

        equation=''
        delim0=0
        delim="$1"

        while IFS= read -rn1 c; do
                if [[ $delim0 -eq 0 ]] && [[ "$c" == "$delim" ]]; then
                        delim0=1
                        equation=''
                elif [[ $delim0 -ne 0 ]] && [[ "$c" == "$delim" ]]; then
                        delim0=0
                        echo -n "$equation" | texmath
                elif [[ $delim0 -ne 0 ]]; then
                        equation="$equation$c"
                else
                        echo -n "$c"
                fi
        done
}

transform_single '$'

input.txt中：

<newlines>
<newlines>
# Hello world!
<newlines>
This is a test string.
<newlines>

调用：

bash script.sh < input.txt

输出：

# Hello world!This is a test string.

例外输出：

与输入文件相同。

Answer 1

工作脚本

#!/bin/bash

transform_single() {
    if (($# != 1)); then echo 'Error: illegal number of args' 1>&2; fi
    equation=''
    delim0=0
    delim="$1"
    while IFS= read -r -d $'\0' -n 1 c; do
        if ((delim0 == 0)) && [[ "$c" == "$delim" ]]; then
            delim0=1
            equation=''
        elif ((delim0 != 0)) && [[ "$c" == "$delim" ]]; then
            delim0=0
            echo -n "$equation" | texmath
        elif ((delim0 != 0)); then
            equation="$equation$c"
        else
            echo -n "$c"
        fi
    done
}

transform_single '$'

问题是您必须设置分隔符以读取空字符，以保留换行符。