Numbers and Arithmetic in bash

Internal

• bash

How to Tell if a Variable has an Integer Value

With printf

printf invoked with %d will attempt to print the argument as an integer. If the argument is not an integer, printf will return a non-zero exit value.

arg="22"
if printf %d "${arg}" >/dev/null 2>&1; then
    echo "${arg} is an int"
else
    echo "${arg} is NOT an int"
fi

With expr

v=10

if expr ${v} + 1 >/dev/null 2>&1; then
    #
    # the value contained by v is an integer
    #
else
    #
    # the value contained by v is NOT an integer
    #
fi

More details:

expr

Integral Number Operations

$((...))

op1=4
op2=2
echo $((op1 + op2))
echo $((op1 - op2))
echo $((op1 * op2)) 
echo $((op1 / op2)) # integral division
echo $((op1 % op2)) # modulo
echo $((op1 += op2))

displays:

6
2
8
2

echo $(($(date '+%s') % 2))

also works.

expr

expr

bc

local t0="$(date -u +%s)"
local t1="$(date -u +%s)"
local delta="$(bc <<<"${t1}-${t0}")"

Floating Point Operations

Multiplication:

local operand1=10.1
local operand2=20.2

result=$(echo ${operand1} ${operand2} | awk '{printf "%5.1f\n",$1*$2}') || exit 1
result=$(echo ${operand1} ${operand2} | awk '{printf "%.1f\n",$1*$2}') || exit 1

Division:

local operand1=10.1
local operand2=20.2

result=$(echo ${operand1} ${operand2} | awk '{printf "%5.1f\n",$1/$2}') || exit 1

Increment an Integer

Increment an Integer with ((...)) Operator

Decrement an Integer

Decrement an Integer with ((...)) Operator

Numeric Evaluation in a Condition

a=1
b=2

if [[ $((a - b)) -lt 0 ]]; then
  ...
fi

if [[ $(expr ${a} - ${b}) < 0 ]]; then
  ...
fi

Random Numbers

Bash script that generates a random MAC address