char指针与c ++中的cout混淆 [英] char pointer confusing with cout in c++

问题描述

当我用 printf（）打印一个字符指针时，它使用转换说明符来决定是打印地址还是根据％u或％ s。

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， / code>决定什么应该打印在地址和整个字符串？这里是一个示例源：

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 尝试输出ch作为字符串。

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

我如何获得 ch 的地址 cptr 使用 cout ？

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

推荐答案

过载解析选择 ostream&用于打印C风格字符串的操作符<<（ostream& o，const char * c）; 你想要另一个 ostream&运算符<<（ostream& o，const void * p）; 。你可能最喜欢这里的演员：

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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