linux - 我如何将这个 echo 语句传递到我的 bash 函数中？

Question

bitToSixtyFour () {

    echo "$((2#$1))"
}
while true
do
    if (( i == ${#newString} || i >= ${#newString} ))
    then
            break
    fi

    echo ${newString:i:i+6} | bitToSixtyFour

    i+=6
done

所以在我的 while 循环中，我试图找出如何将 echo 语句作为参数传递到我的函数 bitToSixtyFour 中。它说找不到 bitToSixtyFour。

Answer 1

根据定义，该函数从命令行获取参数 $1:

bitToSixtyFour arg1

从标准输入读取 read

bitToSixtyFour() {
    read varname
    echo $((2#$varname))
}

echo 1010 | bitToSixtyFour

使用变量来避免创建子 shell 和管道会更有效

bitToSixtyFour() {
    bitToSixtyFour=$((2#$1))
}

newString=1100101010011110101110110001100110
for ((i=0;i<${#newString};i+=6)); do
    substr=${newString:i:6}
    bitToSixtyFour "$substr"
    printf "%3s %s\n" "$bitToSixtyFour" "$substr"
done

和问题定义

bitToSixtyFour() {
    echo "$((2#$1))"
}

newString=1100101010011110101110110001100110
for ((i=0;i<${#newString};i+=6)); do
    substr=${newString:i:6}
    res=$(bitToSixtyFour "$substr")
    printf "%3s %s\n" "$res" "$substr"
done