使用QMake构建时出现以下错误:
LNK2019:未解析的外部符号“public:void __cdecl TimerTodo :: notify(class TodoBaseTask *)”(?notify @ TimerTodo @@ QEAAXPEAVTodoBaseTask @@@ Z)在函数“private:void __cdecl TimerTodo :: timerOver(void)”中引用(?timerOver @TimerTodo @@ AEAAXXZ)LNK2019:未解析的外部符号“public:void __cdecl TimerTodo :: hasNotified(class TimerTodo *)”(?hasNotified @ TimerTodo @@ QEAAXPEAV1 @@ Z)在函数“private:void __cdecl TimerTodo”中引用:: timerOver(void)“(?timerOver @TimerTodo @@ AEAAXXZ)LNK1120:2个未解析的外部
这是我的 Headers :
#ifndef TIMERTODO_H
#define TIMERTODO_H
#include <QTimer>
class TodoBaseTask;
class TimerTodo : public QTimer
{
public:
TimerTodo(TodoBaseTask *timer);
void StartTimer();
private slots:
void timerOver();
signals:
void notify(TodoBaseTask *todo);
void hasNotified(TimerTodo *timer);
private:
TodoBaseTask *m_todo;
};
#endif // TIMERTODO_H
这是我的来源:
#include "timertodo.h"
#include "todobasetask.h"
TimerTodo::TimerTodo(TodoBaseTask *todo)
{
m_todo = todo;
connect(this, SIGNAL(timeout()), this, SLOT(timerOver()));
}
void TimerTodo::StartTimer()
{
QDateTime nextNotify = m_todo->getDeadLine().addDays(-1);
this->start(QDateTime::currentDateTime().msecsTo(nextNotify));
}
void TimerTodo::timerOver()
{
emit notify(m_todo);
emit hasNotified(this);
}
怎么解决?
1 回答
这在_1770833中解释:
(强调我的)
因此,您需要将此宏放在具有自己的信号或插槽的每个类中 .