What is the use of a statement with no effect in C++? [duplicate]

Question

In one library that I am using I saw this code:

template<typename T>
void f(SomeTemplatedClass<T> input)
{
    (void)input;
    ...
    use(input); // code that uses input
}

I have no idea what is the meaning of this code. If I remove the cast to void, I get a

statement has no effect 

warning in gcc. So I suppose someone did it purposefully and purposefully added the cast to get a rid of the warning.

Do you have any experience with a statement that has no effect, yet it is needed for some reason?

EDIT:

Is it safe to assume that this has nothing to do with templates? For example circumventing an old compiler bug or the like?

Answer 1

This construct is a common way of tricking the compiler into not emitting a warning for unused parameters. I have not seen it used for any other purpose.

(void)input;

While it is common, it is also a really bad idea.

• it is highly platform dependent -- it may work on one compiler and not another.
• it is unnecessary. There is always a better way to deal with unused parameters. The modern way is to simply omit the parameter name.
• it can get left behind if the code changes and the parameter is now used (as appears to be the case here).
• it can backfire. Some compilers may treat this as invalid.

According to the C++ standard N3936 S5.4/11:

In some contexts, an expression only appears for its side effects. Such an expression is called a discarded-value expression. The expression is evaluated and its value is discarded.

A compiler would be entitled to observe that there is no side-effect and therefore this construct deserves at least a warning. According to @chris, MSVC is one of those compilers.