bash 如何在bash中将浮点数四舍五入到小数点后3位
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How to round a floating point number upto 3 digits after decimal point in bash
提问by Enamul Hassan
I am a new bashlearner. I want to print the result of an expression given as input having 3 digitsafter decimal point with rounding if needed.
I can use the following code, but it does not round. Say if I give 5+50*3/20 + (19*2)/7as input for the following code, the given output is 17.928. Actual result is 17.92857.... So, it is truncatinginstead of rounding. I want to round it, that means the output should be17.929. My code:
我是一个新的bash学习者。如果需要，我想打印作为输入给出的表达式的结果，3 digits小数点后带有四舍五入。我可以使用以下代码，但它不会四舍五入。假设如果我5+50*3/20 + (19*2)/7将以下代码作为输入，则给定的输出是17.928. 实际结果是17.92857...。所以，它是truncating而不是四舍五入。我想把它四舍五入，这意味着输出应该是17.929. 我的代码：
read a
echo "scale = 3; $a" | bc -l
Equivalent C++code can be(in mainfunction):
等效C++代码可以是（在main函数中）：
float a = 5+50*3.0/20.0 + (19*2.0)/7.0;
cout<<setprecision(3)<<fixed<<a<<endl;
回答by nils
What about
关于什么
a=`echo "5+50*3/20 + (19*2)/7" | bc -l`
a_rounded=`printf "%.3f" $a`
echo "a = $a"
echo "a_rounded = $a_rounded"
which outputs
哪个输出
a = 17.92857142857142857142
a_rounded = 17.929
?
?
回答by anubhava
You can use awk:
您可以使用 awk：
awk 'BEGIN{printf "%.3f\n", (5+50*3/20 + (19*2)/7)}'
17.929
%.3foutput format will round up the number to 3 decimal points.
%.3f输出格式会将数字四舍五入到 3 个小数点。
