undefined reference to the shared library function - c++

I have implemented a shared library in Linux and try to test it, but I get an error "undefined reference to `CEDD(char*)'".
I use Eclipse with following parameters:
Path to include files (here is
everything ok)
Path to the library
and its name. Path is correct and the
name is WISE_C (full name:
libWISE_C.so)
My Code:
Test programm I use for tests:
#include <iostream>
#include <Descriptor.h>
int main() {
char* path = "/export/home/pdmazubi3/workspace/proj1/src/pic.jpg";
double * cedd = CEDD(path); ///// <-ERROR!
std::cout << "!!!Hello World!!!" << cedd[1];
return 0;
}
Header Descriptor.h:
double* CEDD(char* path);
A part of Descriptor.c with desirable function:
#include "Descriptor.h"
#include "highgui.h"
#include "cv.h"
double* CEDD(char* path)
{
IplImage* srcImg;
IplImage* ImageGrid;
...
}
What I am doing wrog? I have raed a lot of articles in the Internet but I didn't found a solution.
undefined reference to `CEDD(char*)' is a compiler or linker error?

It's a linker error (although I don't think it usually includes the 'char*' bit), so it seems that it either cannot find your library or the library does not contain the function. The latter might also mean that it does contain the actual function, but with a different name; make sure both projects a compiled as C and not C++.
Edit: I missed that you program is C++ (which also explains the more detailed linker message). In this case you should add extern "C" (conditionally, so it is only used when using C++) to the declaration of CEDD if the library is in C.

Are you sure that you linked the object code generated for descriptor.c when building?

Related

Obtaining filename at runtime for a shared library c++

I have some code that is compiled as a shared library and used with a universal driver, which can be used with other shared libraries that are specific to a particular application.
My question pertains to obtaining some sort of indicator of the name of the binary containing a code that lives in that shared library.
For example, let's say I have 3 files, the first is driver.cpp, the universal driver:
#include "interface.h"
#include <stdio.h>
int main(int argc, char *argv[]) {
//perform a function from the shared library
std::cout << foobar() << std::endl;
}
The second is sharedlibrary.cpp, the specific implementation for one case of many:
#include "interface.h"
char* foobar() {
return x;
}
Where x is some indicator that this function is defined in sharedlibrary.cpp, or that this function is linked from sharedlibrary.so, or the current stack frame is using the specific binary rather than just being included in driver.cpp.
The last file is interface.h, which provides the interface to the library via extern "C"
extern "C" {
char foobar();
}
I would like to reiterate, for clarity, that I am looking for some indication that this function is being linked from sharedlibrary.so. Many solutions looking for runtime filenames give the executable name using either argv[0] or readlink(), but I have no control over the actual naming of driver.cpp or its executable name. Rather, I can distribute sharedlibrary.so, and would like to be able to use its name from within itself, if possible.
If it helps, I know that a microsoft-specific solution could be to use AfxGetApp()->m_pszAppName to obtain the DLL name. However, I am looking for a linux solution that does not necessarily need to be portable.
EDIT: I do not know or control the names of driver.cpp, sharedlibrary.cpp, or sharedlibrary.h at compile time. I wish to discover the name of sharedlibrary.cpp at run time.
The updated sharedlibrary.cpp with x replaced with the solution looks like this
#include "interface.h"
#include <dlfcn.h>
void func() {
//Some function that is defined in sharedlibrary.cpp
}
char* foobar() {
Dl_info DlInfo;
if(!dladdr((void*)func, &DlInfo)) {
return "default_name";
}
return DlInfo.dli_fname;
}
Obtaining filename at runtime for a shared library c++
My question pertains to obtaining some sort of indicator of the name of the binary containing a code that lives in that shared library.
You can use int dladdr(void *addr, Dl_info *info. It fills a following structure for you:
typedef struct {
const char *dli_fname; /* Pathname of shared object that contains address */
void *dli_fbase;
const char *dli_sname;
void *dli_saddr;
} Dl_info;
You can pass the address of a function exported by the shared library as the argument addr. Or within such function, you could use the instruction pointer value of the current stack frame - if you know how to obtain it.
I believe you must link with the libdl library.
You can use the buildsystem to generate the dynamic library name for linking and preprocess that inside of a header with a function that return a defined macro, in cmake you can see how to do that here.
Then you use the configured-file to return the defined value in a function that's exported from within the dll.
#include "library_name_macro.h"
auto __dllexport libraryName() -> std::string { return LIBRARY_NAME_MACRO; }
I hope, I have understood your question correctly. I hope my answer helps. You know the shared library name, you link that shared library to your program, Later in run time you want to figure out whether a particular function is present in library or not and this logic should be part of shared library itself.
Let's take an example that you have shared library called librandom.so, You have linked this library to your application. You can implement the following function in a librandom.so library, You can pass function name which you want to check whether it is present or not. I have not tested this code, there may be errors. The idea I am proposing is library loads itself again to check whether the function is present when this function is called. May not be ideal method but should serve your purpose.
int isFuncPresent(char funcName[])
{
int isFuncFound = 1;
void *lib_handle;
int x;
char *error;
lib_handle = dlopen("librandom.so", RTLD_LAZY);
if (!lib_handle)
{
fprintf(stderr, "%s\n", dlerror());
isFuncFound = 0;
}
fn = dlsym(lib_handle, funcName);
if ((error = dlerror()) != NULL)
{
fprintf(stderr, "%s\n", error);
isFuncFound = 0;
}
dlclose(lib_handle);
return isFuncFound;
}

How do you link a C++ static library to a C program?

I have the following c++ program:
Client.h
#ifndef Client_Client_h
#define Client_Client_h
#include "Client.h"
class Client {
public:
void f1();
void f2();
};
#endif
Client.cpp
#include <iostream>
#include <stdlib.h>
using namespace std;
#include "Client.h"
void Client::f1(){
cout << "Client.f1()" << endl;
}
void Client::f2() {
cout << "Client.f2()" << endl;
}
compiling the above in XCode 4.3 gives me a static library file called:
libClient.a
Separately, I have a main.c
#include <stdio.h>
//
//using namespace std;
int main(){
// how do I do something like: Client c; c.f1(); c.f2();
// and actually get output ?
printf("hello\n");
return 0;
}
What's steps do I need to take in order to invoke f1() and f2() ? How do I use GCC to link the static library properly?
So far I have tried:
gcc -lClient.a main.c
which gives me :
ld: library not found for -lClient.a
collect2: ld returned 1 exit status
This isn't going to work, or at least is not going to be portable. The one really really obvious thing to do is to make your program C++ so you can access those features.
You can't "natively" use C++ code from C code, for obvious reasons. You don't have access to object-oriented features, so a ton of stuff isn't going to work: constructors, destructors, move/copy semantics and virtual inheritance are probably the biggest things that you'll miss. (That's right: you won't be able to create or destroy objects correctly, unless they have trivial constructors and destructors.)
You'll also run into linkage issues: C++ function names are mangled into a mess that includes their parameter types and return types and classes, which will look like __1cGstrcpy6Fpcpkc_0_. It would be technically feasible to declare the mangled names of the functions in C to use them, or use dlsym to get a pointer to them, but that's plain silly. Don't do that.
If you need to create a function in C++ that needs to be callable from C, you can specify it as extern "C" and its name won't be mangled, and it will be accessible from C, and it will be itself able to use C++ features:
extern "C" void Foo()
{
std::string hello = "Hello world!";
std::cout << hello << std::endl;
}
You will then need to declare it on the C side of your program like this:
void Foo();
And you'll be able to call it.
It's possible for you to wrap all your C++ calls that you want to expose to your C program in extern "C" functions, and return a pointer to your type and deal with it that way, but it's going to get annoying very quickly. You really should just use C++.
As far as linking with the static library is concerned, in Xcode, go to your project settings, pick your target, go to the Build Phases tab, unfold the "Link Binary With Libraries" section, and drop your .a file there.

Do I need to build TinyThread++?

I've extracted the sources from the zip file at the website and put them in Code::Blocks' 'include' folder, but even then it cannot compile the provided 'hello.cpp' example.
(For reference:)
#include <iostream>
#include <tinythread.h>
using namespace std;
using namespace tthread;
// This is the child thread function
void HelloThread(void * aArg)
{
cout << "Hello world!" << endl;
}
// This is the main program (i.e. the main thread)
int main()
{
// Start the child thread
thread t(HelloThread, 0);
// Wait for the thread to finish
t.join();
}
And these are the following errors:
|41|undefined reference to `tthread::thread::thread(void (*)(void*), void*)'|
|44|undefined reference to `tthread::thread::join()'|
|44|undefined reference to `tthread::thread::~thread()'|
|44|undefined reference to `tthread::thread::~thread()'|
The same thing happens with wxDev-C++. Am I missing something; like, do I need to build the libraries or anything? If so, how?
From the readme.txt inside the archive:
Using TinyThread++
To use TinyThread++ in your own project, just add tinythread.cpp and
tinythread.h to your project. In your own code, do:
#include <tinythread.h>
using namespace tthread;
If you wish to use the fast_mutex class, inlude fast_mutex.h:
#include <fast_mutex.h>
Just including the header leads to unresolved symbols because the .cpp does not get compiled.
TinyThread is tiny.
In additional to including the headers, you just add the TinyThread.cpp to your project or ensure that it builds as part of your project.
The particular error happened to me in VC++ until I added the CPP file to my project.
The thread class and methods were not getting compiled otherwise.

Using C Headers in C++

I have searched the googles for this and have found that you use
extern "C" {
#include "header.h"
}
To include a C library inside of a C++ library... however, when I do this. The C++ program seems to pick up all my #defines and struct definitions but none of the function declarations leaving me with undefined reference to `function'.
Here is a minimal amount of src I am using.
json.h
//json.h
typedef struct json_object json_object;
struct json_object {
char key[15][50];
int size;
char value[15][50];
};
void json_parseText(char * text, struct json_object *jo);
test.cpp
//test.cpp
extern "C" {
#include "json.h"
}
int main() {
struct json_object jo;
char * keyVal;
char * text = "{ \"MsgType\": \"article\" }";
json_parseText(text, &jo);
}
g++ yields the following:
test.cpp:(.text+0x2c): undefined reference to `json_parseText'
notice that it is not complaining about the struct definition, so it seems like it got that from the header file. But not the function. This baffles me. I have never used C++ before now, but for my testing framework it must be in C++. Let me know if you have any thoughts on how to fix this. Thanks.
That's a link-time error. In other words, your C++ compiler picked your header all right; you just forgot to link with your library.
Right now, you tell your compiler that such functions and structures exist, but not where it can find them.
For a shared library (.so), you'll have to pass -l[lib name] to G++; you might also have to specify additional folders in the library search path, as -l requires a file name (without the extension) instead of a path. For a static library (.a), you'll have to include its path in the files to compile.

How to use 2 C libs that export the same function names [duplicate]

This question already has answers here:
Closed 13 years ago.
Duplicate of the following question: C function conflict
Hi,
in my current project I have to use some kind of interface lib. The function names are given by this interface, what this functions do is developers choice. As far as I can tell a project shall use this functions and when it comes to compiling you choose the lib and with it the functionality. What I try to do is to use an existing lib and my lib at the same time by wrapping the other and call it in mein functions:
otherlib:
int function1 (int a) {
// do something
}
mylib:
int function1 (int a) {
//my code here
otherlib::function1(a);
}
Problem is I don't have access to the other lib and the other lib doesn't have any namespaces. I already tried
namespace old {
#include "otherlib.h"
}
and then call the old function by old::function1 in my function. This works as long as it's only header file. The lib exports it's symbol back into global space. Also something like
namespace new {
function1 (int a) {
::function1(a);
}
}
didn't work. Last but not least I tried ifdefs and defines suggested here
but I wasn't successful.
Any ideas how to solve this? Thanks in advance.
EDIT: I neither have access to the old lib nor the project both libs shall be used in.
EDIT2: at least the old lib is a static one
Namespaces in C solved using library names prefixes like:
libfoo --> foo_function1
libbar --> bar_function1
These prefixes are actual namespaces. so if you write libbar
int bar_function1(int a) {
function1(a);
}
This is the way to solve problems.
C has namespaces --- they just called prefixes ;)
Another option is to do various dirty tricks with dynamic loading of libraries like:
h1=dlopen("libfoo.so")
foo_function1=dlsym(h1,"function1")
h2=dlopen("libbar.so")
bar_function1=dlsym(h2,"function1")
It seems as if the other lib is C and your code is C++. You can be running into a mangling problem (C++ compilers mangle the symbols -- add extra stuff in the symbol name do differentiate overloads and the like).
If the library is pure C you can try:
extern "C" { // disable mangling of symbol names in the block
#include "otherlib.h"
}
namespace new_lib { // new is a reserved word
int function1( int a ) {
::function1(a);
}
}
I have not tried it. Also consider providing the error messages you are getting.
Another option would be (if the library is dynamic) dynamically loading the lib and calling the function. In linux (I don't know about windows) you can use dlopen to open the library, dlsym to obtain the symbol and call it:
// off the top of my head, not tried:
int function1( int a )
{
int (*f)(int); // define the function pointer
void * handle = dlopen( "library.so" );
f = dlsym( handle, "function1" );
f( a ); // calls function1(a) in the dynamic library
}
In this case, as you are not linking against the library you won't get a symbol conflict, but then again, it is only valid for dynamic libraries and it is quite cumbersome for regular usage.
UPDATE
If your users will not use 'otherlib' directly (they won't include their headers) and they will be only C++, then the first approach could be possible (even if horrible to read):
// newlib.h
namespace hideout {
int f( int a );
}
using namespace hideout; // usually I would not put this on the header
// newlib.cpp
extern "C" { // if otherlib is C, else remove this line
#include "otherlib.h"
}
namespace hideout {
int f( int a ) { return ::f( a*2 ); }
}
// main.cpp
#include "newlib.h"
int main()
{
std::cout << f( 5 ) << std::endl;
}
How does it work? User code will only see a declaration of function1 (in the example f()) as they are not including otherlib.h. Inside your compilation unit you see the two declarations but you differentiate through the use of the namespace. The using statement in the header does not bother you as you are fully qualifying in your cpp. The user main.cpp will include only your header, so the compiler will only see hideout::f, and will see it anywhere due to the using statement. The linker will have no problem as the C++ symbol is mangled identifying the real namespace:
// g++ 4.0 in macosx:
00002dbe T __ZN7hideout9function1Ei // namespace hideout, function1 takes int, returns int
00002db0 T _function1
If user code will include both your header and otherlib.h then it will have to qualify which function it wants to call.
If you're really desperate, you could write a wrapper library that uses namespaces or prefixes or allows for the dlsym trick. This wrapper library would need to be dynamically linked (to avoid symbol conflicts). The dynamic library could then safely have the old static library embedded in it. Just make sure you don't export the symbols from the static library when making the dynamic wrapper library.
You can't resolve this at link time, so you'll need to resolve it at runtime via dynamic libraries. The symbol for those functions is essentially baked it once the library has been generated. If two libraries export the same symbol, they cannot both be linked with statically.