FIX: Linker Errors on Array of Constant Pointers to Constants

WINDOWS             | WINDOWS NT

The information in this article applies to:

• Microsoft Visual C++ for Windows, versions 1.0, 1.5, 1.51, 1.52
• Microsoft Visual C++, 32-bit Edition, version 1.0

SYMPTOMS

Building a project that includes the code listed in the "Sample Code" section of this article results in the following link error or warning:

For Visual C++ for Windows:

    MODULE2.OBJ(module2.cpp) : error L2025: char const __near*
         __near* __near str : symbol defined more than once
    LINK returned error code 2.

    module2.obj: warning LNK4006: xxxxx already defined in
         module1.obj; second definition ignored.

RESOLUTION

To work around the problem, explicitly declare the const as static, as in this example:

   static const char *const str[2] = { "string1", "string2" };

STATUS

Microsoft has confirmed this to be a bug in the Microsoft products listed at the beginning of this article. This bug was corrected in Visual C++, 32- bit Edition, version 2.0.

MORE INFORMATION

In C++, const objects do not have external linkage, unless they are explicitly declared extern. Const objects have internal linkage, meaning that the name of the object is local to the source file in which it is declared.

The following sample code has two source files and one header file. The array of const pointers to const chars is included twice in the program. However, because this is C++, const objects should have internal linkage, the arrays should only be visible in the source file they are declared in. This means that the linker should not fail with an L2025 error, or report the LNK4006 warning.

Sample Code

// Compile options needed: None

// Test.h

// Module1.cpp

#include "test.h"           // included test.h once in project.
#include <iostream.h>

void function1()

   cout << "In Function1" << endl;

// Module2.cpp

#include "test.h"          // included test.h twice in project
#include <iostream.h>

void function2()

  cout << "In Function2" << endl;

void main()

  function1();
  function2();

  return;