Ошибка invalid use of void expression

I have a function int rt_task_start (RT_TASK *task, void(*task_func)(void *arg), void *arg)
where in second argument i am passing a function with argument.

1. When i only pass a function name at that time there is no problem.(as expected it’s working). rt_task_start(&demo_task1, demo, 1);
2. But when i pass rt_task_start(&demo_task1, demo(&val), 1); it’s giving me error error: invalid use of void expression. Variable val is defined before. int val = 0;
3. When i call with this rt_task_start(&demo_task1, demo(val), 1); this is showing error Warning passing argument 1 of 'demo' makes pointer from integer without a cast then error: invalid use of void expression.
4. int *val;
*val = 0;
rt_task_start(&demo_task1, demo(&val), 1); this is also giving me error.

I can’t understand what should i pass, as a void pointer. It’s giving me error. Any Idea Please!

asked Mar 29, 2014 at 13:57

void (*task_func)(void *arg);

The above statement defines task_func to be a pointer to a function which takes a pointer of type void * and returns no value.

Therefore, when you call your function rt_task_start, you should pass a pointer to a function as the second argument. Also, you should pass a pointer of type void * as the third argument, not an integer. A function name evaluates to a pointer to the function, so you can simply pass the function name as the argument value, or you can use the address-of operator & before the function name.

int arg = 4;

// both calls are equivalent

rt_task_start(&demo_task1, demo, &arg);
rt_task_start(&demo_task1, &demo, &arg);

answered Mar 29, 2014 at 14:07

I’m not sure how the code in (1) can possibly compile. But here is what you should be using:

int rt_task_start (RT_TASK *task, void(*task_func)(void *arg), void *arg);
int val = 1;
rt_task_start(&demo_task1, demo, &val);

You cannot pass the function pointer bound to a specific argument, that is something like a closure, which isn’t available in C. You can, however, pass the function pointer and then separately pass the argument you want to apply (which is what the function signature suggests you should do). But you must pass that argument as a pointer, not a literal.

answered Mar 29, 2014 at 14:07

Surely you want:

  int i=1;
  rt_task_start(&demo_task1, demo, (void*) &i);

Just by matching the argument types, remember the second argument is just a function pointer, not a function call with its own argument, it’s own argument is only used when you call it within rt_task_demo. If you then want to use the value ‘1’ in function ‘rt_task_demo’ you would recast it like

int ii = *(int*) arg;

answered Mar 29, 2014 at 14:15

The “invalid use of void expression” error occurs in C/C++ when we try to assign a call to a void function to a variable. It also occurs when a void function is cast to another data type.

Code

In the following code, the function f() has a void return type. This function is passed as an argument to the function g(), which expects an int as its argument. C++ tries to cast the void return type of the function f() to an integer, which causes the “invalid use of void expression” error.

#include <iostream>
using namespace std;



  


                       
  



void f()
{
  cout<<"Hello World"<<endl;
}
int g(int k)
{
	return (k+5);



  


                       
  



}
int main()
{
	int x = g(f());
  cout<<x<<endl;
}

Solution

The solution is to avoid casting void return types to any other data type. Since the function g() can only accept an integer data type, we should only pass it data of integer data type.

#include <iostream>
using namespace std;
void f()



  


                       
  



{
  cout<<"Hello World"<<endl;
}
int g(int k)
{
	return (k+5);
}
int main()
{
  f();



  


                       
  



	int x = g(5);
  cout<<x<<endl;
}

void

expression

invalid

c++

error

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

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#include <stdio.h>

template <typename T>
int foo(T* x, T* y)
{
    *x  = 3;
    *y = *x + 1;
    return 0;
}

template <typename T>
T* foo1(T* x)
{
    *x = *x +1;
    return x;
}
 
int main(void)
{
    int a=7;
    int b = 5;
    char c = 'a';
    foo(&a,&b); // calls foo<int>
    printf("%dn%dn",a,b);

    foo1(&c); // calls foo1<char>
    printf("%cn",c);
}

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/* Some way to signal the real type of an argument.  Such a type is analogous to
   printf's format string -- that's one way to signal type too. */
/* User-defined structures will need to be handled through a proxy function */
typedef enum TypeLabel { Char_T, SChar_T, UChar_T, Int_T, UInt_T, /* ... */ } TypeLabel;
void increment(TypeLabel c, void* x) {
# if ! defined BRANCH
# define BRANCH(label, type) do {                      
    /* Manipulate x here */                            
    case label: *((type*)x) = (*(type*)x) + 1; return; 
  } while(0)

  switch (c) {
    BRANCH(Char_T , char);
    BRANCH(SChar_T, signed char);
    BRANCH(UChar_T, unsigned char);
    BRANCH(Int_T  , int);
    BRANCH(UInt_T , unsigned int);
    /* ... */
  }
# else
# error "BRANCH already defined"
# endif 
}

# include <stdio.h>
int main(void) {
  int bar = 3;
  increment(Int_T, &bar);
  printf("%dn", bar);
}

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