函数指针调用不编译 [英] Function pointer call doesn't compile

问题描述

我只是没有得到它，为什么行22未能编译？

  #include< stdexcept> 
 #include< dlfcn.h> 
 #includeLibrary.h
 
 int main（int argc，char * argv []）
 {
 try 
 {
 void * libHandle = 0; 
 
 libHandle = dlopen（libExpandableTestLibrary.so，RTLD_LAZY）; 
 if（！libHandle）
 throw std :: logic_error（dlerror（））; 
 
 std :: cout<< Libary优雅地打开< std :: endl; 
 
 void * fuPtr = 0; 
 fuPtr = dlsym（libHandle，createLibrary）; 
 if（！fuPtr）
 throw std :: logic_error（dlerror（））; 
 
库* libInstance = static_cast< Library *（）>（fuPtr）（）; 
 //教程：http://www.linuxjournal.com/article/3687 
 //教程代码：shape * my_shape = static_cast< shape *（）>（mkr）（）; 
 //编译器错误消息：Application.cpp：22：56：error：invalid static_cast从类型'void *'到类型'Library *（）'
 libInstance-> Foo 
 
 dlclose（libHandle）; 
 
} catch（std :: exception& ex）
 {
 std :: cerr< ex.what（）<< std :: endl; 
} 
} 
  

欢迎任何帮助
如果您需要

我认为 fuPtr 指向一个函数，该函数应该返回一个指向 Library 对象的指针（假定加载的名称为createLibrary ）。

在这种情况下，包含您的演员的行需要如下：

  Library * libInstance = reinterpret_cast< Library *（*）（）>（fuPtr）（）; 
  

I just dont get it, why line 22 is failing to compile?

#include <stdexcept>
#include <dlfcn.h>
#include "Library.h"

int main(int argc, char *argv[])
{
    try
    {
        void* libHandle = 0;

        libHandle = dlopen("libExpandableTestLibrary.so", RTLD_LAZY);
        if(!libHandle)
            throw std::logic_error(dlerror());

        std::cout << "Libary opened gracefully" << std::endl;

        void* fuPtr = 0;
        fuPtr = dlsym(libHandle, "createLibrary");
        if(!fuPtr)
                throw std::logic_error(dlerror());

        Library* libInstance = static_cast<Library* ()>(fuPtr)();
        // Tutorial: http://www.linuxjournal.com/article/3687
        // Tutorial Code: shape *my_shape = static_cast<shape *()>(mkr)();
        // Compiler error message:  Application.cpp:22:56: error: invalid static_cast from type ‘void*’ to type ‘Library*()’
        libInstance->Foo();

        dlclose(libHandle);

    } catch(std::exception& ex)
    {
        std::cerr << ex.what() << std::endl;
    }
}

Any help is welcome If you need additional information's just let me know.

解决方案

I take it that fuPtr points to a function that is supposed to return a pointer to a Library object (given the name loaded is "createLibrary").

In that case, the line including your cast needs to look like this:

Library* libInstance = reinterpret_cast<Library* (*)()>(fuPtr)();

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