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Fix issue18076: allow -run to work with - (stdin) #7435

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Merged
wilzbach merged 2 commits into from
Dec 15, 2017
Merged

Fix issue18076: allow -run to work with - (stdin) #7435

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91701a4
Fix issue 18076: make -run work with - (stdin)
Dec 14, 2017
a920164
Add test case.
Dec 14, 2017
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5 changes: 4 additions & 1 deletion src/ddmd/mars.d
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Original file line number Diff line number Diff line change
Expand Up @@ -2156,7 +2156,10 @@ private bool parseCommandLine(const ref Strings arguments, const size_t argc, re
error("-run must be followed by a source file, not '%s'", arguments[i + 1]);
break;
}
files.push(arguments[i + 1]);
if (strcmp(arguments[i + 1], "-") == 0)
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@wilzbach wilzbach Dec 15, 2017

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This is too simplistic, it won't work if there's a command after -run, e.g. dmd -run -m64 -

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@quickfur quickfur Dec 15, 2017 •
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Actually, -run takes an argument, so your example is invalid. (I'm not sure why whoever originally implemented -run did it this way, but that's how it currently works, and this PR is just following in its steps.)(And yes, the first time I realized -run took an argument I was surprised too. :-P)

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@wilzbach wilzbach Dec 15, 2017

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Oh sorry. I just looked over this again from my phone and imagined that doing dmd -run -version=Foo myfile.d is legimate, so yeah that seems like bad design too me too, but that's an opportunity for another PR ;-)

files.push("__stdin.d");
else
files.push(arguments[i + 1]);
params.runargs.setDim(length - 1);
for (size_t j = 0; j < length - 1; ++j)
{
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8 changes: 8 additions & 0 deletions test/runnable/test18076.sh
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@@ -0,0 +1,8 @@
#!/usr/bin/env bash

dir=${RESULTS_DIR}${SEP}runnable
output_file=${dir}/test18076.sh.out

echo 'import std.stdio; void main() { writeln("Success"); }' | \
$DMD -m${MODEL} -run - > ${output_file} || exit 1
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@wilzbach wilzbach Dec 14, 2017

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Don't you have to check that it actually prints "Success" here?

[[ $(echo "Success") == "Success" ]] || exit 1

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@quickfur quickfur Dec 14, 2017

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You're right, fixed.

grep -q '^Success$' ${output_file} || exit 1
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@wilzbach wilzbach Dec 14, 2017 •
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FWIW the || exit is superfluous here.
A bash script will yield the last return value as exit code if not explicitly exited
Convince yourself:

sh -c "grep -q 'foo' afile.d" ; echo $?

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@quickfur quickfur Dec 14, 2017

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True. I'm wary of relying on these implicit behaviours, though. Stating || exit 1 makes it more obvious what's happening. Not that it makes a difference, I just prefer to have things up-front than implicit. Implicit things tend to be fragile and unintentionally break later.

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@wilzbach wilzbach Dec 15, 2017 •
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That's why one usually does a set -e at the beginning 😉
https://unix.stackexchange.com/questions/15998/what-does-the-e-do-in-a-bash-shebang