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[bash] Don’t print function definition when checking for existence (#3448)

When just checking whether a function is already defined or not, it’s not
necessary to print out it’s definition (should it be defined).

bash’s `declare` provides the `-F`-option (which implies `-f`), which should
give a minor performance improvement

Signed-off-by: Christoph Anton Mitterer <mail@christoph.anton.mitterer.name>
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Christoph Anton Mitterer 2023-10-02 14:02:35 +02:00 committed by GitHub
parent ee4ba104e7
commit e833823e15
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@ -12,7 +12,7 @@
if [[ $- =~ i ]]; then if [[ $- =~ i ]]; then
# To use custom commands instead of find, override _fzf_compgen_{path,dir} # To use custom commands instead of find, override _fzf_compgen_{path,dir}
if ! declare -f _fzf_compgen_path > /dev/null; then if ! declare -F _fzf_compgen_path > /dev/null; then
_fzf_compgen_path() { _fzf_compgen_path() {
echo "$1" echo "$1"
command find -L "$1" \ command find -L "$1" \
@ -21,7 +21,7 @@ if ! declare -f _fzf_compgen_path > /dev/null; then
} }
fi fi
if ! declare -f _fzf_compgen_dir > /dev/null; then if ! declare -F _fzf_compgen_dir > /dev/null; then
_fzf_compgen_dir() { _fzf_compgen_dir() {
command find -L "$1" \ command find -L "$1" \
-name .git -prune -o -name .hg -prune -o -name .svn -prune -o -type d \ -name .git -prune -o -name .hg -prune -o -name .svn -prune -o -type d \