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I am using the thread safe variable macros in the LabWindows/CVI environment and have observed that it is possible to get a pointer to a thread safe variable before it has been released. (from a previous request)

Because the data I am interesting in protecting is a struct, I am unable to explicitly set nesting level, so I assume that the nesting level stays at 0, i.e. that once a single thread safe pointer has been issued, a request for second will be denied until the first has been released. However, I have observed this not to be true while stepping through a debug session. Execution continues through the DefineThreadSafeVar(CLI, SafeCli); statement by continuing to use the F8 step-into key, and subsequent requests for a pointer to the thread-safe variable are granted without ever having released the original.

My expectations are that these macros should prevent access to a thread-safe variable once a pointer to it has been issued and not yet released.

Are my expectations incorrect?
Or have I implemented the calls incorrectly?

Here is my source code: 

#include <utility.h>

typedef struct  {
    int hndl;
    int connect;
    int sock;
}CLI;

DefineThreadSafeVar(CLI, SafeCli);

void func1(void);
void func2(void);

int main(void)
{
    InitializeSafeCli(); 
    func1();

    return 0;
}

void func1(void)
{
    CLI *safe;
    safe = GetPointerToSafeCli();//original issue
    safe->connect = 2;
    safe->hndl = 3;
    safe->sock = 4;
    func2();
    safe->connect;
    safe->hndl;
    safe->sock;
    ReleasePointerToSafeCli();
}

void func2(void)
{
    CLI *safe;
    safe = GetPointerToSafeCli();//request is granted. previous issue had not been released.  
                                 //shouldn't request have been denied ?
    safe->connect = 5;//variable is modified.
    safe->hndl = 6;
    safe->sock = 7;
}
ryyker
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    Is it possible that since you're calling `func2()` within the same thread it simply already has access to it? Understanding you may have more code beyond this example you're not running multiple threads in this case. `func2()` is in the same call stack since it's being called inside func1() and hence the same thread. – Talaria Jan 04 '16 at 14:35

1 Answers1

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In your case, you are calling func2() within func1() and subsequently is within the same call stack and thus the same thread. You're granted access because you are requesting the pointer from within the same thread that already has access to the pointer.

GetPointerToSafeCli() is a waiting call. Had it been called from thread A and then again in thread B before ReleasePointerToSafeCli() was called back in thread A, thread B would wait until the pointer was released before granting access.

LabWindows/CVI - Programming with DefineThreadSafeScalarVar

Talaria
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