问题描述:

I am a new bash learner. I want to print the result of an expression given as input having 3 digits after 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)/7 as input for the following code, the given output is 17.928. Actual result is 17.92857.... So, it is truncating instead of rounding. I want to round it, that means the output should be 17.929. My code:

read a

echo "scale = 3; $a" | bc -l

Equivalent C++ code can be(in main function):

float a = 5+50*3.0/20.0 + (19*2.0)/7.0;

cout<<setprecision(3)<<fixed<<a<<endl;

网友答案:

What about

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

a         = 17.92857142857142857142
a_rounded = 17.929

?

网友答案:

You can use awk:

awk 'BEGIN{printf "%.3f\n", (5+50*3/20 + (19*2)/7)}'
17.929

%.3f output format will round up the number to 3 decimal points.

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