#!/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 :-)
Related
I have this Bash script and I had a problem in line 16.
How can I take the previous result of line 15 and add
it to the variable in line 16?
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$i-*`; do
echo "$j"
metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15)
num= $num + $metab (line16)
done
echo "$num"
done
For integers:
Use arithmetic expansion: $((EXPR))
num=$((num1 + num2))
num=$(($num1 + $num2)) # Also works
num=$((num1 + 2 + 3)) # ...
num=$[num1+num2] # Old, deprecated arithmetic expression syntax
Using the external expr utility. Note that this is only needed for really old systems.
num=`expr $num1 + $num2` # Whitespace for expr is important
For floating point:
Bash doesn't directly support this, but there are a couple of external tools you can use:
num=$(awk "BEGIN {print $num1+$num2; exit}")
num=$(python -c "print $num1+$num2")
num=$(perl -e "print $num1+$num2")
num=$(echo $num1 + $num2 | bc) # Whitespace for echo is important
You can also use scientific notation (for example, 2.5e+2).
Common pitfalls:
When setting a variable, you cannot have whitespace on either side of =, otherwise it will force the shell to interpret the first word as the name of the application to run (for example, num= or num)
num= 1 num =2
bc and expr expect each number and operator as a separate argument, so whitespace is important. They cannot process arguments like 3+ +4.
num=`expr $num1+ $num2`
Use the $(( )) arithmetic expansion.
num=$(( $num + $metab ))
See Chapter 13. Arithmetic Expansion for more information.
There are a thousand and one ways to do it. Here's one using dc (a reverse Polish desk calculator which supports unlimited precision arithmetic):
dc <<<"$num1 $num2 + p"
But if that's too bash-y for you (or portability matters) you could say
echo $num1 $num2 + p | dc
But maybe you're one of those people who thinks RPN is icky and weird; don't worry! bc is here for you:
bc <<< "$num1 + $num2"
echo $num1 + $num2 | bc
That said, there are some unrelated improvements you could be making to your script:
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in output-$i-* ; do # 'for' can glob directly, no need to ls
echo "$j"
# 'grep' can read files, no need to use 'cat'
metab=$(grep EndBuffer "$j" | awk '{sum+=$2} END { print sum/120}')
num=$(( $num + $metab ))
done
echo "$num"
done
As described in Bash FAQ 022, Bash does not natively support floating point numbers. If you need to sum floating point numbers the use of an external tool (like bc or dc) is required.
In this case the solution would be
num=$(dc <<<"$num $metab + p")
To add accumulate possibly-floating-point numbers into num.
In Bash,
num=5
x=6
(( num += x ))
echo $num # ==> 11
Note that Bash can only handle integer arithmetic, so if your AWK command returns a fraction, then you'll want to redesign: here's your code rewritten a bit to do all math in AWK.
num=0
for ((i=1; i<=2; i++)); do
for j in output-$i-*; do
echo "$j"
num=$(
awk -v n="$num" '
/EndBuffer/ {sum += $2}
END {print n + (sum/120)}
' "$j"
)
done
echo "$num"
done
I always forget the syntax so I come to Google Search, but then I never find the one I'm familiar with :P. This is the cleanest to me and more true to what I'd expect in other languages.
i=0
((i++))
echo $i;
I really like this method as well. There is less clutter:
count=$[count+1]
#!/bin/bash
read X
read Y
echo "$(($X+$Y))"
You should declare metab as integer and then use arithmetic evaluation
declare -i metab num
...
num+=metab
...
For more information, see 6.5 Shell Arithmetic.
Use the shell built-in let. It is similar to (( expr )):
A=1
B=1
let "C = $A + $B"
echo $C # C == 2
Source: Bash let builtin command
Another portable POSIX compliant way to do in Bash, which can be defined as a function in .bashrc for all the arithmetic operators of convenience.
addNumbers () {
local IFS='+'
printf "%s\n" "$(( $* ))"
}
and just call it in command-line as,
addNumbers 1 2 3 4 5 100
115
The idea is to use the Input-Field-Separator(IFS), a special variable in Bash used for word splitting after expansion and to split lines into words. The function changes the value locally to use word-splitting character as the sum operator +.
Remember the IFS is changed locally and does not take effect on the default IFS behaviour outside the function scope. An excerpt from the man bash page,
The shell treats each character of IFS as a delimiter, and splits the results of the other expansions into words on these characters. If IFS is unset, or its value is exactly , the default, then sequences of , , and at the beginning and end of the results of the previous expansions are ignored, and any sequence of IFS characters not at the beginning or end serves to delimit words.
The "$(( $* ))" represents the list of arguments passed to be split by + and later the sum value is output using the printf function. The function can be extended to add scope for other arithmetic operations also.
#!/usr/bin/bash
#integer numbers
#===============#
num1=30
num2=5
echo $(( num1 + num2 ))
echo $(( num1-num2 ))
echo $(( num1*num2 ))
echo $(( num1/num2 ))
echo $(( num1%num2 ))
read -p "Enter first number : " a
read -p "Enter second number : " b
# we can store the result
result=$(( a+b ))
echo sum of $a \& $b is $result # \ is used to espace &
#decimal numbers
#bash only support integers so we have to delegate to a tool such as bc
#==============#
num2=3.4
num1=534.3
echo $num1+$num2 | bc
echo $num1-$num2 | bc
echo $num1*$num2 |bc
echo "scale=20;$num1/$num2" | bc
echo $num1%$num2 | bc
# we can store the result
#result=$( ( echo $num1+$num2 ) | bc )
result=$( echo $num1+$num2 | bc )
echo result is $result
##Bonus##
#Calling built in methods of bc
num=27
echo "scale=2;sqrt($num)" | bc -l # bc provides support for calculating square root
echo "scale=2;$num^3" | bc -l # calculate power
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$i-*`; do
echo "$j"
metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15)
let num=num+metab (line 16)
done
echo "$num"
done
Works on MacOS. bc is a command line calculator
#!/bin/bash
sum=0
for (( i=1; i<=5; i++ )); do
sum=$(echo "$sum + 1.1" | bc) # bc: if you want to use decimal
done
echo "Total: $sum"
Currently stuck in a situation where I ask the user to input a line of numbers with a space in between, then have the program display those numbers with a delay, then add them. I have everything down, but can't seem to figure out a line of code to coherently calculate the sum of their input, as most of my attempts end up with an error, or have the final number multiplied by the 2nd one (not even sure how?). Any help is appreciated.
echo Enter a line of numbers to be added.
read NUMBERS
COUNTER=0
for NUM in $NUMBERS
do
sleep 1
COUNTER=`expr $COUNTER + 1`
if [ "$NUM" ]; then
echo "$NUM"
fi
done
I've tried echo expr $NUM + $NUM to little success, but this is really all I can some up with.
Start with
NUMBERS="4 3 2 6 5 1"
echo $NUMBERS
Your script can be changed into
sum=0
for NUM in ${NUMBERS}
do
sleep 1
((counter++))
(( sum += NUM ))
echo "Digit ${counter}: Sum=$sum"
done
echo Sum=$sum
Another way is using bc, usefull for input like 1.6 2.3
sed 's/ /+/g' <<< "${NUMBERS}" | bc
Set two variables n and m, store their sum in $x, print it:
n=5 m=7 x=$((n + m)) ; echo $x
Output:
12
The above syntax is POSIX compatible, (i.e. works in dash, ksh, bash, etc.); from man dash:
Arithmetic Expansion
Arithmetic expansion provides a mechanism for evaluating an arithmetic
expression and substituting its value. The format for arithmetic expanâ
sion is as follows:
$((expression))
The expression is treated as if it were in double-quotes, except that a
double-quote inside the expression is not treated specially. The shell
expands all tokens in the expression for parameter expansion, command
substitution, and quote removal.
Next, the shell treats this as an arithmetic expression and substitutes
the value of the expression.
Two one-liners that do most of the job in the OP:
POSIX:
while read x ; do echo $(( $(echo $x | tr ' ' '+') )) ; done
bash:
while read x ; do echo $(( ${x// /+} )) ; done
bash with calc, (allows summing real, rational & complex numbers, as well as sub-operations):
while read x ; do calc -- ${x// /+} ; done
Example input line, followed by output:
-8!^(1/3) 2^63 -1
9223372036854775772.7095244707464171953
I want to take the absolute of a number by the following code in bash:
#!/bin/bash
echo "Enter the first file name: "
read first
echo "Enter the second file name: "
read second
s1=$(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.
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.
This is the programme I use.
clear
echo enter a string
read string
length=`echo -n $string | wc -c`
v=0
cons=0
d=0
s=0
len=$length
while [ $len -gt 0 ]
do
h=`echo $string | cut -c $len`
case $h in
[AaEeIiOoUu]) v=`expr $v + 1`
;;
[BbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz]) cons=`expr $cons + 1`
;;
[0-9]) d=`expr $d + 1`
;;
' ') s=`expr $s + 1 `
;;
esac
len=`expr $len - 1 `
done
spl=`expr $len - $v - $cons - $d - $s`
echo "vowels = $v"
echo "consonants = $cons"
echo "digits = $d"
echo "spaces = $s"
echo "special character = $spl"
The program counts all other types of characters except special characters. The output shows a minus value even if there are special characters in the input value. How can I modify the programme to make it display correct number of special characters in the input?
While your script is technically (almost) correct, it is not very pretty (both visually and in terms of code beauty).
The reason your version is not working as expected #devnull has already pointed out, but there is another bug that I will explain further down.
Since you are using bash you could rewrite the whole thing in a more idiomatic, readable and shorter way.
Here is a rewritten version of your script (comments and explanations follow below):
#!/bin/bash
clear
IFS= read -p "enter a string: " string
for ((i = 0; i < ${#string}; i++)); do
case "${string:$i:1}" in
[AaEeIiOoUu]) ((vowels++)) ;;
[[:alpha:]]) ((consonants++)) ;;
[[:digit:]]) ((digits++)) ;;
[[:space:]]) ((whitespace++)) ;;
*) ((others++)) ;;
esac
done
echo "vowels = ${vowels-0}"
echo "consonants = ${consonants-0}"
echo "digits = ${digits-0}"
echo "whitespace = ${whitespace-0}"
echo "other characters = ${others-0}"
indentation
First off, you should always indent your code blocks (if constructs, loops, switch (case) statements, ...) for readability, e.g.
while [ $len -gt 0 ]; do
do_stuff
done
read, whitespace and the prompt
Since you are using bash, read is capable of displaying a prompt for you - the extra echo is not needed. Furthermore, read strips off leading and trailing whitespace which results in an incorrect calculation unless you set IFS to the null string:
IFS= read -p "this is my prompt: " string
iterating over characters in a string
You can use parameter expansion to both get the length of the string as well as slice out one character at a time using a for loop, getting rid of the unnecessary cut and avoiding a subshell.
# ${#string} = length of $string
for ((i = 0; i < ${#string}; i++)); do
c=${string:$i:1} # c is set to the character at position $i in string $string
done
character classes
First off, your consonants statement still includes Ii for the match. Technically it doesn't matter since you cannot fall-through from the vowels match, but if this is an assignment you probably want to remove it.
That being said, you could just use the short character class for readability:
[AaEeIiOoUu]) vowel_stuff ;;
[a-zA-Z]) consonant_stuff ;; # vowels already matched above, so superfluous characters don't matter here
To make your life even easier, there are so called character classes which you can use instead, e.g.
[:digit:] = [0-9]
[:space:] = tabs, newline, form feed, carriage return, space
etc.
Note that your current locale influences certain character classes.
the special characters case
Just use the default case in the switch statement to count them, then you can skip calculating those afterwards:
case ... in
vowels) handle_vowel ;;
[...]
*) handle_other_character ;;
esac
defaults
Using parameter expansion you can also get rid of initializing your variables with 0, you can on-the-fly expand the variables to 0 if they are not set, i.e. they have not been incremented during the loop.
backticks
Unless you are writing code that has to be super-portable and has to work in all kinds of old shells, use the $() syntax in stead of ``.
arithmetic expressions
Same as above, unless you really need it you can use (( )) for arithmetic expressions, e.g.
a=5
((a = a + 10)) # or even ((a += 10))
# $a is now 15
Google and the Advanced Bash-Scripting Guide as well as the bash sections of Greg's Wiki are your friends.
You are using the wrong variable while calculating spl. Using len is clearly wrong as you're decrementing it within the loop and it becomes 0 when the loop is done.
Instead of saying:
spl=`expr $len - $v - $cons - $d - $s`
say:
spl=`expr $length - $v - $cons - $d - $s`