How do I get bc(1) to print the leading zero?

Question

I do something like the following in a Makefile:

echo "0.1 + 0.1" | bc

(in the real file the numbers are dynamic, of course)

It prints .2 but I want it to print 0.2.

I would like to do this without resorting to sed but I can't seem to find how to get bc to print the zero. Or is bc just not able to do this?

Answer 1

this only uses bc, and works with negative numbers:

bc <<< "x=-.1; if(x==0) print \"0.0\" else if(x>0 && x<1) print 0,x else if(x>-1 && x<0) print \"-0\",-x else print x";

try it with:

for y in "0" "0.1" "-0.1" "1.1" "-1.1"; do
  bc <<< "x=$y; if(x==0) print \"0.0\" else if(x>0 && x<1) print 0,x else if(x>-1 && x<0) print \"-0\",-x else print x";
  echo;
done

Answer 2

I cannot find anything about output format in the documentation. Instead of sed, you can also reach for printf:

printf '%3.1f\n' $(bc<<<0.1+0.1)

Answer 3

For positive numbers, it may be as simple as printing (an string) zero:

$ echo '"0";0.1+0.1' | bc
0.2

avoid the zero if the number is bigger (or equal) to 1:

$ echo 'x=0.1+0.1;  if(x<1){"0"};  x' | bc
0.2

It gets a bit more complex if the number may be negative:

echo 'x= 0.3 - 0.5 ; s=1;if(x<0){s=-1};x*=s;if(s<0){"-"};if(x<1) {"0"};x' | bc
-0.2

You may define a function and add it to a library:

$ echo 'define leadzero(x){auto s;
        s=1;if(x<0){s=-1};x*=s;if(s<0){"-"};if(x<1){"0"};
        return(x)};
        leadzero(2.1-12.4)' | bc
-10.3

$ echo 'define leadzero(x){auto s;
        s=1;if(x<0){s=-1};x*=s;if(s<0){"-"};if(x<1){"0"};
        return(x)};
        leadzero(0.1-0.4)' | bc
-0.3

Answer 4

This one will also handle negative numbers:

echo "0.1 - 0.3" | bc | sed -r 's/^(-?)\./\10./'

Answer 5

This might work for you:

echo "x=0.1 + 0.1; if(x<1) print 0; x" | bc

Answer 6

You can also resort to awk to format:

 echo "0.1 + 0.1" | bc | awk '{printf "%f", $0}'

or with awk itself doing the math:

 echo "0.1 0.1" | awk '{printf "%f", $1 + $2}'