Error 089 casting a void function is illegal

I’m getting an «invalid use of void expression» error regarding the following function:

void remove_data(char* memory, char* bitmap, int pos)
{
    int i;

    i = (int)test_bit(bitmap, pos-1);

    if(i != 0)
    {
        memory[pos-1] = NULL;
        toggle_bit(bitmap, pos-1);
    }
}

Now from what I’ve read on other threads regarding this similar issue this error pops up when the programmer uses something within the function that will generate an output. However, I’m failing to see what would cause that in the code I’ve written.

Edit to add:

The toggle bit function used in it is also type void and doesn’t return a value.

void test_bit(char * bitmap, int pos)
{
    int bit;

    bit = bitmap[pos/8] & (1<<(pos%8));
    printf("%dn", &bit);
}

I commented the printf line in test_bit but the error is still there.

asked Oct 8, 2012 at 16:55

However, I’m failing to see what would cause that in the code I’ve written.

Me too, because I don’t think you have posted the relevant code. Regardless, I think this is probably your problem:

i = (int)test_bit(bitmap, pos-1);

Please show us the signature for test_bit. I’m guessing it returns void or you have forward declared it somewhere and accidentally wrote:

void test_bit(char*, int);

Since the function returns void (i.e., nothing) you cannot then proceed to cast the return value (which is nothing) to an int as you are doing, it makes no sense and is illegal.

EDIT: You have verified in the comments that test_bit is in fact declared to return void, so that’s your problem.

answered Oct 8, 2012 at 16:58

answered Oct 8, 2012 at 16:59

Obvious Solution: Don’t assign results of function returning void.

Non-Obvious Solution: Don’t nest function calls, and you are less likely
to be looking in all of the wrong spots, thinking the compiler is broken.

#include <stdio.h> //:for: printf(...)

void YOUR_MISTAKE_ACTUALLY_HERE();
void  MyFunc( void );
void* WRAPPED_MyFunc( int, int, int );


int main( void ){
    printf("[BEG:main]n");


    void* HW = NULL;
    HW = WRAPPED_MyFunc(  //:<---So you check this, but it looks fine.
        1
    ,   YOUR_MISTAKE_ACTUALLY_HERE()
    ,   3
    );; //:<--- compiler says "Invalid Use Of Void Expression" Here.

    if( HW ){ /** NOOP **/ };


    printf("[END:main]n");
} 

    void YOUR_MISTAKE_ACTUALLY_HERE(){ }

    //:Pretend this function is fetched from DLL.
    //:Casting all DLL functions initially to void(*pfn)void
    //:in a lookup table, then re-casting later.
    void MyFunc( void ){

        return;

    }//[;]//

    typedef  
        void* 
        (* MyFunc_TYPEDEF)(            

            int a
        ,   int b
        ,   int c

        );

    void*
    WRAPPED_MyFunc(

        int a
    ,   int b
    ,   int c

    ){


               MyFunc_TYPEDEF
        fun =( MyFunc_TYPEDEF)
               MyFunc;

        void* ret=( fun(

            a
        ,   b
        ,   c

        ));

        return( 
            (void*)
            ret 
        );;

    }//:Func

answered Dec 9, 2019 at 3:39

asked Jun 17, 2021 at 22:58

What happened is you wrote a void expression where an expression of type ‘pointer to function of void returning void’ (i.e., no arguments and no return value), was expected. The name of a function with no parentheses for an argument list is an expression of type ‘pointer to function’ taking whatever arguments & types that named function was declared to expect, and returning whatever type that named function was declared to return.

But if you write the parentheses with it, now you’ve written an expression that 1) would (if it were valid) get evaluated during the call to attachInterrupt() — much earlier that you want it to be! — and return a value of that return type rather than the type you expected.

@timemage has offered two ways to solve this.

answered Jun 17, 2021 at 23:34