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![]() In the following snippet: if [ ! -x `which $prereq` ]; then When $prereq does not exist, `which $prereq` evaluates to the empty string, which results in *no* argument being passed to the -x operator, which then evaluates to true, which is the equivalent of the prereq having been found. In order for this to fail as expected, we must pass an empty argument, which then causes -x to fail. Do this by wrapping the `` in quotes so there's always an argument to -x, even if the value of the argument is zero-length. Signed-off-by: Stephen Warren <swarren@wwwdotorg.org> Reviewed-by: Simon Glass <sjg@chromium.org> Tested-by: Simon Glass <sjg@chromium.org> |
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fat-noncontig-test.sh | ||
fs-test.sh |