格式化输出运算插入器 [英] Formatted output arithmetic inserters

问题描述

我有一个关于算术插入器的基本问题; §27.7.3.6.2 / 1 [ostream.inserters.arithmetic]：

I have a basic question about the arithmetic inserters; § 27.7.3.6.2/1 [ostream.inserters.arithmetic]:

当val类型为bool，long，unsigned long， long long，unsigned long long，double，long double或const void *，格式化转换的执行方式如同执行以下代码片段：

When val is of type bool, long, unsigned long, long long, unsigned long long, double, long double, or const void*, the formatting conversion occurs as if it performed the following code fragment:

bool failed = use_facet<
  num_put<charT,ostreambuf_iterator<charT,traits> >
    > (getloc()).put(*this, *this, fill(), val).failed()

问题是什么确切函数执行从指针到类型的转换，正如Matt McNabb更正， const void * ？例如：

The question is what exact function performs the conversion from a pointer to type to, as Matt McNabb corrected, const void*? For instance:

int *ip = new int(1);
std::cout << ip; //0xaa33fa67

我不关心实现细节，我只是想知道什么函数从指针产生算术结果。在上面的例子中是 put 吗？

推荐答案

从任何非指针到成员/成员函数到 void * 。在这被传递到流之后，它将它传递给 std :: num_put :: put（），将其打印为通用指针，就像使用％p格式标志。

There's an implicit coversion from any non-pointer to member/member function to void*. After this is passed to the stream, it passes it off to std::num_put::put() which prints it out as a generic pointer as if by using the "%p" format flag.

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