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.

如何在bash中进行浮点计算:

不同于在bc命令中使用"here strings"(<<<)，这是我最喜欢的bc浮点示例，来自bc手册页的示例部分(参见man bc手册页)。

在我们开始之前，知道pi的方程是:pi = 4*atan(1)。下面的A()是atan()的BC数学函数。

This is how to store the result of a floating point calculation into a bash variable--in this case into a variable called pi. Note that scale=10 sets the number of decimal digits of precision to 10 in this case. Any decimal digits after this place are truncated. pi=$(echo "scale=10; 4*a(1)" | bc -l) Now, to have a single line of code that also prints out the value of this variable, simply add the echo command to the end as a follow-up command, as follows. Note the truncation at 10 decimal places, as commanded: pi=$(echo "scale=10; 4*a(1)" | bc -l); echo $pi 3.1415926532 Finally, let's throw in some rounding. Here we will use the printf function to round to 4 decimal places. Note that the 3.14159... rounds now to 3.1416. Since we are rounding, we no longer need to use scale=10 to truncate to 10 decimal places, so we'll just remove that part. Here's the end solution: pi=$(printf %.4f $(echo "4*a(1)" | bc -l)); echo $pi 3.1416

下面是上述技术的另一个非常棒的应用程序和演示:测量和打印运行时。

(参见我的另一个答案)。

注意，dt_min从0.01666666666…0.017:

start=$SECONDS; sleep 1; end=$SECONDS; dt_sec=$(( end - start )); dt_min=$(printf %.3f $(echo "$dt_sec/60" | bc -l)); echo "dt_sec = $dt_sec; dt_min = $dt_min"
dt_sec = 1; dt_min = 0.017

相关:

(我的回答)https://unix.stackexchange.com/questions/52313/how-to-get-execution-time-of-a-script-effectively/547849#547849 [我的问题]三个左尖括号(' <<< ')在bash中是什么意思? https://unix.stackexchange.com/questions/80362/what-does-mean/80368#80368 https://askubuntu.com/questions/179898/how-to-round-decimals-using-bc-in-bash/574474#574474

我知道很老了，但太诱人了。所以，答案是:你不能……但你可以。让我们试试这个:

$IMG_WIDTH=1024
$IMG2_WIDTH=2048

$RATIO="$(( IMG_WIDTH / $IMG2_WIDTH )).$(( (IMG_WIDTH * 100 / IMG2_WIDTH) % 100 ))"

这样，在纯bash(不需要启动其他进程)中，在该点之后得到2位截断的数字(称为舍入到下位)。当然，如果你只需要这一点后面的一位数，你可以乘以10，然后对10取模。

它的作用:

First $((…))执行整数除法; 第二个$((…))对一个大100倍的数进行整数除法，本质上是将2位数字移到点的左边，然后(%)通过做模只得到这2位数字。

奖励轨道:bc版本× 1000在我的笔记本电脑上花了1.8秒，而纯bash版本花了0.016秒。

你不能。Bash只处理整数;您必须委托给bc之类的工具。

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.

** bash/shell中的注入安全浮点数学

注意:这个回答的重点是为在bash(或其他shell)中执行数学的注入安全解决方案提供思路。当然，同样也可以使用，只需稍作调整即可执行高级字符串处理等。

所提出的大多数解决方案都是使用外部数据(变量、文件、命令行、环境变量)动态构造小脚本。外部输入可用于向引擎中注入恶意代码，其中很多是恶意代码

下面是使用各种语言进行基本数学计算的比较，其中结果为浮点数。它计算A + B * 0.1(作为浮点数)。

所有的解决方案都试图避免创建难以维护的动态脚本，而是使用静态程序，并将参数传递给指定的变量。它们将安全地处理具有特殊字符的参数-减少代码注入的可能性。例外是'BC'，它不提供输入/输出功能

例外是'bc'，它不提供任何输入/输出，所有数据都来自stdin中的程序，所有输出都到stdout。所有计算都在沙箱中执行，不允许出现副作用(打开文件等)。理论上，注射是安全的设计!

A=5.2
B=4.3

# Awk: Map variable into awk
# Exit 0 (or just exit) for success, non-zero for error.
#
awk -v A="$A" -v B="$B" 'BEGIN { print A + B * 0.1 ; exit 0}'

# Perl
perl -e '($A,$B) = @ARGV ; print $A + $B * 0.1' "$A" "$B"

# Python 2
python -c 'import sys ; a = float(sys.argv[1]) ; b = float(sys.argv[2]) ; print a+b*0.1' "$A" "$B"

# Python 3
python3 -c 'import sys ; a = float(sys.argv[1]) ; b = float(sys.argv[2]) ; print(a+b*0.1)' "$A" "$B"

# BC
bc <<< "scale=1 ; $A + $B * 0.1"