我正在使用 C 创建程序.但是，我需要使用许多具有仅用于 C++ 的 API 的库.那么，我可以在 C++ 中创建一个共享对象，然后使用 C 访问它的功能吗?
I am creating programs using C. However, I require to use a lot of libraries that have API's only for C++. So, is it possible that I can create a shared object in C++ and then access its functionality using C?
如果无法将这些代码接口，我该如何从 C++ 代码到 C 代码获取信息?我尝试从 C 调用 C++ 函数，但是当我包含 <string> 时，在链接过程中出现错误.所以当我从 C 调用 C++ 函数时，我应该只使用与 C 编译器兼容的代码吗?
If it is not possible to interface these codes, how do I get information from C++ code to C code?
I tried calling C++ functions from C, but I get errors during linking when I include <string>. So when I call C++ functions from C, should I only use that code which will be C compiler compatible?
#ifndef CPPFILE_H
#define CPPFILE_H
#ifdef __cplusplus
extern "C" {
#endif
extern int myfunction(const char *filename);
#ifdef __cplusplus
}
#endif
#endif
#include "cppfile.hpp"
#include <string>
int myfunction(const char *filename) {
String S(filename);
return 0;
}
#include "cppfile.hpp"
int main(int argc, char **argv)
{
int i = myfunction(argv[1]);
printf("%d
", i);
return 0;
}
gcc -c cmain.c
g++ -fPIC -shared -o cppfile.so cppfile.cpp
你想要更像这样的东西(这里我将使用一个更有意义的例子):
You want something more like this (and here I will use a slightly more meaningful example):
#ifndef ANIMAL_H
#define ANIMAL_H
#ifdef __cplusplus
class Animal {
public:
Animal() : age(0), height(0) {}
Animal(int age, float height) : age(age), height(height) {}
virtual ~Animal() {}
int getAge();
void setAge(int new_age);
float getHeight();
void setHeight(float new_height);
private:
int age;
float height; // in metres!
};
#endif /* __cplusplus */
#ifdef __cplusplus
extern "C" {
#endif
struct animal; // a nice opaque type
struct animal *animal_create();
struct animal *animal_create_init(int age, float height);
void animal_destroy(struct animal *a);
void animal_setage(struct animal *a, int new_age);
void animal_setheight(struct animal *a, float new_height);
int animal_getage(struct animal *a);
float animal_getheight(struct animal *a);
#ifdef __cplusplus
}
#endif
#endif /* ANIMAL_H */
#include "animal.h"
#define TO_CPP(a) (reinterpret_cast<Animal*>(a))
#define TO_C(a) (reinterpret_cast<animal*>(a))
void Animal::setAge(int new_age) { this->age = new_age; }
int Animal::getAge() { return this->age; }
void Animal::setHeight(float new_height) { this->height = new_height; }
float Animal::getHeight() { return this->height; }
animal *animal_create() {
animal *a = TO_C(new Animal);
return a;
}
animal *animal_create_init(int age, float height) {
animal *a = TO_C(new Animal(age, height));
return a;
}
void animal_destroy(animal *a) {
delete TO_CPP(a);
}
void animal_setage(animal *a, int new_age) {
TO_CPP(a)->setAge(new_age);
}
void animal_setheight(animal *a, float new_height) {
TO_CPP(a)->setHeight(new_height);
}
int animal_getage(animal *a) {
TO_CPP(a)->getAge();
}
float animal_getheight(animal *a) {
TO_CPP(a)->getHeight();
}
#include "animal.h"
#include <stdio.h>
int main()
{
// 6'0" 25yo (perhaps a human? :P)
struct animal *a = animal_create(25, 1.83);
animal_setage(a, 26); // birthday
printf("Age: %d
Height: %f", animal_getage(a), animal_getheight(a));
animal_destroy(a);
return 0;
}
#include "animal.h"
#include <iostream>
int main()
{
// 6'0" 25yo (perhaps a human? :P)
Animal* a = new Animal(25, 1.83);
a->setAge(26); // birthday
std::cout << "Age: " << a->getAge() << std::endl;
std::cout << "Height: " << a->getHeight();
delete a;
return 0;
}
因此，当您编译库时，您使用 C++ 编译器编译 animal.cpp.然后您可以使用 C 代码链接到它，并使用 animal_xxx 函数.
So when you compile the library, you compile animal.cpp with a C++ compiler. You can then link to it with C code, and use the animal_xxx functions.
注意struct animal 和Animal 的使用.Animal 是一个普通的 C++ 类型.这正是它的样子.struct animal 另一方面，是一种不透明"类型.这意味着您的 C 程序可以看到它在那里，并且可以拥有一个，但它不知道里面有什么.它只知道它有一个接受 struct animal* 的函数.
Note the use of struct animal and Animal. Animal is a normal C++ type. It's exactly what it looks like. struct animal, on the other hand, is an "opaque" type. That means that your C program can see it's there, and can have one, but it doesn't know what is inside it. All it knows is that it has a function that takes a struct animal*.
在一个真正的库中，你会希望有内存分配的自定义点.因此，假设这是库 libjungle，您可能至少需要具有合理默认值的 jungle_setmalloc 和 jungle_setfree.然后，您可以在 libjungle 的 C++ 代码中设置全局 new 和 delete 以使用这些用户定义的函数.
In a real library you will want to have customisation points for memory allocation. So assuming this is the library libjungle, you probably want at least jungle_setmalloc and jungle_setfree with sensible defaults. You can then set up the global new and delete in libjungle's C++ code to use these user-defined functions.
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