Warning in C++ QT creator Linux

shashi prasad

Hi,
i used function as

int fun(int x,int y){

return 0;
}

i did not use variable x and y.

but getting warning as
warning: unused parameter 'x' [-Wunused-parameter]

i don not want to remove this variable from this function.

please suggest me how to resolve this issue.

Thanks
Shashi

tomasz3dk

@shashi-prasad Q_UNUSED

joeQ

@shashi-prasad hi, friend, welcome.

you can reference Q_UNUSED

#define UNUSED(x) (void)x;

int fun(int x,int y){
UNUSED(x);
UNUSED(y);
return 0;
}

or just said

int fun(int ,int ){
return 0;
}
``

you can add x,y when you want to use x,y. at any time.

SGaist

Hi,

An alternative to the good solution proposed by @tomasz3dk: comment the name of the parameters.

Rohith

@shashi-prasad

Hi Shashi - prasad,

You are seeing the warning because, you are not using the variable parameter you are just passing value.
If you even print the value of variable with the help of cout or qDebug i hope you will not see the warning.

More over it's just a warning, most of the cases we can ignore Warnings.

SGaist

@Rohith never ignore warnings blindly. If there's a warning there's a reason for it and you should always do your best to fix them because some innocuous looking warnings might be source of strange errors.

See here for an example.

Rohith

@SGaist

More over it's just a warning, most of the cases we can ignore Warnings.

Thanks for sharing the link and my intention in mention the above words are not to totally ignore the warnings there are cases which can be ignored and which can't be ignored. I was unable to explain perfectly in post any way thanks for correcting my mistake.