为什么std :: transform和类似函数将'for'循环增量转换为(void)?

为什么std :: transform和类似函数将'for'循环增量转换为(void)?

问题描述:

以下代码中(void)++ __ result 的目的是什么?

What is the purpose of (void) ++__result in the code below?

std的实现:: transform:

Implementation for std::transform:

// std::transform
template <class _InputIterator, class _OutputIterator, class _UnaryOperation>
inline _LIBCPP_INLINE_VISIBILITY
_OutputIterator
transform(_InputIterator __first, _InputIterator __last, _OutputIterator __result, _UnaryOperation __op)
{
    for (; __first != __last; ++__first, (void) ++__result)
        *__result = __op(*__first);
    return __result;
}


可以重载 operator,。将任何一个操作数强制转换为 void 可以防止调用任何重载运算符,因为重载运算符不能采用 void 参数。

It is possible to overload operator,. Casting either operand to void prevents any overloaded operator from being called, since overloaded operators cannot take void parameters.