gtk_container_get_children

Question

当在预设列表中打印工作站列表并且您处于列表的开头或末尾时,您会看到许多空项目.我创建了大约20个站的列表并尝试显示它们,转到列表的开头和结尾.我认为问题在于这段代码:

static void add_button_clicked_cb(GtkWidget *widget, gpointer data)
{
    preset *ps;
    gchar *buffer;
    GtkTreeIter iter = {0};
    GtkAdjustment* v_scb;
    GtkTreePath *path = NULL;
    GList* menuitems;
    GtkWidget *menuitem;

    ps = malloc(sizeof(preset));
    ps->title = g_strdup(_("unnamed"));
    ps->freq = rint(gtk_adjustment_get_value(adj)) / STEPS;
    settings.presets = g_list_append(settings.presets, (gpointer) ps);
    buffer = g_strdup_printf("%.2f", ps->freq);

    gtk_list_store_append(list_store, &iter);
    gtk_list_store_set(list_store, &iter, 0, ps->title, 1, buffer, -1);

    g_free(buffer);
    gtk_tree_selection_unselect_all(selection);

    v_scb = gtk_scrollable_get_vadjustment(GTK_SCROLLABLE(list_view));
    gtk_adjustment_set_value(v_scb, gtk_adjustment_get_upper(v_scb));

    if (main_visible) {
        gtk_combo_box_text_append_text(GTK_COMBO_BOX_TEXT(preset_combo), ps->title);
        mom_ps = g_list_length(settings.presets) - 1;
        preset_combo_set_item(mom_ps);

        menuitems = gtk_container_get_children(GTK_CONTAINER(tray_menu));
        menuitem = gtk_menu_item_new_with_label(ps->title); 

        gtk_menu_shell_insert(GTK_MENU_SHELL(tray_menu), menuitem, mom_ps);     
        g_signal_connect(G_OBJECT(menuitem), "activate", (GCallback)preset_menuitem_activate_cb, (gpointer)mom_ps);
        gtk_widget_show(menuitem);
    }

    buffer = g_strdup_printf("%d", g_list_length(settings.presets) - 1);
    path = gtk_tree_path_new_from_string(buffer);
    g_free(buffer);
    gtk_tree_view_set_cursor(GTK_TREE_VIEW(list_view), path, NULL, FALSE);
    gtk_tree_path_free(path);
}

你能建议我怎么纠正它？

Answer 1

在里面if (main_visible),你有:

menuitems = gtk_container_get_children(GTK_CONTAINER(tray_menu));

也就是说,您将值设置menuitems为该函数调用的返回值.但是,其余的代码实际上都没有对menuitems做任何事情.所以,编译器会给你那个警告.

修复它可能就像简单地删除menuitems一样简单:

gtk_container_get_children(GTK_CONTAINER(tray_menu));

那肯定会摆脱警告.但你必须考虑一下:也许你应该用该函数调用的返回值做一些事情,而你没有做任何事情的原因是你忘了(或者其他).