Syntax error: operand expected when using Bash

Question

I have two arrays that I want to loop in. I construct those properly and before going into for loop, I do echo them to be sure everything is ok with arrays. But when I run the script, it outputs an error:

l<=: syntax error: operand expected (error token is "<="

I consulted the mighty Google and I understood it suffers from the lack of the second variable, but I mentioned earlier I do echo the values and everything seems to be OK. Here is the snippet..

#!/bin/bash
    k=0
    #this loop is just for being sure array is loaded
    while [[ $k -le ${#hitEnd[@]} ]] 
      do
      echo "hitEnd is: ${hitEnd[k]} and hitStart is: ${hitStart[k]}"
      # here outputs the values correct 
      k=$((k+1))
    done
    k=0
    for ((l=${hitStart[k]};l<=${hitEnd[k]};l++)) ; do //this is error line..
        let array[l]++      
        k=$((k+1))
done

The variables in the for loop are echoed correctly but for loop won't work.. where am I wrong?

#

as gniourf_gniourf answered:

"... At some point, k will reach the value ${#hitEnd[@]}, and this is exactly when hitEnd[k] is not defined and expands to an empty string! Bang!"

meaning error output is displayed not at the beginning of the loop, but when k has a greater value than array's indices, pointing an index that array does not include...

gniourf_gniourf · Accepted Answer · 2012-12-02 18:21:09Z

4

That's because at some point ${hitEnd[k]} expands to nothing (it is undefined). I get the same error with ((l<=)). You should write your for loop as:

k=0
for ((l=${hitStart[0]};k<${#hitEnd[@]} && l<=${hitEnd[k]};l++)); do

so as to always have an index k that corresponds to a defined field in the array ${hitEnd[@]}.

Also, instead of

k=$((k+1))

you can just write

((++k))

Done!

Your script revised using better modern bash practice:

#!/bin/bash
k=0
#this loop is just for being sure array is loaded
while ((k<=${#hitEnd[@]})); do
    echo "hitEnd is: ${hitEnd[k]} and hitStart is: ${hitStart[k]}"
    # here outputs the values correct 
    ((++k))
done
k=0
for ((l=hitStart[0];k<${#hitEnd[@]} && l<=hitEnd[k];++l)); do
    ((++array[l]))
    ((++k))
done

Now, I'm not too sure the for loop does exactly what you want it to... Don't you mean this instead?

#!/bin/bash
# define arrays hitStart[@] and hitEnd[@]...
# define array array[@]

#this loop is just for being sure array is loaded
for ((k=0;k<${#hitEnd[@]};++k)); do
    echo "hitEnd is: ${hitEnd[k]} and hitStart is: ${hitStart[k]}"
    # here outputs the values correct 
    ((++k))
done

for ((k=0;k<${#hitEnd[@]};++k)); do
    for ((l=hitStart[k];l<=hitEnd[k];++l)); do
        ((++array[l]))
    done
done

edited Dec 2, 2012 at 18:21

answered Dec 2, 2012 at 18:15

gniourf_gniourf

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5 Comments

teutara Over a year ago

still i do not understand clearly why the situation that i mention in the question does not work, but thank you for your effort..

gniourf_gniourf Over a year ago

I thought you would understand it! watch out how your variable k is always incremented, and remember that the test part in the for loop (i.e., the l<=hitEnd[k] part) is evaluated at each pass!

teutara Over a year ago

of course, i want it that way. i mean at first run (k=0) it will increment array once, and at second run (k=1) it will increment the array again..

gniourf_gniourf Over a year ago

All right, then at some point, k will reach the value ${#hitEnd[@]}, and this is exactly when hitEnd[k] is not defined and expands to an empty string! Bang!

teutara Over a year ago

aaahh, right! i do not know the reason but i was thinking that error is outputted at the first run of the for-loop. but what you say makes sense.. thank you again for your effort..

sampson-chen · Accepted Answer · 2012-12-02 18:15:23Z

1

A bit bandaid-y, but you rewrite your for-loop into a while loop:

l="${hitStart[k]}"
while [[ "$l" -le "${hitEnd[k]}" ]]; do
        let array[l]++      
        k=$((k+1))
        l=$((l+1))
done

answered Dec 2, 2012 at 18:15

sampson-chen

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