How to supress unused (void **arg) parameter? [duplicate]

Question

In the function below I'm not using the parameter (void **arg). But since it's unused inside the function compiler gives me the error below:

error: unused parameter 'arg' [-Werror=unused-parameter] bool decodeString(pb_istream_t *stream, const pb_field_t *field, void **arg)

I tried to suppress it by writing void(arg) inside the function without any luck. Can anyone help me with the correct way?

Leave out the variable name: bool decodeString(pb_istream_t *stream, const pb_field_t *field, void **arg) becomes bool decodeString(pb_istream_t *stream, const pb_field_t *field, void **) — user4581301
– user4581301, Commented Oct 27, 2021 at 18:15
Side note: I prefer to comment out the name bool decodeString(pb_istream_t *stream, const pb_field_t *field, void ** /*arg*/ ) because the name may hold useful information about how the parameter should be used, should that information become important later. — user4581301
– user4581301, Commented Oct 27, 2021 at 18:16
Does this answer your question? How to suppress "unused parameter" warnings in C? — lulle2007200
– lulle2007200, Commented Oct 27, 2021 at 18:17
@MadPhysicist only viable in C++. dbush's answer below works in C and C++. Mind you, my C is pretty stale. Might have changed. — user4581301
– user4581301, Commented Oct 27, 2021 at 18:20

dbush · Accepted Answer · 2021-10-27 18:16:05Z

4

Use the parameter in an expression casted to void. Then the parameter is "used".

bool decodeString(pb_istream_t *stream, const pb_field_t *field, void **arg) { (void)arg; ... }

answered Oct 27, 2021 at 18:16

dbush

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8 Comments

razz Over a year ago

This does not work, found similar answer in stackoverflow.com/questions/3599160/… and tried it before. error: variable or field 'arg' declared void void(arg);

Bathsheba Over a year ago

Be very careful with doing this as conceptually you have induced a read of the value. Personally I'd comment arg - you don't lose the self-commenting nature of the function and you can always uncomment it if you need it. This is a good idiomatic (especially in C) technique though. Although I do remember reading something about a nasty bug emanating from this, with gcc and optimisations turned up to the maximum. Might have dreamt it though.

user4581301 Over a year ago

@razz, if this doesn't work in C you have a second problem.

0___________ Over a year ago

you can also int decodeString(void **) { /* ... */ }

dbush Over a year ago

@razz That's a separate issue that should be in a separate question.

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