使用QMetaType的反射(按名称进行类实例化)

Question

我想根据它的名称实例化一个类。我发现Qt允许QMetaType::create。
我还发现，我们必须注册类使用Q_DECLARE_METATYPE。但我做了所有需要做的(我认为)，但它没有工作。

这是我的基本类:

#ifndef     __BLOCK_HH__
# define    __BLOCK_HH__
# include   <QMetaType>
# include   <QObject>
# include   <string>
# include   <vector>
# include   "Util.hpp"
# include   "Common.hh"
class       Block
{
protected:
    std::string _name;
    uint32      _u32StartAddr;
    uint32      _u32EndAddr;
    uint8       *_pu8Content;
    bool        _bSure;
    bool        _bVirgin;
    std::string _type;
    std::vector<std::string>    _errorCauses;
    Block(const std::string&, uint32, uint32, uint8 *, const std::string&);
public:
    Block(const std::string& = "", uint32 = 0, uint32 = 0, uint8 * = 0);
    Block(const Block&);
    virtual ~Block();
    Block&  operator=(const Block&);
    void        setName(const std::string&);
    void        setStartAddr(uint32);
    void        setEndAddr(uint32);
    void        setContent(uint8 *);
    void        sure();
    void        notSure();
    void        erase();
    const std::string&  getName() const;
    uint32              getStartAddr() const;
    uint32              getEndAddr() const;
    uint8               *getContent() const;
    bool                isSure() const;
    bool                isVirgin() const;
    const std::string&  getType() const;
    void            addError(const std::string&);
    void            dumpError(std::ostream& = std::cerr) const;
    virtual void    dump(std::ostream& = std::cout) const;
    virtual void    parse();
    virtual bool    operator<(const Block&) const;
};
Q_DECLARE_METATYPE(Block)
#endif

这是继承类之一:

#ifndef     __CSS_HH__
# define    __CSS_HH__
# include   "Block.hh"
typedef struct
{
    uint32  boot_lma;
    uint32  boot_vma;
    uint32  boot_length;
    uint32  boot_entry;
    uint32  nb_of_param;
} t_css;
class       CSS : public Block
{
    uint32  _u32BootFlashAddr;
    uint32  _u32BootSdramAddr;
    uint32  _u32BootSize;
    uint32  _u32Bep;
    uint32  _u32CpuClock;
    uint32  _u32SdramSize;
    bool    _bEncram;
public:
    CSS(uint32 = 0, uint32 = 0, uint8 * = 0);
    virtual ~CSS();
    virtual void    dump(std::ostream& = std::cout) const;
    virtual void    parse();
    uint32  getBootFlashAddr() const;
    uint32  getBootSdramAddr() const;
    uint32  getBootSize() const;
    uint32  getBEP() const;
    uint32  getCpuClock() const;
    uint32  getSdramSize() const;
    bool    encramEnabled() const;
};
Q_DECLARE_METATYPE(CSS)
#endif

这就是我用来实例化的代码:

Block       *Mapper::constructBlock(const std::string& name, uint32 u32Start, uint32 u32End, uint8 *pu8Content)
{
    Block   *block = 0;
    int32   i32Id = QMetaType::type(name.c_str());
    if (i32Id != QMetaType::UnknownType)
    {
        block = static_cast<Block *>(QMetaType::create(i32Id));
        if (block)
        {
            block->setStartAddr(u32Start);
            block->setEndAddr(u32End);
            block->setContent(pu8Content);
            block->parse();
        }
    }
    return (block);
}

我在这里给的名字是Block类的"Block"和CSS类的"CSS"。但QMetaType::type总是返回QMetaType::UnknownType。
也许我忘记了注册课程的东西。你有什么主意吗?

谢谢

Answer 1

您还需要向qRegisterMetaType()注册该类型，以使该名称在运行时可用。Q_DECLARE_METATYPE使基于模板的Qt函数在编译时知道类型。