C++ 显示字符串的地址
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Displaying the address of a string
提问by Greko2009
I have this code:
我有这个代码：
char* hello = "Hello World";
std::cout << "Pointer value = " << hello << std::endl;
std::cout << "Pointer address = " << &hello << std::endl;
And here is the result:
结果如下：
Pointer value = Hello World
Pointer address = 0012FF74
When I debug to my program using OllyDbg, I see that the value of 0x0012FF74 is e.g. 0x00412374.
当我使用 OllyDbg 调试我的程序时，我看到 0x0012FF74 的值是例如 0x00412374。
Is there any way I can print the actual address that hellopoints to?
有什么办法可以打印hello指向的实际地址吗？
回答by
If you use &helloit prints the address of the pointer, not the address of the string. Cast the pointer to a void*to use the correct overload of operator<<.
如果你使用&hello它打印指针的地址，而不是字符串的地址。将指针强制转换为 avoid*以使用 的正确重载operator<<。
std::cout << "String address = " << static_cast<void*>(hello) << std::endl;
回答by Michel Keijzers
I don't have a compiler but probably the following works:
我没有编译器，但可能以下工作：
std::cout << "Pointer address = " << (void*) hello << std::endl;
Reason: using only hello would treat is as a string (char array), by casting it to a void pointer it will be shown as hex address.
原因：仅使用 hello 会将 is 视为字符串（字符数组），通过将其转换为 void 指针，它将显示为十六进制地址。
回答by triclosan
or so:
或者：
std::cout << "Pointer address = " << &hello[0] << std::endl;
回答by Rich Walter
This also works:
这也有效：
std::cout << "Pointer address = " << (int *)hello << std::endl;
