This question already has answers here:
Why does "extern const int n;" not work as expected?
(5 answers)
Closed 4 years ago.
Consider the following a.cpp and b.cpp files:
ebra#him:/tmp$ cat a.cpp
const int i = 5;
ebra#him:/tmp$ cat b.cpp
int main()
{
extern int i;
return i;
}
ebra#him:/tmp$ g++ *.cpp
/tmp/ccqBWi4e.o: In function `main':
b.cpp:(.text+0x6): undefined reference to `i'
collect2: error: ld returned 1 exit status
The question is how I can use the i variable that is declared in a.cpp file inside b.cpp?
Note that
I added the keyword const inside b.cpp too, but nothing changed.
I have the same problem with static and static const variables too!
In C++, when you declare a variable to be const at namespace scope it automatically has internal linkage. Adding static will also yield in internal linkage with or without const
Therefore they are not available outside the translation unit, hence the linker error.
Put it in a header file, that way they both have the same value (different variables, same name, same value).
I think you can also override to make it external linkage, but this will get you nothing: no behaviour changes; No efficiency improvement, for int.
Related
This question already has answers here:
Why does const imply internal linkage in C++, when it doesn't in C?
(7 answers)
Closed 1 year ago.
I have this cpp file which contains 2 arrays:
const float arr[123]={/*..*/};
const float ave[213]={/*..*/};
And in my main.cpp I have:
int main()
{
extern float arr[123];
float temp[123];
for(unsigned i=0; i<123;i++)
temp[i]=arr[i];
return 0;
}
When I build my project I got an error saying that there’s no definition for arr.
It’s a linker error.
What could be the problem?
Your const variable has internal linkage by default.
You need to declare it as extern as well to counteract that.
I read some answers about this topic, but I am still not sure:
In C++ a global const variable definition is automatically static. However I can access it from another cpp-file through extern:
// module.cpp
const int i = 0;
and
// main.cpp
extern const int i;
int main ()
{
if (i > 10)
return 0;
else
return 1;
}
Why is this possible (accessing an object with internal linkage from another module)? Normally I should be forced to define i as extern const int i = 0 in module.cpp, to have an explicitely global but non-static const, or?
In comparison, this is not possible:
// module.cpp
static int i = 0;
and
// main.cpp
extern int i;
int main ()
{
i = 10; // but read-only access like (i > 10) would be possible!
return 0;
}
So is the answer that: yes, you can access internal linked objects from other modules, but only for read (so always with const)?
Edit:
Sorry, but I made a mistake: in my original code I just tried an expression without effect (in both examples):
extern const int i; // or extern int i for second example
int main ()
{
i>10;
return 0;
}
I thought, that it behaves the same, as if program flow or data was dependent of this expression, but in fact it does not! The compiler seems to just cut out this effectless expression, so that the linker does not see it! So all is alright: in the first example i must be indeed defined extern const int i = 0 in module.cpp, and in the second example i cannot be accessed at all (unless in effectless expression). Compiler is VC++2010.
Edit2:
However, now I don't understand why this is possible:
// module.cpp
extern const int i = 0;
and
// main.cpp
int i = 99;
int main ()
{
bool b = i>10;
return 0;
}
i have both external linkage. But no error. When in module.cpp I define int i = 0, then error (multiple symbols). Why?
As for me it looks like a compiler bug. Constant variable i is not defined It is only declared in main.cpp. As for variable i in module.cpp then it has internal linkage and shall not be accessible outside the module.
Relative to your addition to the original post then the compiler has nothing common with this situation. It is linker that checks whether there are duplicate external symbols. I think it decided that if one variable has qualifier const and other has no it then there are two different variables. I think that it is implementation defined whether the linker will issue an error or not. Moreover it can have some options that could control the behaviour of the linker in such situations.
This question already has answers here:
How do I use extern to share variables between source files?
(19 answers)
Closed 9 years ago.
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I'm writing little student project and stuck with the problem that I have a few global variables and need to use it in a few source files, but I get the error undefined reference to variable_name. Let's create three source files for example:
tst1.h:
extern int global_a;
void Init();
tst1.cpp:
#include "tst1.h"
void Init(){
global_a = 1;
}
tst2.cpp:
#include "tst1.h"
int main(){
Init();
}
When I compile and link, that's what I get:
$ g++ -c tst1.cpp
$ g++ -c tst2.cpp
$ g++ tst2.o tst1.o
tst1.o: In function `Init()':
tst1.cpp:(.text+0x6): undefined reference to `global_a'
collect2: error: ld returned 1 exit status
If I remove the extern statement, then I get the other problem, let me show:
$ g++ -c tst1.cpp
$ g++ -c tst2.cpp
$ g++ tst2.o tst1.o
tst1.o:(.bss+0x0): multiple definition of `global_a'
tst2.o:(.bss+0x0): first defined here
collect2: error: ld returned 1 exit status
I really need some variables to be global, for example my little project works with assembly code, and have a variables like string rax = "%rax %eax %ax %ah %al"; which should be referenced through different source files.
So, how to properly initialize the global variables?
You only declared the variable but not defined it. This record
extern int global_a;
is a declaration not a definition. To define it you could in any module to write
int global_a;
Or it would be better to define function init the following way
int Init { /* some code */; return 1; }
and in main module before function main to write
int global_a = Init();
tst1.cpp should read instead:
#include "tst1.h"
int global_a = 1;
void Init(){
}
You can also write the initializer line as:
int global_a(1);
Or in C++11:
int global_a{1};
A global should only be defined (i.e. written without the extern prefix) in one source file, and not in a header file.
you need to to add
#ifndef TST1_H
#define TST1_H
.....
#endif
to tst1.h. it included twice in tst2.cpp
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What is an undefined reference/unresolved external symbol error and how do I fix it?
I recently started working on an interpreter in C++, but I got annoyed that vectors or arrays could not be passed to external class methods no matter what I tried and so I deleted everything I had worked on. As it turns out, I can't pass even an int to another class. I decided to give C++ another chance before resorting to C or Java, but the compiler still doesn't work as I would expect. Maybe I'm forgetting something simple about C++, as I haven't used it in a while, but this seems simple enough. My problem is: I can't pass arguments to methods in other classes when they're not defined in the same file. Here's what I'm trying to do:
Main: main.cpp
#include "myclass.h"
int main() {
MyClass test;
int n = test.add(25, 30);
return n;
}
Header: myclass.h
class MyClass {
public:
int add(int a, int b);
};
Class implementation: myclass.cpp
#include "myclass.h"
int MyClass::add(int a, int b) {
return a + b;
}
Compiling this with g++ main.cpp yields
/tmp/ccAZr6EY.o: In function main':
main.cpp:(.text+0x1a): undefined reference toMyClass::add(int, int)'
collect2: error: ld returned 1 exit status
What the heck am I doing wrong? Also, the compiler yells at me for the same thing even if my functions aren't parameterized, so it must be a problem with the header.
Any help is much appreciated - thanks!
You need to compile both files
g++ main.cpp myclass.cpp
If you only compile main.cpp, the compiler finds the declaration of MyClass::add in your header but the linker later fails to find an implementation of MyClass::add to jump to.
Is there any chance that a function defined with static can be accessed outside the file scope?
It depends upon what you mean by "access". Of course, the function cannot be called by name in any other file since it's static in a different file, but you have have a function pointer to it.
$ cat f1.c
/* static */
static int number(void)
{
return 42;
}
/* "global" pointer */
int (*pf)(void);
void initialize(void)
{
pf = number;
}
$ cat f2.c
#include <stdio.h>
extern int (*pf)(void);
extern void initialize(void);
int main(void)
{
initialize();
printf("%d\n", pf());
return 0;
}
$ gcc -ansi -pedantic -W -Wall f1.c f2.c
$ ./a.out
42
It could be called from outside the scope via function pointer.
For example, if you had:
static int transform(int x)
{
return x * 2;
}
typedef int (*FUNC_PTR)(int);
FUNC_PTR get_pointer(void)
{
return transform;
}
then a function outside the scope can call get_pointer() and use the returned function pointer to call transform.
No, unless there's a bug in the compiler. Normally the static function code is not tagged with a name used for exporting the function in the object file, so it is not presented to the linker and it just can't link to it.
This of course only applies to calling the function by name. Other code within the same file can get the function address and pass it into a non-static function in another file and then the function from another file can call your static function.
It cannot be accessed outside a file by it's name. But, you can as well assign it to function pointer and use it wherever you want.
"Accessed"? It depends on what you mean by this term. I assume when you say "static function" you are talking about standalone function declared static (i.e. declared with internal linkage) as opposed to static class member functions in C++, since the latter are obviosly and easily accessible from anywhere.
Now, a standalone function declared static has internal linkage. It cannot be linked to from any other translation unit. Or, putting it differently, it cannot be referred to by name from any other translation unit. If that's what you meant by "access from outside the file scope", then no, it can't be done.
However, if the other translation units somehow get a pointer to that function (i.e. if you somehow allow that pointer to "leak" into the ouside world), then anybody can still call that function by making an idirect call and thus "access" it. For example, if you declare
static void foo_static(void) {
}
extern void (*foo_ptr)(void) = foo_static;
then in any other translation unit the user will be able to do
extern void (*foo_ptr)(void);
foo_ptr();
and the call will go to your foo_static function. I don't know if that kind of access qualifies as "access" in your question.
Following the standard, a static function cannot be accessed outside of the scope of the file by name because it is subject to internal linkage. It's name is not exported, and not provided to the linker. However, it can still be accessed and called by function pointer, like any other function.
Only with trickery. The function is generally not visible to the linker so it won't let you do it.
But, if you provide a function inside the same compilation unit (as the static function) which returns the address of that function:
In main.c:
#inclde <stdio.h>
int (*getGet7(void))(void);
int main (void) {
int (*fn)(void) = getGet7();
printf ("Result is: %d\n", fn());
return 0;
}
In hidden.c:
static int get7 (void) {
return 7;
}
int (*getGet7(void)) (void) {
return get7;
}
This will result in the static function get7 being called.
pax> gcc -o demo main.c hidden.c ; ./demo
Result is: 7
No, the purpose of the keyword static is to limit the scope of the function name to the file.