我试图在Bash脚本中划分两个图像宽度,但Bash给我0作为结果:

RESULT=$(($IMG_WIDTH/$IMG2_WIDTH))

我确实研究了Bash指南,我知道我应该使用bc,在互联网上的所有例子中,他们都使用bc。在echo中,我试图把同样的东西放在我的SCALE中,但它不起作用。

以下是我在教程中找到的例子:

echo "scale=2; ${userinput}" | bc 

我怎么能让巴斯给我0.5这样的浮点数呢?


当前回答

稍微改进一下马文的回答:

RESULT=$(awk "BEGIN {printf \"%.2f\",${IMG_WIDTH}/${IMG2_WIDTH}}")

BC并不总是安装包。

其他回答

让我们举个例子来理解,如果你想找到n个数组元素的平均值(当然平均值将是浮点数/小数)

declare -a arr
echo "How many numbers you want to enter?"
read n
echo "Enter the Array Elements"
for(( i=0 ; i<$n ; i++))
do
      read array_elements
      arr[$i]="$array_elements"
done
sum=0
for i in "${arr[@]}"
do
     #sum and avg
     sum=$(($sum + $i))
     #average will come in decimals
     avg=`echo $sum / $n | bc -l`
done
# Output results:
printf "Average of Array Elements %.2f:" $avg

因此,我们将使用"| bc -l"来进行浮动计算

如果你找到了你喜欢的变体,你也可以把它包装到一个函数中。

这里我将一些bashism包装到div函数中:

一个衬套:

function div { local _d=${3:-2}; local _n=0000000000; _n=${_n:0:$_d}; local _r=$(($1$_n/$2)); _r=${_r:0:-$_d}.${_r: -$_d}; echo $_r;}

或多行:

function div {
  local _d=${3:-2}
  local _n=0000000000
  _n=${_n:0:$_d}
  local _r=$(($1$_n/$2))
  _r=${_r:0:-$_d}.${_r: -$_d}
  echo $_r
}

现在你得到了这个函数

div <dividend> <divisor> [<precision=2>]

然后像这样使用它

> div 1 2
.50

> div 273 123 5
2.21951

> x=$(div 22 7)
> echo $x
3.14

更新 我添加了一个小脚本,为您提供了bash的基本浮点数操作:

用法:

> add 1.2 3.45
4.65
> sub 1000 .007
999.993
> mul 1.1 7.07
7.7770
> div 10 3
3.
> div 10 3.000
3.333

这里是脚本:

#!/bin/bash
__op() {
        local z=00000000000000000000000000000000
        local a1=${1%.*}
        local x1=${1//./}
        local n1=$((${#x1}-${#a1}))
        local a2=${2%.*}
        local x2=${2//./}
        local n2=$((${#x2}-${#a2}))
        local n=$n1
        if (($n1 < $n2)); then
                local n=$n2
                x1=$x1${z:0:$(($n2-$n1))}
        fi
        if (($n1 > $n2)); then
                x2=$x2${z:0:$(($n1-$n2))}
        fi
        if [ "$3" == "/" ]; then
                x1=$x1${z:0:$n}
        fi
        local r=$(($x1"$3"$x2))
        local l=$((${#r}-$n))
        if [ "$3" == "*" ]; then
                l=$(($l-$n))
        fi
        echo ${r:0:$l}.${r:$l}
}
add() { __op $1 $2 + ;}
sub() { __op $1 $2 - ;}
mul() { __op $1 $2 "*" ;}
div() { __op $1 $2 / ;}

您可以通过-l选项使用bc (L字母)

RESULT=$(echo "$IMG_WIDTH/$IMG2_WIDTH" | bc -l)

There are scenarios in wich you cannot use bc becouse it might simply not be present, like in some cut down versions of busybox or embedded systems. In any case limiting outer dependencies is always a good thing to do so you can always add zeroes to the number being divided by (numerator), that is the same as multiplying by a power of 10 (you should choose a power of 10 according to the precision you need), that will make the division output an integer number. Once you have that integer treat it as a string and position the decimal point (moving it from right to left) a number of times equal to the power of ten you multiplied the numerator by. This is a simple way of obtaining float results by using only integer numbers.

用calc,这是我发现的最简单的方法 例子:

calc 1 + 1

 2

calc 1/10

 0.1