带参数的Void *函数(Void* function with a parameter)

带参数的Void *函数(Void* function with a parameter)

对于我正在尝试编写的应用程序，我需要能够在interface编写GLEnable(GL_REPEAT)工作）。 一旦用户这样做，系统应该使用正确的参数调用该函数。

为此，我使用以下变量来获取正确的函数：

std::map<std::string, void(*)(GLenum)> glEnableDisable;

并通过以下方式存储数据：

glEnableDisable["glEnable"] = glEnable; glEnableDisable["glDisable"] = glDisable;

（注意，上面给出警告“'='不能从'void（__ stdcall *）（GLenum）'转换为'void（__ cdecl *）（GLenum）'”）

最后，通过以下方式调用该函数：

((void (*)(GLenum)) Main.glEnableDisable[SplittedUp[0]])(Parameter);

现在，我想知道我做错了什么。 我没有参数这个工作正常，但我真的需要参数。

提前致谢，

乔伊

For an application I'm trying to write I need to be able to write GLEnable(GL_REPEAT) in an interface (got this working). Once the user does that, the system should call the function with the correct parameter.

For this, I am using the following variable to get the correct function:

std::map<std::string, void(*)(GLenum)> glEnableDisable;

And storing data in it through:

glEnableDisable["glEnable"] = glEnable; glEnableDisable["glDisable"] = glDisable;

(Note that the above gives a warning "'=' cannot convert from 'void (__stdcall *)(GLenum)' to 'void (__cdecl *)(GLenum)'")

Finally, the function is called through:

((void (*)(GLenum)) Main.glEnableDisable[SplittedUp[0]])(Parameter);

Now, I wonder what I'm doing wrong. I had this working fine without parameters, but I really need parameters for this.

Thanks in advance,

Joey

最满意答案

改变这个：

std::map<std::string, void(*)(GLenum)> glEnableDisable;

对此：

std::map<std::string, std::function<void(GLenum)>> glEnableDisable;

并且再也不用担心召唤约定了。 std::function能够存储任何函数或可调用对象，无论其调用约定如何。

不需要演员：

Main.glEnableDisable[SplittedUp[0]])(Parameter);

Change this:

std::map<std::string, void(*)(GLenum)> glEnableDisable;

to this:

std::map<std::string, std::function<void(GLenum)>> glEnableDisable;

and never worry about calling conventions again. An std::function is able to store any function or callable object, whatever its calling convention is.

Cast is not needed:

Main.glEnableDisable[SplittedUp[0]])(Parameter);