BASH Shell 编程 经典教程 《高级SHELL脚本编程》中文版

				#!/bin/bash				# empty-array.sh								#  感谢Stephane Chazelas制作这个例子的原始版本, 				#+ 同时感谢Michael Zick对这个例子所作的扩展. 												# 空数组与包含有空元素的数组, 这两个概念不同. 								array0=( first second third )				array1=( '' )   # "array1"包含一个空元素. 				array2=( )      # 没有元素 . . . "array2"为空. 								echo				ListArray()				{				echo				echo "Elements in array0:  ${array0[@]}"				echo "Elements in array1:  ${array1[@]}"				echo "Elements in array2:  ${array2[@]}"				echo				echo "Length of first element in array0 = ${#array0}"				echo "Length of first element in array1 = ${#array1}"				echo "Length of first element in array2 = ${#array2}"				echo				echo "Number of elements in array0 = ${#array0[*]}"  # 3				echo "Number of elements in array1 = ${#array1[*]}"  # 1  (惊奇!)				echo "Number of elements in array2 = ${#array2[*]}"  # 0				}								# ===================================================================								ListArray								# 尝试扩展这些数组. 								# 添加一个元素到这个数组. 				array0=( "${array0[@]}" "new1" )				array1=( "${array1[@]}" "new1" )				array2=( "${array2[@]}" "new1" )								ListArray								# 或				array0[${#array0[*]}]="new2"				array1[${#array1[*]}]="new2"				array2[${#array2[*]}]="new2"								ListArray								# 如果你按照上边的方法对数组进行扩展的话; 数组比较象'栈' 				# 上边的操作就是'压栈' 				# 栈'高'为: 				height=${#array2[@]}				echo				echo "Stack height for array2 = $height"								# '出栈'就是: 				unset array2[${#array2[@]}-1]	#  数组从0开始索引, 				height=${#array2[@]}            #+ 这意味着第一个数组下标为0. 				echo				echo "POP"				echo "New stack height for array2 = $height"								ListArray								# 只列出数组array0的第二个和第三个元素.				from=1		# 从0开始索引. 				to=2		#				array3=( ${array0[@]:1:2} )				echo				echo "Elements in array3:  ${array3[@]}"								# 处理方式就像是字符串(字符数组). 				# 试试其他的"字符串"形式. 								# 替换: 				array4=( ${array0[@]/second/2nd} )				echo				echo "Elements in array4:  ${array4[@]}"								# 替换掉所有匹配通配符的字符串. 				array5=( ${array0[@]//new?/old} )				echo				echo "Elements in array5:  ${array5[@]}"								# 当你开始觉得对此有把握的时候 . . .				array6=( ${array0[@]#*new} )				echo # 这个可能会让你感到惊奇. 				echo "Elements in array6:  ${array6[@]}"								array7=( ${array0[@]#new1} )				echo # 数组array6之后就没有惊奇了. 				echo "Elements in array7:  ${array7[@]}"								# 看起来非常像 . . .				array8=( ${array0[@]/new1/} )				echo				echo "Elements in array8:  ${array8[@]}"								#  所以, 让我们怎么形容呢? 								#  对数组var[@]中的每个元素				#+ 进行连续的字符串操作. 				#  因此: 如果结果是长度为0的字符串, 				#+ Bash支持字符串向量操作, 				#+ 元素会在结果赋值中消失不见. 								#  一个问题, 这些字符串是强引用还是弱引用? 								zap='new*'				array9=( ${array0[@]/$zap/} )				echo				echo "Elements in array9:  ${array9[@]}"								# 当你还在考虑, 你身在Kansas州何处时 . . .				array10=( ${array0[@]#$zap} )				echo				echo "Elements in array10:  ${array10[@]}"								# 比较array7和array10. 				# 比较array8和array9. 								# 答案: 必须是弱引用. 								exit 0