I want to print prime numbers using for loop in shell script. Please provide some suggestions.
actually I could able to print odd numbers from range of 1 to 100, but now I am unable to print prime numbers. Provide some script to print prime numbers
# cat prime.sh
#!/bin/bash
for ((i=1; i<=100; i++))
do
output=$(( $i % 2))
if [ $output -ne 0 ]
then
echo "We got odd numbers: $i"
fi
done
The below code returns the list of prime numbers within a specified range.
the below command returns the prime numbers between 10 to 30
./filename.sh 10 30
#!/bin/bash
# Check if a number is prime
# function to return whether prime or not
check_prime() {
current_number=$1
flag=0
i=2
while test $i -le `expr $current_number / 2`
do
if test `expr $current_number % $i` -eq 0
then
flag=1
fi
i=`expr $i + 1`
done
if test $flag -eq 0
then echo $current_number
fi
}
# Assign From and to number
from_number=$1
to_number=$2
for (( number=$from_number; number<=$to_number; number++ ))
do
check_prime $number
done
This code will Print all prime numbers upto given number & also print count of it
hope it helps !!
#!/bin/bash
echo "enter a number upto which you want the prime numbers"
read num
count=0
for (( n=2; n<=$num; n++ ))
do
t=1
if [ $n -lt 2 ]
then echo "Please give other numbers than 0 and 1"
else
#echo $n
for (( i=2; i<$n; i++ ))
do
#echo $n
if (($n%i==0))
then
#echo "$n is not prime number"
t=0
break;
fi
done
#echo $t
if [ $t == 1 ]
then
echo "$n is a prime number"
count=`expr $count + 1`
fi
fi
done
echo "Total prime numbers upto $num are $count "
Related
How to keep track of how many guesses the user makes and print a message that tells them they lose? Exit after 5 guesses.
num=$(( $RANDOM % 100 + 1 ))
while [ "$input" -ne "$num" ]; do
if [ "$input" -gt "$num" ]; then
echo "The number is too high."
read input
elif [ "$input" -lt "$num" ]; then
echo "The number is too low."
read input
fi
done
echo "Great, you picked the right number."
Would you please try the following:
#!/bin/bash
num=$(( $RANDOM % 100 + 1 ))
for (( i = 0; i < 5; i++ )); do
read -p "Enter a number: " input
if (( input == num )); then
echo "Great, you picked the right number."
break
elif (( input > num )); then
echo "The number is too high."
elif (( input < num )); then
echo "The number is too low."
fi
done
BTW I tested the script by modifying the random number generator to $RANDOM % 10 + 1 but still not easy :).
This is my code to find happy prime number, I don't know why it is giving error.
#!/bin/bash
happy(){
rem = $0
sum = $0
while [ $result -gt 0 ];
do
rem = $((result%10))
p = $((rem*rem))
sum = $((sum + p))
result = $((result/10))
done
return $sum
}
echo "Enter a number"
read num
for (( i=1; i<101; i++ ))
do
result=$i
while [ $result -ne 0 && $result -ne 4 ];
do
happy $result
done
if [ $?==1 ]
then echo "happy number"
else
echo "not a happy number"
fi
done
I see lots of syntax errors and some logical errors in your script.
This should be a fixed version (works at least for 13 and 4 :))
Errors I've found:
rem = $0 and similar: the spaces around assignments are not allowed in bash,
rem = $0: you assign a value that is never used,
rem = $0 and sum = $0 the first argument passed to the function is $1 not $0,
the input number is assigned to num but never used,
the exit condition from the for and while loops is broken,
...
#!/bin/bash
happy()
{
result=$1
sum=0
while [ $result -gt 0 ]; do
rem=$(( result % 10 ))
p=$(( rem * rem ))
sum=$(( sum + $p ))
result=$(( result / 10 ))
done
echo "$sum"
}
echo "Enter a number"
read num
result=$num
for (( i=1; i<101; i++ )) do
result=$(happy $result)
if [ $result == 1 ]; then
echo "$num is a happy number"
exit
fi
done
echo "$num is not a happy number"
I am new to scripting. The script below checks if a number inputted is a prime number or divisible by 2. Why is the third line from the bottom needed (i=expr $i + 1). I commented that line out to see what the script would do and it hangs the script up. Please advise.
#! /bin/bash
echo -n "Enter a number: "
read num
i=2
while [ $i -lt $num ]
do
if [ `expr $num % $i` -eq 0 ]
then
echo "$num is not a prime number"
echo "Since it is divisible by $i"
exit
fi
i=`expr $i + 1`
done
echo "$num is a prime number "
If you don't increment i, then the test [ $i -lt $num ] will never be false, assuming i starts out less than num.
I have this concern, when getting the factors of a number. My code gives me the total numbers of all factors of a number.What I want is it should only display the factors of that number, not the total. For example I input 9 the output should be 1,3 and 9.
My code
echo -n " enter a no. "
read n
i=1
mul=1
until [ $i -gt $n ]
do
mul=`expr $mul \* $i `
i=`expr $i + 1 `
done
echo " factorial of $n is $mul "
Assuming that the user enters the number in its simplest form, e.g. 9 instead of 09.0, this works:
echo -n " enter a no. "
read n
for i in $(seq 1 $n)
do
[ $(expr $n / $i \* $i) == $n ] && echo $i
done
though a bit slowly for large numbers.
check even number - OK
if [ $(( $n % 2 )) -eq 0 ]
then
echo "$n is even number"
fi
how to check odd number ?
if [ $(( $n % ????? )) -eq 0 ]
then
echo "$n is odd number"
fi
Thank
Use "not equal 0":
if [ $(( $n % 2)) -ne 0 ]
then
echo "$n is odd"
fi
See also: http://tldp.org/LDP/abs/html/comparison-ops.html
You can also use "n%2 equals 1" since the remainder of an odd number divided by two is one:
if [ $(( $1 % 2)) -eq 1 ]
then
echo "$1 is odd"
fi
But the former (not equal 0) is the more general case, so I would prefer it.
All the answers above use a single square bracket [ which is outdated in bash (we're talking about bash, right?). The best practice to achieve the determination of an odd or even number n is:
if (( n%2==0 )); then
printf "%d is even\n" $n
else
printf "%d is odd\n" $n
fi
or, as the OP wants it, i.e., check if n is odd:
if (( n%2 )); then
printf "%d is odd\n" $n
fi
echo -n "Enter numnber : "
read n
rem=$(( $n % 2 ))
if [ $rem -eq 0 ]then
echo "$n is even number"
else
echo "$n is odd number"
fi
I prefer the simplicity of:
x=8; ((x%2)) || echo even
x=7; ((x%2)) && echo odd