guidupas
14th February 2014, 11:46
Hello All!
I am trying to declare a signal as the code below shows:
bancodados.h
#ifndef BANCODADOS_H
#define BANCODADOS_H
#include <QObject>
class bancoDados : public QObject
{
Q_OBJECT
public:
explicit bancoDados(QObject *parent = 0);
signals:
void conectar();
public slots:
};
bancodados.cpp
#include "bancodados.h"
#include <QSqlDatabase>
bancoDados::bancoDados(QObject *parent) :
QObject(parent)
{
}
void bancoDados::conectar()
{
bool retorno = true;
emit retorno;
}
Well, when I run this code it show the error:
:-1: error: 1 duplicate symbol for architecture x86_64
But, if I change "void conectar();" from signals to public, it runs, but I need to it stays as a signal
Can anyone help me?
Thanks a lot.
I am trying to declare a signal as the code below shows:
bancodados.h
#ifndef BANCODADOS_H
#define BANCODADOS_H
#include <QObject>
class bancoDados : public QObject
{
Q_OBJECT
public:
explicit bancoDados(QObject *parent = 0);
signals:
void conectar();
public slots:
};
bancodados.cpp
#include "bancodados.h"
#include <QSqlDatabase>
bancoDados::bancoDados(QObject *parent) :
QObject(parent)
{
}
void bancoDados::conectar()
{
bool retorno = true;
emit retorno;
}
Well, when I run this code it show the error:
:-1: error: 1 duplicate symbol for architecture x86_64
But, if I change "void conectar();" from signals to public, it runs, but I need to it stays as a signal
Can anyone help me?
Thanks a lot.