使用g ++进行动态共享库编译 [英] Dynamic Shared Library compilation with g++

问题描述

我正在尝试从 Program-Library-HOWTO中编译以下简单的DL库示例代码与g ++。这只是一个例子，所以我可以学习如何使用和写共享库。我正在开发的库的实际代码将用C ++编写。

I'm trying to compile the following simple DL library example code from Program-Library-HOWTO with g++. This is just an example so I can learn how to use and write shared libraries. The real code for the library I'm developing will be written in C++.

#include <stdlib.h>
#include <stdio.h>
#include <dlfcn.h>

int main(int argc, char **argv) {
    void *handle;
    double (*cosine)(double);
    char *error;

    handle = dlopen ("/lib/libm.so.6", RTLD_LAZY);
    if (!handle) {
        fputs (dlerror(), stderr);
        exit(1);
    }

    cosine = dlsym(handle, "cos");
    if ((error = dlerror()) != NULL)  {
        fputs(error, stderr);
        exit(1);
    }

    printf ("%f\n", (*cosine)(2.0));
    dlclose(handle);
}

如果我用gcc编译程序工作正常。

If I compile the program with gcc it works fine.

gcc -o foo foo.c -ldl

当我将文件名和编译器更改为以下

When I change the filename and compiler to the following

g++ -o foo foo.cpp -ldl

我得到以下错误：

foo.cpp：16：错误：从'void *'到'double（*）（double）'的无效转换

foo.cpp:16: error: invalid conversion from 'void*' to 'double (*)(double)'

我理解（我我的理解，纠正我，如果这是错误的），我不能做一个隐式转换从C ++中的void指针，但C让我和这个是为什么上面的代码将使用gcc编译而不是使用g ++。所以我试图通过改变上面的行16显式转换：

I understand (I think I understand, correct me if this is wrong) that I can't do an implicit cast from a void pointer in C++, but C lets me, and this is why the above code will compile using gcc but not using g++. So I tried an explicit cast by changing line 16 above to:

cosine = (double *)dlsym(handle, "cos");

有了这个，我得到以下错误：

With this in place, I get the following error:

foo.cpp：16：错误：无法在赋值

foo.cpp:16: error: cannot convert 'double*' to 'double (*)(double)' in assignment

这些问题可能与我自己对正确的C ++编码标准的一般忽略有关。任何人都可以指点一个关于开发使用C ++示例代码的Linux动态库的好教程。

These problems probably have more to do with my own general ignorance of proper C++ coding standards than anything else. Can anyone point me to a good tutorial on developing dynamic libraries for Linux that uses C++ example code?

推荐答案

void * 到任何指针类型（包括函数指针）; C ++需要显式转换。正如leiflundgren所说，你需要把 dlsym（）的返回值转换成你需要的函数指针类型。

C allows implicit casts from void * to any pointer type (including function pointers); C++ requires explicit casting. As leiflundgren says, you need to cast the return value of dlsym() to the function pointer type you need.

很多人发现C的函数指针语法尴尬。一个常见的模式是typedef函数指针：

Many people find C's function pointer syntax awkward. One common pattern is to typedef the function pointer:

typedef double (*cosine_func_ptr)(double);

您可以定义函数指针变量 cosine 作为您的类型的成员：

You can define your function pointer variable cosine as a member of your type:

cosine_func_ptr cosine;

使用类型而不是awkward函数指针语法进行转换：

And cast using the type instead of the awkward function pointer syntax:

cosine = (cosine_func_ptr)dlsym(handle, "cos");

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