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echo 10**2 which prints 10**2. How to calculate the right result, 100?
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$1 means an input argument and -z means non-defined or empty. You're testing whether an input argument to the script was defined when running the script.
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The += and -= Operators These operators are used to increment/decrement the value of the left operand with the value specified after the operator. ((i+=1)) let "i+=1"
You can use the let builtin:
let var=10**2 # sets var to 100. echo $var # prints 100 or arithmetic expansion:
var=$((10**2)) # sets var to 100. Arithmetic expansion has the advantage of allowing you to do shell arithmetic and then just use the expression without storing it in a variable:
echo $((10**2)) # prints 100. For large numbers you might want to use the exponentiation operator of the external command bc as:
bash:$ echo 2^100 | bc 1267650600228229401496703205376 If you want to store the above result in a variable you can use command substitution either via the $() syntax:
var=$(echo 2^100 | bc) or the older backtick syntax:
var=`echo 2^100 | bc` Note that command substitution is not the same as arithmetic expansion:
$(( )) # arithmetic expansion $( ) # command substitution Various ways:
echo $((10**2)) awk 'BEGIN{print 10^2}' # POSIX standard awk 'BEGIN{print 10**2}' # GNU awk extension echo '10 ^ 2' | bc dc -e '10 2 ^ p' If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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