Advanced Bash−Scripting Guide

# Replace all matching wildcarded string
declare −a array5=( ${array0[@]//new?/old} )
echo
echo "Elements in array5: ${array5[@]}"
# Just when you are getting the feel for this...
declare −a array6=( ${array0[@]#*new} )
echo # This one might surprise you
echo "Elements in array6: ${array6[@]}"
declare −a array7=( ${array0[@]#new1} )
echo # After array6 this should not be a surprise
echo "Elements in array7: ${array7[@]}"
# Which looks a lot like...
declare −a array8=( ${array0[@]/new1/} )
echo
echo "Elements in array8: ${array8[@]}"
# So what can one say about this?
# The string operations are performed on
#+ each of the elements in var[@] in succession.
# Therefore : BASH supports string vector operations
# If the result is a zero length string, that
#+ element disappears in the resulting assignment.
# Question, are those strings hard or soft quotes?
zap='new*'
declare −a array9=( ${array0[@]/$zap/} )
echo
echo "Elements in array9: ${array9[@]}"
# Just when you thought you where still in Kansas...
declare −a array10=( ${array0[@]#$zap} )
echo
echo "Elements in array10: ${array10[@]}"
# Compare array7 with array10
# Compare array8 with array9
# Answer, must be soft quotes.
exit 0
Advanced Bash−Scripting Guide
The relationship of ${array_name[@]} and ${array_name[*]} is analogous to that between $@ and $*. This
powerful array notation has a number of uses.
# Copying an array.
array2=( "${array1[@]}" )
# or
array2="${array1[@]}"
# Adding an element to an array.
array=( "${array[@]}" "new element" )
# or
array[${#array[*]}]="new element"
# Thanks, S.C.
Chapter 26. Arrays 297