使用 g++ 编译动态共享库 [英] Dynamic Shared Library compilation with g++
问题描述
我正在尝试从 编译以下简单的 DL 库示例代码使用 g++ 的程序库-HOWTO.这只是一个示例,因此我可以学习如何使用和编写共享库.我正在开发的库的真实代码将用 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
", (*cosine)(2.0));
dlclose(handle);
}
如果我用 gcc 编译程序,它可以正常工作.
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内容时
When I change the filename and compiler to the following
g++ -o foo foo.cpp -ldl
我收到以下错误:
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(我认为我理解,如果有错请纠正我)我不能从 C++ 中的 void 指针进行隐式转换,但 C 允许我这样做,这就是为什么上面的代码将使用 gcc 但不使用 g++ 编译.所以我通过将上面的第 16 行更改为:
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:不能在赋值中将 'double*' 转换为 'double (*)(double)'
foo.cpp:16: error: cannot convert 'double*' to 'double (*)(double)' in assignment
这些问题可能更多地与我自己对适当的 C++ 编码标准的普遍无知有关,而不是其他任何事情.谁能给我指出一个关于使用 C++ 示例代码为 Linux 开发动态库的好教程?
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
C 允许从 void *
到任何指针类型(包括函数指针)的隐式强制转换;C++ 需要显式转换.正如 leiflundgren 所说,您需要将 dlsym()
的返回值转换为您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.
很多人觉得 C 的函数指针语法很尴尬.一种常见的模式是 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);
您可以将函数指针变量 cosine
定义为您的类型的成员:
You can define your function pointer variable cosine
as a member of your type:
cosine_func_ptr cosine;
并使用类型而不是笨拙的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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