In some C project, I have seen this code:
static void *l_alloc (void *ud, void *ptr, size_t osize, size_t nsize) { (void)ud; (void)osize; /* some code not using `ud` or `osize` */ return ptr; } Do the two casts to void serve any purpose?
2 Answers
It is there to avoid warnings from the compiler because some parameters are unused.
4 Comments
Ciro Santilli OurBigBook.com
What is the best way to suppress the warnings: stackoverflow.com/questions/3417837/…
Tan Wang
@Benoit what does casting to void actually do? Is its only function to show the compiler that you're intentionally ignoring something or does (void) actually do something and when the compiler sees it, it'll just count it as having done something with the variable and therefore not issue a warning?
zwol
@TanWang Its only function is to show the compiler that you're intentionally ignoring something. It will not do anything at runtime.
Kamil
Anyone knows where this is documented? Language specyfication? Compiler manual? Does it have a name (such statement)?
The reason for having unused parameters in the prototype is usually because the function needs to conform to some external API - perhaps it is a library function, or a pointer to that function is passed to another function that expects this calling convention. However not all arguments used by the calling convention are actually needed in the function itself.
The reason for mentioning the parameter name in the body is to avoid warnings like
unused.c: In function ‘l_alloc’: unused.c:3:22: warning: unused parameter ‘ud’ [-Wunused-parameter] void *l_alloc (void *ud, void *ptr, size_t osize, size_t nsize) { ^~ This warning can be suppressed with using the actual parameter in the function body. For example if you do have the following statement:
ud; This warning is now suppressed. However now GCC will produce another warning:
unused.c:5:5: warning: statement with no effect [-Wunused-value] ud; ^~ This warning tells that the statement ud;, while being syntactically valid C, does not affect anything at all, and is possibly a mistake, not unlike the statement
abort; which should perhaps have been written as abort(); instead for it to do something.
And that's where the (void) cast comes in - it will tell the compiler unambiguously and explicitly that the statement is supposed to have absolutely no effect at all.
(void)to somewhat different effect. This question is about C