"Integer expression expected error" in bc

Question

I'm getting a "bc: integer expression expected" in this line:

numV=$numVtest

while running the script. How can I solve this error?

firstv=1200         #first velocity value
lastv=5000          #last velocity value
increment=200       #velocity increment
numVtest=100        #use to limit number of velocity panels
                #otherwise, use very large value (100)
#================================================
# Compute number of velocity panels

 numV='bc -l << -END
 ( ( $lastv - $firstv ) / $increment ) + 1
 END'
 if [ $numVtest -lt $numV ] ; then
 numV=$numVtest
 fi

chepner · Accepted Answer · 2017-02-24 13:06:36Z

2

numV is not a number; it's a string that looks like a call to bc, so it cannot be used as an argument to -lt. I suspect you meant to use backquotes, not single quotes, in its definition.

numV=`bc -l << -END
...
`

However, it is better practice to use $(...) for command substitution instead of backquotes.

numV=$(bc -l << -END
( ( $lastv - $firstv ) / $increment ) + 1
END
)
if [ "$numVtest" -lt "$numV" ] ; then
    numV=$numVtest
fi

The next problem is that -lt only works with integers, though, while the result of bc could be a floating-point value. You need to do the comparison inside bc as well. bc will output a 1 if the comparison is true, and a 0 if it is false.

result=$(bc -l << -END
( ( $lastv - $firstv ) / $increment ) + 1 > $numVtest
END
)
if [ "$result" = 1 ] ; then
    numV=$numVtest
fi

edited Feb 24, 2017 at 13:06

answered Feb 24, 2017 at 12:50

chepner

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4 Comments

Aserre Over a year ago

note that the test command will still fail as $numV is a float, and bash can only handle integers. You'll have to cast $numV as an int using either string manipulation (i.e. ${numV%.*}) or printf

chepner Over a year ago

Ugh, you're right. I missed the forest for the trees.

clt60 Over a year ago

... or just use the bc without -l and you will get integers... ;) e.g. bc -l <<<"10/3" vs bc <<<"10/3"

chepner Over a year ago

I'm assuming the OP actually wants floating-point values, or else there is no need for bc. (If integers are fine, I recommend SLePort's answer.)

SLePort · Accepted Answer · 2017-02-24 13:13:39Z

1

If the result is expected to be an integer, you don't need bc. You can use the $(()) syntax instead:

numV=$(( ($lastv - $firstv) / $increment + 1 ))
if [ "$numVtest" -lt "$numV" ] ; then
    numV=$numVtest
fi

edited Feb 24, 2017 at 13:13

answered Feb 24, 2017 at 13:04

SLePort

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1 Comment

chepner Over a year ago

That depends on whether the result of the division is expected to be an integer (which does, however, seem likely, or at least sufficient) or a floating-point value.

Community · Accepted Answer · 2017-05-23 12:00:10Z

1

Your problem is, as pointed out in chepner's answer, that you just assign a string to numV.

You can alternatively use one less operation: if bc evaluates a boolean statement, it prints 0 or 1, and to map that to a Bash conditional, we can wrap the bc command substitution into (( )):

if (( $(bc -l <<< "$numVtest < ($lastv - $firstv) / $increment + 1") )); then
    numV=$numVtest
fi

edited May 23, 2017 at 12:00

CommunityBot

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answered Feb 24, 2017 at 14:30

Benjamin W.

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