warning: variable set but not used [-Wunused-but-set-variable]

Question

I have been getting following warning while compiling the C source code in the gcc 4.6.1.

   warning: variable set but not used [-Wunused-but-set-variable]

I refered to this link Wunused but could get exactly what is causing this warning.Would anybody tell me in more details what is causing this warning and how can We get rid of it?

[EDIT] I have a following snippet of code. The compile shows the above mentioned warning. Could you please suggest me how can correct it?

   test_function(){
   BOOL BoolTest;
   BoolTest = test_fucntion2();

   #ifdef CHECK
   if (!BoolTest) {
   misc_StartErrorReport();
   misc_ErrorReport("\n test_function2: Input not indexed.\n");
   misc_FinishErrorReport();
          }
   #endif
   // 
    BoolTest is no more used below it.
   // } 

Solution 1

You need to include the preprocessor guards around the declaration and initialisation of BoolTest:

test_function()
{
#ifdef CHECK
    BOOL BoolTest = test_function2();
#else
    test_function2();
#endif


#ifdef CHECK
    if (!BoolTest) {
        misc_StartErrorReport();
        misc_ErrorReport("\n test_function2: Input not indexed.\n");
        misc_FinishErrorReport();
    }
#endif

(this assumes that you still want to call test_function2() even if CHECK is not defined, presumably for its side-effects - if not, then you don't need the #else section and you can combine the two #ifdef blocks into one).

Solution 2

Setting a variable is assigning it a value (maybe implicitly)

int main(void) {
    int local1, local2;
    local1 = 0; /* local1 set to 0 */
    local2 = 0; /* local2 set to 0 */
    return 0;
}

In the program above, both variables were set to a value but they weren't used. If I replace the second line with

    int local2 = local1;

now I have used the local1 variable -- and the warnings should be only 1.

To get rid of the warning, delete the assignment from your code. This may, in turn create other warnings ... :)

Solution 3

It means that you assign a value to a variable, but then you never read that value later in your code (hence the verbage, "set but not used"). For example:

int useful = 10;
int useless = 3;
if (useful) {
    //Do stuff
}

Notice that you give both useful and useless values, but you only read the value in useful. Usually, when I get this message it means that I forgot about a variable or found a way to inline a statement that no longer needs that variable.

Solution 4

With g++ 7.x and higher and clang++ 4.x and higher (using c++11 or higher), as well as with Visual Studio 2017 version 15.3 and later (available with /std:c++17), you can use the standarized [[maybe_unused]] attribute.

For example,

int main()
{
    int x [[maybe_unused]] = 5;
}

will not give a warning, not even with -Wunused-variable and the likes.

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