Syntax Error: Operand Expected When Using Bash

Syntax error: operand expected (error token is +)

Combination of ceving and Tomek's:

#!/bin/bash
read num1 num2 num3
while [ $num1 -lt $num3 ]
do
    echo $num1
    num1=$((num1+num2))
done

syntax error: operand expected (error token is + ) in bash

The actual error reason is because of uninitialized variable sum going through the first iteration of the loop. Initialize the variable before entering the loop.

Also a major logical flaw is that you are not even iterating over the input arguments, but just over the counter i which will produce incorrect results if you pass arguments other than 1 2 from the command-line.

You need to pass over the actual arguments argc and argv (arg count and arg vector: for understanding purposes only) and you don't need bc at all

argc=$#
argv=("$@")

sum=0
for ((i=0; i<${argc}; i++)); do
    sum=$((${argv[i]} + $sum))
done

Bash script syntax error: operand expected (error token is =)

bash variable assignment can not have whitespace around =. Drop the spaces around =:

let br=${br}/10

Do the same for all such cases.

Syntax error: operand expected ... in script that was working right

The easiest way to write those assignment is to providean arithmetic context:

((SIZEKB=(SIZE+1023)/1024))
((REQUIREDKB=SIZEKB+MINFREEKB))

You can use also

SIZEKB=$[(SIZE+1023)/1024]
REQUIREDKB=$[SIZEKB+MINFREEKB]

but this is deprecated and provided for compatibility with old scripts (perhaps those written for bash version 1?).

(Linux) syntax error: operand expected (error token is **)

Use $(( )) to evaluate arithmetic expressions in bash.

volume=$(($1 * $1 * $1))

You can also use the let command:

let volume=$1*$1*$1

or you can use (( )) to execute an arithmetic statement

((volume = $1 * $1 * $1))

What you wrote was being treated as a filename wildcard.

You also need to provide an argument to the script, this will be used as $1.

$ bash testarith.sh 5
The volume of the cube is 125

Bash script error for syntax operand expected =

tl;dr:

Double-quote your command substitution:

let initialLines="$(grep '' "$1" | wc -l || true)"

That said, as explained in chepner's helpful answer, you could just use a simple assignment without needing to double-quote the RHS, although to get the same outcome as with let here, be sure to declare the variable with -i first; e.g., declare -i initialLines

It looks like the RHS of your assignment - the command substitution ($(...)) - evaluates either to the empty string or a whitespace-prefixed string in the course of the shell expansions that happen before the command is executed, so that bash ends up trying to parse one of the following commands:

let initialLines=

let initialLines=   2

If you try these in isolation, you'll get the syntax error you've experienced.

The most likely explanation is that you're using BSD wc, such as on macOS, which outputs the line count requested with -l with leading whitespace.

A simple fix is to double-quote the command substitution:

let initialLines="$(grep '' "$1" | wc -l || true)"

As for the command in your command substitution (it may just be an example, but it's worth commenting on):

grep '' will always return all input lines, so it's nothing more than a less efficient cat.
There is no point in using || true:
- The exit code of the subshell in which the command substitution runs is not relevant, even with set -e in effect.
- Also, by using true as the alternative command, you'll break the let command as well, because builtin true has no stdout output.
In summary, your entire command substitution could be reduced to wc -l < "$1".