Error: Invalid use of void expression

Question

#include <gtk/gtk.h>
#include <string>

using namespace std;

class WIN
{    
    protected:
    GtkWidget *window;

    public:

    GtkWidget* get_window(){ return window; }
    void set_window(GtkWidget* w){ window = w; }
    void set_title(string s) 
    {
        gtk_window_set_title (GTK_WINDOW(window), s.c_str());
    }

};

int main (int argc, char *argv[])
{
    /* Initialize GTK+ and all of its supporting libraries. */
    gtk_init (&argc, &argv);
    WIN obj1;
    obj1.set_window(gtk_window_new (GTK_WINDOW_TOPLEVEL));
    obj1.set_title("Hello World");
    GtkWidget *w = obj1.get_window();
    obj1.set_window(gtk_widget_show(w));

    /* Hand control over to the main loop. */
    gtk_main();
    return 0;
}

@Tomalak Geret'kal: Thanks for advices I will take care of these.. — Umair Khan, Mar 15 '11 at 20:26

score 2 · Accepted Answer · answered Mar 15 '11 at 19:19

2

Seems like gtk_widget_show() returns void. That's the void expression you are using in an invalid way.

http://library.gnome.org/devel/gtk/2.99/GtkWidget.html

answered Mar 15 '11 at 19:19

Bo Persson

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Erik · Answer 2 · 2011-03-15T19:21:41.723

1

gtk_widget_show() returns void.

You're calling obj1.set_window(void)

Change:

obj1.set_window(gtk_widget_show(w));

To

gtk_widget_show(w);

edited Mar 15 '11 at 19:21

answered Mar 15 '11 at 19:15

Erik

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1

In C++, `(void)` is unnecessary, it's just `()`. I edited your post accordingly. – Puppy Mar 15 '11 at 19:19
3

@DeadMG: Please don't. The void was there to illustrate what he was doing, passing void to a function that expected something else. – Erik Mar 15 '11 at 19:21
1

@DeadMG: It's also rude to edit things that aren't broken, especially when your edit itself is broken. – Lightness Races in Orbit Mar 15 '11 at 19:29

Error: Invalid use of void expression

2 Answers2

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