C++ 打印数组元素内存地址C和C++，为什么输出不同？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/6239833/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Printing array element memory adresses C and C++, why different output?
提问by W3ctor
Possible Duplicate:
How to simulate printf's %p format when using std::cout?
I try to print out the array element memory addresses in C and C++.
我尝试打印出 C 和 C++ 中的数组元素内存地址。
In C:
在 C：
char array[10];
int i;
for(i =0; i<10;i++){
printf(" %p \n", &array[i]);
}
I got the memory addresses: 0xbfbe3312, 0xbfbe3313, 0xbfbe3314, ....
我得到了内存地址：0xbfbe3312, 0xbfbe3313, 0xbfbe3314, ...。
But if I try to make the same in C++:
但是如果我尝试在 C++ 中做同样的事情：
char array[10];
for(int i =0; i<10;i++){
std::cout<<&array[i]<<std::endl;
}
I got this output:
我得到了这个输出：
?
P??
??
?k?
?
?
??
??
?克？
?
Why is it different? Should I use the coutdifferently in C++ to print out the memory addresses? How should I print out the memory addresses?
为什么不一样？我应该使用coutC++ 中的不同来打印出内存地址吗？我应该如何打印出内存地址？
回答by Xeo
Cast the address to void*before printing, in C++ the operator<<of ostreamis overloaded for (const) char*so that it thinks it's a c-style string:
void*在打印之前将地址强制转换为，在 C++ 中，operator<<ofostream被重载，(const) char*以便它认为它是一个 c 样式的字符串：
char array[10];
for(int i =0; i<10;i++){
std::cout << static_cast<void*>(&array[i]) << std::endl;
}
Also see this answer of mine.
另请参阅我的这个答案。
回答by Neil
The type of &array[i]is char*, and so cout<<thinks that you want to print a string.
的类型&array[i]是char*，因此cout<<认为您要打印一个字符串。
回答by RC Howe
std::coutdoes not necessarily treat pointers as pointers. A char*pointer could be a string, for instance. Taking the address of an element in a chararray basically outputs the substring from that point.
std::cout不一定将指针视为指针。例如，char*指针可以是 a string。获取char数组中元素的地址基本上是从该点输出子字符串。
回答by ?ukasz Milewski
You must cast &array[i]to void*
你必须投射&array[i]到void*
for(int i =0; i<10;i++){
std::cout<<(void*)&array[i]<<std::endl;
}
This is because C++ streams work differently for different types. For example when you pass char*to it, your data is treated as a C-string - thus it is printed as a list of characters.
这是因为 C++ 流对不同类型的工作方式不同。例如，当你传递char*给它时，你的数据被视为一个 C 字符串 - 因此它被打印为一个字符列表。
You must explicitly tell C++ that you want to print an address by casting.
您必须明确告诉 C++ 您要通过强制转换来打印地址。
By the way (void*)is not the best way to do that as you should avoid C-like casting. Always use C++-style casting (static_cast, dynamic_cast, reinterpret_cast). In this case static_castwould do the job.
顺便说一句，(void*)这不是最好的方法，因为您应该避免类似 C 的强制转换。始终使用 C++ 风格的强制转换 ( static_cast, dynamic_cast, reinterpret_cast)。在这种情况下static_cast将完成这项工作。
