Setting NoOptDefVal forces using = syntax

[loderunner]

Pflags accepts the following syntaxes for an integer argument:

--verbose=0
--verbose 0
-v 0
-v=0

Using this program:

v := pflag.IntP("verbose", "v", 1, "make output verbose")
pflag.Parse()
fmt.Printf("verbosity: %d\n", *v)

I get the following results:

$ ./test
verbosity: 1
$ ./test --verbose=0
verbosity: 0
$ ./test --verbose 2
verbosity: 2
$ ./test -v 0
verbosity: 0
$ ./test -v=2
verbosity: 2

If I set NoOptDefVal, I get inconsistent behavior.

v := pflag.IntP("verbose", "v", 1, "make output verbose")
verboseFlag := pflag.Lookup("verbose")
// In case of --verbose or -v with no arg, set to 2
verboseFlag.NoOptDefVal = "2"
pflag.Parse()

fmt.Printf("verbosity: %d\n", *v)

Here are the results:

$ ./test
verbosity: 1
$ ./test --verbose=0
verbosity: 0
$ ./test --verbose 2
verbosity: 2
$ ./test --verbose 3
verbosity: 2
$ ./test -v=0
verbosity: 0
$ ./test -v=1
verbosity: 1
$ ./test -v=2
verbosity: 2
$ ./test -v 1
verbosity: 2

I expected -v 3 and -v=3 to work the same way.

Is this a bug, or am I missing the point of NoOptDefVal ?

[loderunner]

I created a Pull request #135

Here's the output I get (and expected).

$ ./test
verbosity: 1
$ ./test --verbose=0
verbosity: 0
$ ./test --verbose 3
verbosity: 3
$ ./test --verbose
verbosity: 2
$ ./test -v=0
verbosity: 0
$ ./test -v
verbosity: 2
$ ./test -v 1
verbosity: 1

[n10v]

It's not possible to solve this issue unequivocally. Just consider if user will call three bool flags in the row like this: -abc. How it should be parsed? That's why your tests are failed.
And actually it will considerably change the behaviour, that we do not want.

[ghost]

I do not fully understand the issue.

It's not possible to solve this issue unequivocally. Just consider if user will call three bool flags in the row like this: -abc

If 3 bool flags are passed, shouldn't pflag know that -a is a bool. if -abc is shorthand for a string flag, then that would override -abc as a multiple bool flags possibility.

Am I missing something?

[eparis]

Imagine a is a string flag shorthand with NoOptDefVal="a_default"
Imagine there are bool shorthand flags b and c

What does -abc mean? Does it mean:
a = "bc" and b=false c=false
or does it mean:
a=a_default and b=true c=true

[ghost]

Ah I understand what is meant by the above now.

What does -abc mean? Does it mean:
a = "bc" and b=false c=false
or does it mean:
a=a_default and b=true c=true

I don't understand why this is an issue. Yes it is a case against the idea, but if pflag follows a set of rules it seems obvious to me that a="bc" would be the correct answer. Since -a is being passed the string value "bc" it would overwrite it default and use "bc". If I wanted the seperate functionality the correct solution would be to use:

-a -bc

Are there any other examples where this would actually be a problem?

EDIT:

Actually it appears that this is handled in NodeJS as parsed as @eparis second option:

a=a_default and b=true c=true

I digress, Go isn't Node, but it seems every library handles it differently.