bash: bc command's evaluation of *

Question

I am using the bc command in UNIX to parse some math expressions. I have two small functions:

function bashcalc {
      echo $@ | bc -l
}

function2 {
      ...
      catMinusMouse=$(bashcalc "$cat_angle - $mouse_angle")
      cos=$(cosine $catMinusMouse)
      val=$(bashcalc "$cat_radius * $cos")   ##PARSE ERROR
      ...
}

When I tried to run the expression following val, I got quite a few "(standard_in) 1: parse error"s.

My first thought was that the asterisk was the issue so I tried escaping it. That however gave me an illegal character error.

The solution ended up being removing all of the whitespace

   val=$(bashcalc "$cat_radius*$cos")

QUESTION: Why did calculating catMinusMouse (with spaces around the subtraction operator) work while the same format with multiplication did not work?

Try running `echo 2 * 3` in a non-empty directory and see if you can figure it out. — Mat, May 01 '13 at 09:54
That's interesting. However, could I make this work with the use of escaping? If I do `$(bashcalc $cat_radius \* $cos)`, wouldn't I be effectively calling `echo 2 \* 2`, which properly escapes the * and doesn't list the directory? — bobbyjoe93, May 01 '13 at 10:21
Problem isn't at the call site, it's in the expansion of `$@` in the function. jm666's answer covers it. — Mat, May 01 '13 at 10:25

score 3 · Answer 1 · answered May 01 '13 at 10:25

3

you need escape the * or enclose it into "quotes"

3 variants:

#!/bin/bash

function bashcalc {
    echo "$@" | bc -l
}

function2() {
    cat_radius=0.9
    catMinusMouse=0.4

    val=$(bashcalc "$cat_radius" "*" "c($catMinusMouse)")
    echo $val

    #or
    val=$(bashcalc "$cat_radius * c($catMinusMouse)")
    echo $val

    #or
    val=$(bc -l <<EOF
$cat_radius * c($catMinusMouse)
EOF
)
    echo $val
}

function2

answered May 01 '13 at 10:25

clt60

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1

You should properly double-quote all variable expansions here, but the beef is the use of `"$@"` pro bare `$@`. – tripleee May 01 '13 at 10:44

score 2 · Answer 2 · answered May 01 '13 at 10:51

2

The real problem here is that you have not quoted $@ in your bashcalc function.

Change it to:

function bashcalc {
      echo "$@" | bc -l
}

Even better, don't use echo. Change it to:

bashcalc() {
    bc -l <<< "$@"
}

answered May 01 '13 at 10:51

dogbane

266,786
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score 1 · Answer 3 · answered May 01 '13 at 09:54

1

Try with the following:

val=$(echo "$cat_radius * $cos"| bc)

that is, pipe to bc (bashcalc) and it will make the calculation.

answered May 01 '13 at 09:54

fedorqui

275,237
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That's new. The program `bc` predates Bash by a large margin, so whatever the name stands for, it's certainly not "Bash Calculator". [Wikipedia](http://en.wikipedia.org/wiki/Bc_programming_language) says it's "basic calculator". – tripleee May 01 '13 at 10:42

bash: bc command's evaluation of *

3 Answers3