What is purpose of using static_cast<void>()? [duplicate] - c++

Question

This question already has answers here:
what is the use of "static_cast<void>" in macro?
(3 answers)
Why cast unused return values to void?
(10 answers)
Closed 4 years ago.
static_cast<void>() is the 'C++ way' of writing void conversion
In the en.cppreference.com website mentioned as discards the value of the expression. In below link four points on Explanation section
http://en.cppreference.com/w/cpp/language/static_cast
Where and Why should we use static_cast<void>()? give a example..

Answer 1

This is a way to tell that it is ok for a variable to be unused suppressing corresponding compiler warning. This approach has been deprecated with introduction of [[maybe_unused]] attribute in C++17.

Answer 2

The usual purpose of casting to void is to “use” the result of a computation. In relatively strict build environments it is common to output warnings, or even errors, when a variable is declared, maybe even written to, but the result is never used. If, in your code, you know you do not need a result somewhere, you can use the static_cast<void> method to mark the result as discarded – but the compiler will consider the variable used then and no longer create a warning or error.
An example:
#include <iostream>
int myFunction() __attribute__ ((warn_unused_result));
int myFunction()
{
return 42;
}
int main()
{
// warning: ignoring return value of 'int myFunction()',
// declared with attribute warn_unused_result [-Wunused-result]
myFunction();
// warning: unused variable 'result' [-Wunused-variable]
auto result = myFunction();
// no warning
auto result2 = myFunction();
static_cast<void>(result2);
}
When compiled with g++ -std=c++14 -Wall example.cpp, the first two function calls will create warnings.
As VTT pointed out in his post, from C++17 you have the option of using the [[maybe_unused]] attribute instead.