Why Cast a Function's Return Value to `void`?

Why Explicitly Ignoring Return Values with Void Casting?

In programming, occasionally you may encounter code like this:

int fn();

void whatever()
{
    (void) fn();
}

You might wonder why the return value of fn() is cast to void.

Motivation

The primary reason for this cast is to explicitly indicate that the return value is being intentionally ignored. It serves as a signal to other developers that this function's purpose is to perform an action and not retrieve data, even though it returns something.

Handling Potential Errors

In some cases, casting to void can ensure that error codes are handled appropriately. By explicitly ignoring the return value, the code can avoid potential runtime errors or unexpected behavior caused by unhandled return values.

Style Considerations

While it's not a strict requirement, some coders prefer to cast unused return values to void in certain situations:

• When reviewing code standards or writing one, it's advisable to exempt calls to overloaded operators that don't use function call notation from the need for explicit casting.
• Others opt for using C-style casts ((void)) in this context, considering the full static cast notation to be unnecessary.

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