cout <用char *参数打印字符串，而不是指针值

问题描述

这：

  const char * terry =hello; 
 cout<< terry; 
  

打印 hello ，而不是内存地址'h'。为什么会发生这种情况？

解决方案

原因是 std :: cout 会将 char * 作为指向C风格字符串（的第一个字符）的指针，并将其打印。如果您想改用地址，则只需将其转换为不是的指针，例如：

  cout<< （void *）terry; 
  

（或使用 const void *

---

如果你在这个特殊情况下，更多的是纯粹主义者而不是实用主义者，你也可以使用C ++ static_cast ：

  cout<< static_cast< const void *> （terry）; 
 

虽然在这种情况下没有必要，但是转换为 void * 将工作正常。以下示例代码显示了操作中的所有这些选项：

  #include< iostream& 
 int main（void）{
 const char * terry =hello; 
 std :: cout<< terry<< '\\\
'; 
 std :: cout<< （void *）terry<< '\\\
'; 
 std :: cout<< （const void *）terry<< '\\\
'; 
 std :: cout<< static_cast< const void *> （毛圈） '\\\
'; 
 return 0; 
} 
  



输出（地址在您的环境中可能不同）：

  hello 
 0x8048870 
 0x8048870 
 0x8048870 
   static_cast 时，你应该确保你不要试图丢弃constness与 static_cast< void *> （这是 const_cast 用于）。这是较新的C ++类型转换所做的检查之一，而旧样式的转换没有这个限制。
 
This:
const char * terry = "hello";
cout<<terry;
prints hello instead of the memory address of the 'h'. Why is this happening?
 解决方案 
The reason for that is that std::cout will treat a char * as a pointer to (the first character of) a C-style string and print it as such. If you want the address instead, you can just cast it to a pointer that isn't treated that way, something like:
cout << (void *) terry;
(or use the const void * cast if you're worried about casting away constness, something that's not an issue in this particular case).
If you're more of a purist than pragmatist, you can also use the C++ static_cast, along the lines of:
cout << static_cast <const void *> (terry);
though it's unnecessary in this particular case, the cast to a void * will work fine. The following sample code shows all these options in action:
#include <iostream>
int main (void) {
    const char *terry = "hello";
    std::cout << terry << '\n';
    std::cout << (void *) terry << '\n';
    std::cout << (const void *) terry << '\n';
    std::cout << static_cast<const void *> (terry) << '\n';
    return 0;
}
outputting (the address may be different in your environment):
hello
0x8048870
0x8048870
0x8048870
Note that, when using the static_cast, you should ensure you don't try to cast away the constness with static_cast <void *> (that's what const_cast is for). This is one of the checks done by the newer C++ casts and the old-style cast does not have this limitation.
                        
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