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POWhy isn't my new operator called
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copied!<p>I wanted to see that a dynamically loaded library (loaded with dlopen etc.) really uses its own new an delete operators and not these ones defined in the calling program. So I wrote the following library.cpp</p> <pre><code>#include <exception> #include <new> #include <cstdlib> #include <cstdio> #include "base.hpp" void* operator new(size_t size) { std::printf("New of library called\n"); void *p=std::malloc(size); if (p == 0) // did malloc succeed? throw std::bad_alloc(); // ANSI/ISO compliant behavior return p; } void operator delete(void* p) { std::printf("Delete of library called\n"); std::free(p); } class Derived : public Base { public: Derived() : Base(10) { } }; extern "C" { Base* create() { return new Derived; } void destroy(Base* p) { delete p; } } </code></pre> <p>and compiled it with</p> <pre><code>g++ -g -Wall -fPIC -shared library.cpp -o library.so </code></pre> <p>or as Employed Russian suggested to try (but in the end nothing changed)</p> <pre><code>g++ -g -Wall -fPIC -shared -Wl,-Bsymbolic library.cpp -o library.so </code></pre> <p>The class Base is only holding an int value and a function get_value() to get this value. After that I wrote client.cpp like this</p> <pre><code>#include <exception> #include <new> #include <iostream> #include <cstdlib> #include <cstdio> #include <dlfcn.h> #include "base.hpp" void* operator new(size_t size) { std::printf("New of client called\n"); void *p=std::malloc(size); if (p == 0) // did malloc succeed? throw std::bad_alloc(); // ANSI/ISO compliant behavior return p; } void operator delete(void* p) { std::printf("Delete of client called\n"); std::free(p); } typedef Base* create_module_t(); typedef void destroy_module_t(Base *); int main() { void* handle = dlopen("./library.so", RTLD_LAZY); if (handle == NULL) { std::cout << dlerror() << std::endl; return 1; } create_module_t* create_module = NULL; void* func = dlsym(handle, "create"); if (func == NULL) { std::cout << dlerror() << std::endl; return 1; } else create_module = (create_module_t *)func; destroy_module_t* destroy_module = NULL; func = dlsym(handle, "destroy"); if (func == NULL) { std::cout << dlerror() << std::endl; return 1; } else destroy_module = (destroy_module_t *)func; Base* a = create_module(); std::cout << "Value: " << a->get_value() << std::endl; destroy_module(a); return 0; } </code></pre> <p>and compiled it with</p> <pre><code>g++ -Wall -g -o client -ldl client.cpp </code></pre> <p>Executing client I only get a "New of client called" and a "Delete of client called". Even if I use the compiler switch -Bsymbolic for the library like Employed Russian suggested.</p> <p>Now: What went wrong? I thought shared library are using their own new/delete and therefore you have to provide next to the factory create a destructor destroy in the library code.</p> <p>Supplementary question: Why do I need the destroy(Base* p) function? If this function only calls the delete-operator of the client I could also do it by myself, i.e "delete a" instead of destroy_module(a) in the next to last line.</p> <p>Answer I found: The library can also provide a new/delete-operator pair. So if I use first the library's new and later the client's delete I can probably step into a pitfall. Sadly until now I never saw my library using it's own new or delete... So the original question still isn't answered.</p> <p>Supplement: I'm only referring to the Linux platform.</p> <p>Edit: The important parts are in the comments to Employed Russian's Answer. So I'm giving the main clue in a nutshell: If one calls the gcc this way</p> <pre><code>g++ -Wall -g -fPIC -shared library.cpp -o library.so -Wl,-Bsymbolic </code></pre> <p>the library will use it's own new/delete operators. Otherwise results</p> <pre><code>g++ -Wall -g -fPIC -shared library.cpp -o library.so </code></pre> <p>in a library that's using the new/delete operators of the calling program. Thanks to Employed Russian!</p>

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