为什么将 char 指针流式传输到 cout 不打印地址?

Why does streaming a char pointer to cout not print an address?(为什么将 char 指针流式传输到 cout 不打印地址?)

问题描述

当我用 printf() 打印一个 char 指针时，它会根据 %u 或 %s 使用转换说明符来决定是打印地址还是打印整个字符串.

When I print a char pointer with printf(), it makes the decision with conversion specifier whether the address should be printed or the whole string according to %u or %s.

但是当我想用 cout 做同样的事情时，cout 将如何决定应该在地址和整个字符串之间打印什么?这是一个示例来源:

But when I want to do the same thing with cout, how will cout decide what should be printed among address and whole string? Here is an example source:

int main()
{
  char ch='a';
  char *cptr=&ch;
  cout<<cptr<<endl;
  return 0;
}

这里，在我的 GNU 编译器中，cout 试图将 ch 输出为字符串.

Here, in my GNU compiler, cout is trying to output ch as a string.

如何使用 cout 通过 cptr 获取 ch 的地址?

How I can get address of ch via cptr using cout?

推荐答案

重载分辨率选择ostream&operator<<(ostream& o, const char *c); 用于打印 C 风格的字符串.你想要另一个 ostream&operator<<(ostream&o, const void *p); 被选中.您可能最好在此处使用演员表:

Overload resolution selects the ostream& operator<<(ostream& o, const char *c); which is used for printing C-style strings. You want the other ostream& operator<<(ostream& o, const void *p); to be selected. You are probably best off with a cast here:

 cout << static_cast<void *>(cptr) << endl;

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