我目前在访问 dll 文件中的方法时遇到问题。我试图访问的方法是类型。
int dstoch (float,float,float,float,float,float,float,float,float);
这是我正在使用的代码
typedef int (*LPMyfunct)(float,float,float,float,float,float,float,float,float);
HINSTANCE hDLL = NULL;
LPMyfunct lpdstoch;
hDLL = LoadLibrary("c:\\myfile.dll");
if(hDLL!=NULL)
{
std::cout << "Library loaded \n";
lpdstoch = (LPMyfunct)GetProcAddress((HMODULE)hDLL, "dstoch");
int res = LPMyfunct(1,2,3,4,5,6,7,8,9); //this is where I am getting an error
}
编译时错误状态:
a value of type LPMyfunct cannot be used to initialize an entity of type int
关于它为什么不接受 9 个参数的任何建议?