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tests: don't expect the shell to support 'local'
It doesn't work on Solaris 11.3 x86/64. Reported by Kiyoshi Kanazawa. http://lists.gnu.org/archive/html/bug-bison/2018-10/msg00051.html * examples/test: Don't use 'local'.
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@@ -45,7 +45,7 @@ fi
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# -------
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# -------
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cleanup ()
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cleanup ()
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{
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{
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local status=$?
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status=$?
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if test -z "$DEBUG"; then
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if test -z "$DEBUG"; then
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cd $cwd
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cd $cwd
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rm -rf $$.dir
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rm -rf $$.dir
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@@ -61,22 +61,22 @@ cd $$.dir
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# -noerr: ignore stderr, otherwise merge it into effective output.
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# -noerr: ignore stderr, otherwise merge it into effective output.
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run ()
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run ()
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{
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{
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local noerr=false
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noerr=false
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case $1 in
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case $1 in
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(-noerr) noerr=true; shift;;
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(-noerr) noerr=true; shift;;
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esac
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esac
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# Expected exit status.
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# Expected exit status.
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local sta_exp="$1"
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sta_exp=$1
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shift
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shift
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# Expected output.
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# Expected output.
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local out_exp="$1"
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out_exp=$1
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shift
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shift
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# Effective exit status.
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# Effective exit status.
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local sta_eff=0
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sta_eff=0
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$prog "$@" - <input >out_eff 2>err_eff || sta_eff=$?
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$prog "$@" - <input >out_eff 2>err_eff || sta_eff=$?
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# Combine effective output and error streams.
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# Combine effective output and error streams.
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local out_eff="$(cat out_eff && $noerr || sed -e 's/^/err: /g' err_eff)"
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out_eff=$(cat out_eff && $noerr || sed -e 's/^/err: /g' err_eff)
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if test $sta_eff -eq $sta_exp; then
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if test $sta_eff -eq $sta_exp; then
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if test "$out_eff" = "$out_exp"; then
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if test "$out_eff" = "$out_exp"; then
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echo "$me: PASS: $number"
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echo "$me: PASS: $number"
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