I am getting following runtime error.
anyfile.cpp (60) : E_FATAL : Could not
start process libprocbase_so.so
(/opt/company/processes/sharedbase_so.so:
undefined symbol:
_ZTV16CResourceManager)
I found that meaning of _ZTV16CResourceManager is virtual table of Resource Manager,
Any idea why it is coming? how to solve it?
The problem is that while loading the dynamic libraries (sharedbase_so.so) it has not been able to resolve the symbol. Try to locate in what library (if any) the symbol is defined and ensure that ld can load it (add it to the path). If it is not present in any dynamic library, go back to the object files and try to determine where it is and why it did not make it to the binary.
A common situation where this can be an issue is with classes that are mostly defined inlined, like:
struct test {
virtual void foo() {}
//virtual void bar();
};
If the line with bar is commented out, then the compiler will generate the vtable as a weak symbol in all translation units that include it. If bar is uncommented, the compiler will not generate the vtable in each translation unit, but only in the translation unit that defines test::bar. Make sure that even if you do not use test::bar, the object file gets linked in the library.
Where did you declare the class ResourceManager ?
It seems that this library is not loaded - when you use runtime linking, you have to load all dependencies by hand in the correct order.
Related
If I defined a function twice, I'll get a redefinition error message, but
I'm confused that redefinition happened in compile or link time?
and why you can override malloc in libc without a redefinition error?
You get a function redefinition error when you have two function with the same prototype or signature (function signature is made of the function name number of parameters and parameter types, does NOT include the return type).
This is a compile time error if the compiler see two functions with the same signature:
int foo(int a);
double foo(int b);
Why you can override function calls in libraries? Let's look at how the code is build into an executable:
the compiler is called for each source file and outputs an object file: any function call which cannot be resolved (i.e. calling a function in a different file) is an external symbol which the linker will have to resolve.
the linker take all the object files and tries to resolve all the symbols; but it does this on a first come first served manner. For a external symbol it will consider the first definition it finds and not worry about the fact that there may be more definitions of the same symbol available.
So, the linker actually allows you to override a function's behavior. And it all depends on the order the files are linked - the first function definition it finds is the one used to resolve the symbol.
Hope this sheds some light on the matter.
Either or both. It can also result from the programmer editing source code or modifying build scripts.
"Redefinition" errors are emitted when the linker find two things (symbols) with the same name.
There are many reasons the linker might find two symbols with the same name. Some possibilities (there are many permutations) include;
An object file is specified twice in the link command. This usually results from an error in a build script.
Two object files that contain the same function definition. This results from code duplication - for example, a function definition being copied into different source files, which are then compiled and linked. It can also result from monkey business with the preprocessor (e.g. #includeing a file that contains the definition of a global variable by two source files).
The causes of things like the above are generally programmer error (e.g. supplying a bad linker command in a build script, misuse of the preprocessor, copying and pasting code between projects.
The reason functions in libraries like libc can often be "overridden" is that the linker typically only looks for symbols in libraries if it can't find them in object files. So, if an object file defines malloc() the linker will resolve all calls to that, and not attempt to resolve using the version in a library. This sort of thing is quite dangerous, because some other functions within libraries (e.g. even within libc) may resolve directly to the original malloc() (e.g. some calls may be inlined) which can unpredictable behaviours. This sort of behaviour is also linker specific: although this sort of thing is common with unix/linux variants, there are systems where the linkers do things differently.
For Example
#include <iostream>
int add(int x, int y);
int main()
{
cout << add(5, 5) << endl;
}
This would compile but not link. I understand the problem, I just don't understand why it compiles fine but doesn't link.
Because the compiler doesn't know whether that function is provided by a library (or other translation unit). There's nothing in your prototype that tells the compiler the function is defined locally (you could use static for that).
The input to a C or C++ compiler is one translation unit - more or less one source code file. Once the compiler is finished with that one source code, it has done its job.
If you call/use a symbol, such as a function, which is not part of that translation unit, the compiler assumes it's defined somewhere else.
Later on, you link together all the object files and possibly the libraries you want to use, all references are tied together - it's only at this point, when pulling together everything that's supposed to create an executable, one can know that something is missing.
When a compiler compiles, it generates the output (object file) with the table of defined symbols (T) and undefined symbols (U) (see man page of nm). Hence there is no requirement that all the references are defined in every translation unit. When all the object files are linked (with any libraries etc), the final binary should have all the symbols defined (unless the target in itself is a library). This is the job of the linker. Hence based on the requested target type (library or not), the linker might not or might give an error for undefined functions. Even if the target is a non-library, if it is not statically linked, it still might refer to shared libraries (.so or .dll), hence if on the target machine while the binary is run, if the shared libraries are missing or if any symbols missing, you might even get a linker error. Hence between compiler, linker and loader, every one is trying to best provide you with the definition of every symbol needed. Here by giving declaring add, you are pacifying the compiler, which hopes that the linker or loader would do the required job. Since you didnt pacify the linker (by say providing it with a shared library reference), it stops and cribs. If you have even pacified the linker, you would have got the error in the loader.
This may be minor, but I'm curious about the reason.
This comes from a practice code of my friend:
#include <iostream>
using namespace std ;
extern int* PPPP;
void main(){
cout<<"*PPPP"<<*PPPP<<endl;
}
By mistake, the PPPP is actually declared nowhere.
But curiously we can compile this into a static lib.
However, we can't make this into a dll, there are link errors (unresolved external sysmbol pppp)
We are guessing it's because that when making a static lib, the name PPPP (though extern) do has a space in memory anyhow, so, no problem occur in this.
We are not sure about this at all. We hope to hear some more and accurate information about this.
Thanks in advance.
A static library is intended to be linked to another set of files, so it can contain undefined symbols as these would be resolved at a later stage (or not, in which case you get a linker error).
However a DLL, just like an executable, needs to be fully linked and so can't contain any undefined references.
I'm guessing that when it's made into a static library, the linker assumes that any unresolved symbols will be available when fully linked.
If you link that static library to some program without a symbol called PPPP defined, it'll fail with a linker error.
When you say:
extern int* PPPP;
you are promising the compiler that PPPP is located in another translation unit.
The linker will try to find PPPP in the object files and libraries it's given and, if it can't, it'll issue an error.
Description :
a. Class X contains a static private data member ptr and static public function member getptr()/setptr().
In X.cpp, the ptr is set to NULL.
b. libXYZ.so (shared object) contains the object of class X (i.e libXYZ.so contains X.o).
c. libVWX.so (shared object) contains the object of class X (i.e libVWX.so contains X.o).
d. Executable a.exe contains X.cpp as part of translation units and finally is linked to libXYZ.so, libVWX.so
PS:
1. There are no user namespaces involved in any of the classes.
2. The libraries and executable contain many other classes also.
3. no dlopen() has been done. All libraries are linked during compile time using -L and -l flags.
Problem Statement:
When compiling and linking a.exe with other libraries (i.e libXYZ.so and libVWX.so), I expected a linker error (conflict/occurance of same symbol multiple times) but did not get one.
When the program was executed - the behavior was strange in SUSE 10 Linux and HP-UX 11 IA64.
In Linux, when execution flow was pushed across all the objects in different libraries, the effect was registered in only one copy of X.
In HPUX, when execution flow was pushed across all the objects in different libraries, the effect was registered in 3 differnt copies of X (2 belonging to each libraries and 1 for executable)
PS : I mean during running the program, the flow did passed thourgh multiple objects belonging to a.exe, libXYZ.so and libVWX.so) which interacted with static pointer belonging to X.
Question:
Is Expecting linker error not correct? Since two compilers passed through compilation silently, May be there is a standard rule in case of this type of scenario which I am missing. If so, Please let me know the same.
How does the compiler (gcc in Linux and aCC in HPUX) decide how many copies of X to keep in the final executable and refer them in such scenarios.
Is there any flag supported by gcc and aCC which will warn/stop compilation to the users in these kind of scenarios?
Thanks for your help in advance.
I'm not too sure that I've completely understood the scenario. However,
the default behavior on loading dynamic objects under Linux (and other
Unices) is to make all symbols in the library available, and to only use
the first encountered. Thus, if you both libXYZ.so and libVWX.so
contain a symbol X::ourData, it is not an error; if you load them in
that order, libVWX.so will use the X::ourData from libXYZ.so,
instead of its own. Logically, this is a lot like a template definition
in a header: the compiler chooses one, more or less by chance, and if
any of the definitions is not the same as all of the others, it's
undefined behavior. This behavior can be
overridden by passing the flag RTLD_LOCAL to dlopen.
With regards to your questions:
The linker is simply implementing the default behavior of dlopen (that which you get when the system loads the library implicitely). Thus, no error (but the logical equivalent of undefined behavior if any of the definitions isn't the same).
The compiler doesn't decide. The decision is made when the .so is loaded, depending on whether you specify RTLD_GLOBAL or RTLD_LOCAL when calling dlopen. When the runtime calls dlopen implicitly, to resolve a dependency, it will use RTLD_GLOBAL if this occurs when loading the main executable, and what ever was used to load the library when the dependency comes from a library. (This means, of course, that RTLD_GLOBAL will propagate until you invoke dlopen explicitly.)
The function is "public static", so I assume it's OOP-meaning of "static" (does not need instance), not C meaning of static (file-static; local to compilation unit). Therefore the functions are extern.
Now in Linux you have explicit right to override library symbols, both using another library or in the executable. All extern symbols in libraries are resolved using the global offset table, even the one the library actually defines itself. And while functions defined in the executable are normally not resolved like this, but the linker notices the symbols will get to the symbol table from the libraries and puts the reference to the executable-defined one there. So the libraries will see the symbol defined in the executable, if you generated it.
This is explicit feature, designed so you can do things like replace memory allocation functions or wrap filesystem operations. HP-UX probably does not have the feature, so each library ends up calling it's own implementation, while any other object that would have the symbol undefined will see one of them.
There is a difference beetween "extern" symbols (which is the default in c++) and "shared libary extern". By default symbols are only "extern" which means the scope of one "link unit" e.g. an executable or a library.
So the expected behaviour would be: no compiler error and every module works with its own copy.
That leads to problems of course in case of inline compiling etc...etc...
To declare a symbol "shared library extern" you have to use a ".def" file or an compiler declaration.
e.g. in visual c++ that would be "_declspec(dllexport)" and "_declspec(dllimport)".
I do not know the declarations for gcc at the moment but I am sure someone does :-)
Under Solaris 10, I'm creating a library A.so that calls a function f() which is defined in library B.so. To compile the library A.so, I declare in my code f() as extern.
Unfortunately, I "forgot" to declare in A's makefile that it has to link with B.
However, "make A" causes no warning, no error, and the library A.so is created.
Of course, when executing A's code, the call of f() crashes because it is undefined.
Is there a way (linker option, code trick...) to make the compilation of library A fail ?
How can I be sure that all symbols refered to in library A are defined at compile time ?
Thanks for any suggestions.
Simplest way: add a "test_lib" target in the makefile, that will produce a binary using all the symbols expored from libraryA. (doesn't have to be anything meaningful... just take the address, no need to call the function or anything, it just needs to be referenced).
Tanks
I think I found something interesting and even simpler in the linker's manual (d'ho)
The -z defs option and the --no-undefined option force a fatal error if any undefined symbols remain at the end of the link. This mode is the default when an executable is built. For historic reasons, this mode is not the default when building a shared object. Use of the -z defs option is recommended, as this mode assures the object being built is self-contained. A self-contained object has all symbolic references resolved internally, or to the object's immediate dependencies.