Bash Commands - Arithmetic Tests using (( ... ))

来源：互联网发布：前端面试题js继承编辑：程序博客网时间：2024/05/19 11:45

The (( )) construct expands and evaluates an arithmetic expression. If the expression evaluates as zero, it
returns an exit status of 1, or "false". A non-zero expression returns an exit status of 0, or "true". This is in
marked contrast to using the test and [ ] constructs previously discussed.

#!/bin/bash

# arith-tests.sh
# Arithmetic tests.
# The (( ... )) construct evaluates and tests numerical expressions.

# Exit status opposite from [ ... ] construct!

(( 0 )) # 1

echo "Exit status of \"(( 0 ))\" is $?."

(( 1 )) # 0

echo "Exit status of \"(( 1 ))\" is $?."

(( 5 > 4 )) # true

echo "Exit status of \"(( 5 > 4 ))\" is $?." # 0

(( 5 > 9 )) # false

echo "Exit status of \"(( 5 > 9 ))\" is $?." # 1

(( 5 == 5 )) # true

echo "Exit status of \"(( 5 == 5 ))\" is $?." # 0

# (( 5 = 5 )) gives an error message.
(( 5 - 5 )) # 0

echo "Exit status of \"(( 5 - 5 ))\" is $?." # 1

(( 5 / 4 )) # Division o.k.

echo "Exit status of \"(( 5 / 4 ))\" is $?." # 0

(( 1 / 2 ))                                                                                    # Division result < 1.
echo "Exit status of \"(( 1 / 2 ))\" is $?."                            # Rounded off to 0.
                                         # 1
(( 1 / 0 )) 2>/dev/null # Illegal division by 0.
#
^^^^^^^^^^^
echo "Exit status of \"(( 1 / 0 ))\" is $?."
                                                                                                         # 1
# What effect does the "2>/dev/null" have?
# What would happen if it were removed?
# Try removing it, then rerunning the script.
# ======================================= #
# (( ... )) also useful in an if-then test.
var1=5
var2=4
if (( var1 > var2 ))
then                                                                      #Note: Not $var1, $var2. Why?
echo "$var1 is greater than $var2"
fi                                                                            # 5 is greater than 4

exit 0