what is the reason behind getting the "invalid use of void expression" error in C?

Question

I have this piece of C code here :-

#include<stdio.h>
void message();
int main()
{
    message(message());
    return 0;
}
void message()
{
    printf("hello\n");
}

The compiler throws an error message reading " Invalid use of void expression" for the statement block "message(message))".

I expected the output to be two times the printf() statement as according to me, the statement message(message()); indicates that once the inner function call executes and the control returns to main and after that again the outer call executes. However I am getting the error message "Invalid use of void expression" error here.

I have read some explanations but still i'm unable to understand.

Answer 1

The return type of the function message is void

void message();

So calling the function passing an argument of the incomplete type void invokes an error.

message(message());

If you want to call the function twice then write

message(); 
message();

or as one expression

message(), message();

Pay attention to that it is better to declare the function like

void message( void );

providing to the compiler the function prototype.

If you want to call the function specifying its argument as a call of it itself then the function should be declared and defined the following way

const char * message( const char *s )
{
    puts( s );

    return s;
}

and the function can be called like

message( message( "hello" ) );