Is there a built-in way to check if Go code is already formatted?

[mrush...@pivotallabs.com]

I want to set up Travis CI to fail a build if the code wasn't formatted correctly.  It doesn't appear that gofmt has a flag that will cause the command to pass or fail based on whether any files were or need to be reformatted.

My imaginary shell session with this imaginary command would be something like:

$ gofmt -checkFormatting ./... # some files need formatting, command exits 1

$ go fmt ./...

$ gofmt -checkFormatting ./... # all files already formatted, command exits 0

I just found codereview.py which has its own CheckGoFmt function that does something similar to what I want to do, so that makes it look like gofmt does not currently have the feature I'm looking for.  Is this something that belongs in gofmt?  Or does an external script for this functionality make more sense from the Go authors' perspectives?

Mark

[andrey mirtchovski]

you can use a combination of gofmt -l file and the exit code of gofmt. if you had parsing errors exit code would be 2 (generally), and if you didn't have parsing errors but you had a differing formatting the -l flag would force some output.

for example, try this in $GOROOT/test/:

% for i in $GOROOT/test/*; do gofmt -l $i 2>/dev/null; echo $?; done

remove the redirect of stderr to /dev/null to see what the errors were.

​HTH

[aro...@gmail.com]

gofmt -d path/to/file.go 2>&1 | read; [ $? == 1 ]

should do it:

If the file has formatting issues, gofmt will output to stdout.  If the file has parsing errors, gofmt will output to stderr.  2>&1 combines stderr and stdout.  | read will succeed if there's any output.

So if there's any formatting/syntax issues, the read returns 0.  Otherwise it returns 1.

[ $? == 1 ] inverts read's return code.

[Matt Harden]

Instead of [ $? == 1 ], just use the not (!) prefix:

! gofmt -d path/to/file.go 2>&1 | read

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