[0]-bash: [: 0*1%8: integer expression expected - bash

Can someone point out what is wrong with the output.
for i in {0..127} ; do
echo -n [$i]
if [ $i*$j%8 -eq 0 ]; then
echo "\n"
fi
mytool -c "read 0x1540:0xa0:$i*$j"
done
I am trying to format the output into rows containing 8 items each.
I tried the suggestion below and modified my code to
for i in {0..8} ; for j in {0..16}; do
echo -n [$i*$j]
if [[ $i*$j%8 == 0 ]]; then
echo
fi
mytool -c "read 0x1540:0xa0:$i*$j"
done
Above with for i in {0..8} ; for j in {0..16}
I am expecting this to be a nested for loop.I am not very sure if this is how I do a nested loop in bash.
Still the output is not as I expect it.
My output looks like
[0]0x3
[1]0x4
[2]0x21
[3]0x1
[4]0x0
[5]0x0
[6]0x4
[7]0x41
[8]0x84
[9]0x80
[10]0x0
[11]0x0
[12]0x3
[13]0x0
[14]0x43
[15]0x49
[16]0x53
[17]0x43
[18]0x4f
[19]0x2d
[20]0x49
[21]0x4e
[22]0x43
[23]0x20
[24]0x0
[25]0x0
[26]0x9
[27]0x3a
[28]0x37
[29]0x34
[30]0x39
[31]0x34
I want [0] to [7] in ROW1
[8] to [15] in ROW2
and so on.

Use (( )) if you want to do math.
if ((i * j % 8 == 0)); then
Given your problem description I suggest a bit of a rewrite.
for i in {0..15}; do
for j in {0..7}; do
echo -n "[$((i * 8 + j))]"
mytool -c "read 0x1540:0xa0:$i*$j"
done
echo
done

The test command ([ is an alias for test, not syntax) requires the expression to be built up from multiple arguments. This means spaces are critical to separate operators and operands. Since each part is a separate argument, you also need to quote the * so that the shell does not expand it as a file glob prior to calling test/[.
if [ "$i" "*" "$j" % 8 -eq 0 ]; then
The test command expects 7 separate arguments here: $i, *, $j, %, -eq, and 0, which it then assembles into an expression to evaluate. It will not parse an arbitrary string into an expression.
As noted by John Kugelman, there are easier ways to accomplish such arithmetic in bash.

Related

If-statements in Bash Syntax issues produce right answers, while right code shows wrong answers [duplicate]

This question already has answers here:
Why should there be spaces around '[' and ']' in Bash?
(5 answers)
Closed 5 years ago.
I am fairly new to bash scripting and am struggling with some if-statement syntax.
I have currently written up the following loop:
for (( i = 2; i < $# - 1; i++)); do
if [ $i -ne 0]; then
if [ $i -ne 1]; then
echo "$i was not 1 or 0. Please correct this then try again."
exit 1;
fi
fi
done
This code is supposed to test whether any arguments after the first are either a 1 or a 0.
While the following errors are printed:
./blink.sh: line 36: [: missing `]'
./blink.sh: line 36: [: missing `]'
...the code actually runs fine afterwards (so the errors don't kill the program).
My understanding, however, is that in bash, you put spaces before and after the expression inside the if statement. So this:
if [ $i -ne 0]; then
Becomes:
if [ $i -ne 0 ]; then
However, running this code produces the following:
2 was not 1 or 0. Please correct this then try again.
The main issue I am having with this stems from not understanding how to indirectly reference the positional arguments provided by the execution command. As such, I am confused as to what syntax must be altered to call the objects the arguments point to (in this case, hopefully either a 1 or a 0) rather than the position of the arguments themselves (argument 1, 2, 3...).
Thanks!
EDIT: Altering the question to better fit the advice #randomir provided and clear up what the actual question entails
Based on:
This code is supposed to test whether any arguments after the first are either a 1 or a 0.
I'm assuming you're trying to access positional arguments $2, $3, etc. To make your for loop solution work, you would have to use an indirect reference: ${!i} (see shell parameter expansion). For example, this should work:
#!/bin/bash
for (( i = 2; i <= $#; i++ )); do
if [[ ${!i} -ne 0 ]]; then
if [[ ${!i} -ne 1 ]]; then
echo "$i was not 1 or 0. Please correct this then try again."
exit 1;
fi
fi
done
Note the i running from 2 to number of arguments $#. Also, note the use of recommended and less error-prone [[ .. ]] instead of [ .. ] (otherwise you would have to write [ "${!i}" -ne 0 ], etc).
A simpler solution which avoids the unnecessary indirect referencing looks like this:
#!/bin/bash
while [[ $2 ]]; do
if (( $2 != 0 && $2 != 1 )); then
echo "$2 is neither 0, nor 1"
exit 1
fi
shift
done
We start checking the second argument ($2), use the arithmetic expression (( expr )) testing of value of the second argument, and shift positional arguments to the left by 1 at each iteration (now $3 becomes $2, etc).

Integer expression expected BASH scripting IF statement

I am trying to insert names and numbers in a text file. I have wrote a short script for the learning purpose.
v= expr $# % 2
echo $v
if [ "$v" -eq 0 ]; then
i=1
while [ "$i" -lt $# ]
do
echo "$i $i+1" >> database
i=$((i+1))
done
echo "User(s) successfully added \n\v"
else
echo "Arguments are not complete";
fi
When i enter two arguments, the shell output is as follows
0 # (The value of variable v)
./myscript: line 3: [: : integer expression expected
Arguments are not complete # (else statement is executed)
When i replace -eq to == in line 3 (if statement), error msg is gone but still the IF statement doesn't execute as i expect.
0 # (output of variable v)
Arguments are not complete # (else statement is executed)
You need to enclose the variable assignment in $(...) ("command substitution"):
v=$(expr $# % 2)
In the if statement, -eq should be correct. Also, to make sure it works, I would use double square brackets (this might depend on the shell you use):
if [[ ${v} -eq 0 ]]; then
The immediate problem is the failure to use command substitution to capture the result of the expr command: v=$( expr $# % 2 ). However,
expr is no longer needed for arithmetic; use an arithmetic expression just as you did to increment i.
v=$(( $# % 2 ))

I am getting expr syntax errors in bash shell for a simple program

#!/bin/bash
clear
echo "Enter a number"
read a
s = 0
while [ $a -gt 0 ]
do
r = ` expr $a % 10 `
s = ` expr $s + $r `
a = ` expr $a / 10 `
done
echo "sum of digits is = $s"
This is my code guys .
I am getting a bunch of expr syntax errors.
I am using the bash shell.
Thanks!
Your error is caused by the spaces surrounding the = in the assignments, the following replacements should work (I prefer $() to using backticks since they're much easier to nest):
s=0
r=$(expr $a % 10)
s=$(expr $s + $r)
a=$(expr $a / 10)
For example, s = 0 (with the spaces) does not set the variable s to zero, rather it tries to run the command s with the two arguments, = and 0.
However, it's not really necessary to call the external expr1 to do mathematical manipulation and capture the output to a variable. That's because bash itself can do this well enough without resorting to output capture (see ARITHMETIC EVALUATION in the bash man page):
#!/bin/bash
clear
read -p "Enter a number: " number
((sum = 0))
while [[ $number -gt 0 ]]; do
((sum += number % 10))
((number /= 10))
done
echo "Sum of digits is $sum"
You'll notice I've made some other minor changes which I believe enhances the readability, but you could revert back to the your original code if you wish and just use the ((expression)) method rather than expr.
1 If you don't mind calling external executables, there's no need for a loop in bash, you could instead use sneakier methods:
#!/bin/bash
clear
read -p "Enter a number: " number
echo "Sum of digits is $(grep -o . <<<$number | paste -sd+ | bc)"
But, to be brutally honest, I think I prefer the readable solution :-)

Absolute value of a number

I want to take the absolute of a number by the following code in bash:
#!/bin/bash
echo "Enter the first file name: "
read first
echo "Enter the second file name: "
read second
s1=$(stat --format=%s "$first")
s2=$(stat -c '%s' "$second")
res= expr $s2 - $s1
if [ "$res" -lt 0 ]
then
res=$res \* -1
fi
echo $res
Now the problem I am facing is in the if statement, no matter what I changes it always goes in the if, I tried to put [[ ]] around the statement but nothing.
Here is the error:
./p6.sh: line 13: [: : integer expression expected
You might just take ${var#-}.
${var#Pattern} Remove from $var the shortest part of $Pattern that matches the front end of $var. tdlp
Example:
s2=5; s1=4
s3=$((s1-s2))
echo $s3
-1
echo ${s3#-}
1
$ s2=5 s1=4
$ echo $s2 $s1
5 4
$ res= expr $s2 - $s1
1
$ echo $res
What's actually happening on the fourth line is that res is being set to nothing and exported for the expr command. Thus, when you run [ "$res" -lt 0 ] res is expanding to nothing and you see the error.
You could just use an arithmetic expression:
$ (( res=s2-s1 ))
$ echo $res
1
Arithmetic context guarantees the result will be an integer, so even if all your terms are undefined to begin with, you will get an integer result (namely zero).
$ (( res = whoknows - whocares )); echo $res
0
Alternatively, you can tell the shell that res is an integer by declaring it as such:
$ declare -i res
$ res=s2-s1
The interesting thing here is that the right hand side of an assignment is treated in arithmetic context, so you don't need the $ for the expansions.
I know this thread is WAY old at this point, but I wanted to share a function I wrote that could help with this:
abs() {
[[ $[ $# ] -lt 0 ]] && echo "$[ ($#) * -1 ]" || echo "$[ $# ]"
}
This will take any mathematical/numeric expression as an argument and return the absolute value. For instance: abs -4 => 4 or abs 5-8 => 3
A workaround: try to eliminate the minus sign.
with sed
x=-12
x=$( sed "s/-//" <<< $x )
echo $x
12
Checking the first character with parameter expansion
x=-12
[[ ${x:0:1} = '-' ]] && x=${x:1} || :
echo $x
12
This syntax is a ternary opeartor. The colon ':' is the do-nothing instruction.
or substitute the '-' sign with nothing (again parameter expansion)
x=-12
echo ${x/-/}
12
Personally, scripting bash appears easier to me when I think string-first.
I translated this solution to bash. I like it more than the accepted string manipulation method or other conditionals because it keeps the abs() process inside the mathematical section
abs_x=$(( x * ((x>0) - (x<0)) ))
x=-3
abs_x= -3 * (0-1) = 3
x=4
abs_x= 4 * (1-0) = 4
For the purist, assuming bash and a relatively recent one (I tested on 4.2 and 5.1):
abs() {
declare -i _value
_value=$1
(( _value < 0 )) && _value=$(( _value * -1 ))
printf "%d\n" $_value
}
If you don't care about the math and only the result matters, you may use
echo $res | awk -F- '{print $NF}'
The simplest solution:
res="${res/#-}"
Deletes only one / occurrence if - is at the first # character.

solaris simple bash script

I'm trying to execute this simple script in solaris.
I want to sort(numeric) the filenames of the files in source directory and copy the file one by one to another directory. And, I want to print a message after copying every 100 files.
#!/bin/bash
count=0
for i in `ls | sort -n`
do
cp $i ../target
count = $((count+1))
if[ $count%100 -eq 0 ]
then
echo $count files copied
sleep 1
fi
done
this is not working. I tried different things after searching in net.
I get errors like these -
syntax error at line 8: '(' unexpected.
syntax error at line 10: 'then' unexpected.
syntax error at line 13: 'fi' unexpected etc.
What is the problem with this script?
bash version - GNU bash, version 3.00.16(1)-release (sparc-sun-solaris2.10)
The basic problem with the script is spacing. You have spaces where you shouldn't have them:
(wrong) count = $((count+1))
(right) count=$((count+1))
(better) ((count++))
and you're missing spaces where you need them:
(wrong) if[ $count%100 -eq 0 ]
(right) if [ $((count % 100)) -eq 0 ]
(better) if ((count % 100 == 0))
count = $((count+1)) tries to run the command count passing it two arguments, = and the value of count+1. if[ ... tries to run the command if[ because [ is a valid word character; it doesn't automatically start a new token.
Having said all that, I'm puzzled by the unexpected ( error message. Could your bash be too old to recognize $(( syntax? Possibly. It's a very old bash.
count=$((count+1))
if [ `echo $count % 100 | bc` -eq 0 ]
Make these corrections.
Edit: Please try
count=`expr $count + 1`
I see a few errors here. First, you need double quotes around $i in case they have special characters.
Second, you shouldn't ever use
for i in `ls | sort -n`
Instead, try the following
ls -1 | sort -n | while read i
Third, change your if statement to
if ((count%5 == 0))
The (( syntax is bash is made just for integer math.
Fourth, change your counter increment to
((count++))
This is more concise. Also, the space in your version may break things. Remember, spaces matter.

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