C++ 为什么 cout 打印 char 数组与其他数组不同？
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Why does cout print char arrays differently from other arrays?
提问by Luca Matteis
I'm using C++ to understand how exactly pointers work. I have this piece of code using arrays, which I'm using just to understand how the equivalent works with pointers.
我正在使用 C++ 来理解指针的工作原理。我有一段使用数组的代码，我使用它只是为了了解等效项如何与指针一起使用。
int main() {
int arr[10] = {1,2,3};
char arr2[10] = {'c','i','a','o','cout << static_cast<const void*>(arr2) << endl;
'};
cout << arr << endl;
cout << arr2 << endl;
}
However when I run this, arroutputs the address of the first element of the array of ints (as expected) but arr2doesn't output the address of the first element of the array of chars; it actually prints "ciao".
但是，当我运行它时，arr输出整数数组的第一个元素的地址（如预期），但arr2不输出字符数组的第一个元素的地址；它实际上打印“ciao”。
What is it that I'm missing or that I haven't learned yet about this?
我遗漏了什么或我还没有学到什么？
回答by Johannes Schaub - litb
It's the operator<< that is overloaded for const void*and for const char*. Your char array is converted to const char*and passed to that overload, because it fits better than to const void*. The int array, however, is converted to const void*and passed to that version. The version of operator<< taking const void*just outputs the address. The version taking the const char*actually treats it like a C-string and outputs every character until the terminating null character. If you don't want that, convert your char array to const void*explicitly when passing it to operator<<:
重载了 forconst void*和 for的是运算符<< const char*。您的 char 数组被转换const char*并传递给该重载，因为它比 to 更适合const void*。然而，int 数组被转换const void*并传递给那个版本。operator<< 的版本const void*只输出地址。采用 的版本const char*实际上将其视为 C 字符串并输出每个字符，直到终止的空字符。如果您不希望那样，请在将字符数组const void*传递给 operator<< 时将其显式转换为：
cout << &arr2 << endl;
回答by crashmstr
Because cout's operator <<is overloaded for char*to output strings, and arr2matches that.
因为 cout'soperator <<被重载char*以输出字符串，并arr2匹配它。
If you want the address, try casting the character array as a void pointer.
如果需要地址，请尝试将字符数组转换为空指针。
回答by Darron
There is a standard overload for char* that outputs a NUL terminated string.
char* 有一个标准重载，它输出一个 NUL 终止的字符串。
回答by Mystic
While casting is probably a more meaningful approach, you could also use the addressof operator:
虽然强制转换可能是一种更有意义的方法，但您也可以使用 addressof 运算符：
##代码##