C++ GCC – How to Disable ‘May Be Used Uninitialized’ Warning for a Variable

c++compiler-warningsgcc

I'm getting this warning on a stack variable:

warning: object.member may be used uninitialized in this function

In this case I do not wish to force initialization to just to get rid of the warning as it consumes CPU cycles. The variable is a POD structure so a memset on it is not zero cost. I can verify that the variable is never used uninitialized, so I'd just like to suppress the warning for it.

In general I do want the warning, just not on this particular variable in this particular scenario. How can I suppress the warning?

Looks like the pragma diagnostics are the correct way to go but they require quite a recent version of GCC (4.6)

No acceptable solution prior that version is known.

Best Answer

The accepted answer has two big problems that requires more than a comment. First, it deactivates the warning for the whole file. If that pragma resides in a header, probably for more. Warnings are useful and if it is indeed a false positive, one should disable the warning for a bunch of code as small as possible.

Then the warning in the OP is "maybe uninitialized" which is deactivated by -Wmaybe-uninitialized, as opposed to -Wuninitialized.

#pragma GCC diagnostic push                             // save the actual diag context
#pragma GCC diagnostic ignored "-Wmaybe-uninitialized"  // disable maybe warnings
function() or int variable;                             // impacted section of code
#pragma GCC diagnostic pop                              // restore previous diag context

Related Solutions

C++ Warnings – How to Silence a Warning About Unused Variables

You can put it in "(void)var;" expression (does nothing) so that a compiler sees it is used. This is portable between compilers.

E.g.

void foo(int param1, int param2)
{
    (void)param2;
    bar(param1);
}

Or,

#define UNUSED(expr) do { (void)(expr); } while (0)
...

void foo(int param1, int param2)
{
    UNUSED(param2);
    bar(param1);
}

C++ GCC – Variable Not Detected as Not Used in GCC

In theory, the default constructor for std::vector<int> could have arbitrary side effects, so the compiler cannot figure out whether removing the definition of a would change the semantics of the program. You only get those warning for built-in types.

A better example is a lock:

{
    lock a;
    // ...
    // do critical stuff
    // a is never used here
    // ...
    // lock is automatically released by a's destructor (RAII)
}

Even though a is never used after its definition, removing the first line would be wrong.

Best Answer

Related Solutions

C++ Warnings – How to Silence a Warning About Unused Variables

C++ GCC – Variable Not Detected as Not Used in GCC

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