在字符串中[0]和[0]有什么区别

string a = "asdf";
cout<<&a[0];
cout<<a[0];

Why are these two outputs different? Why is &a[0] not the address but the whole string?

为什么这两个输出不同?为什么&a[0]不是地址而是整个字符串?

3 个解决方案

#1

&a[0] has type char *. Stream operator << is deliberately overloaded for const char * arguments to output zero-terminated string (C-style string) that begins at that address. E.g. if you do

bb&a b0有char类型*。流操作符< <被故意重载，以便const char *参数输出从该地址开始的零终止字符串(c风格的字符串)。例如,如果你< p>

const char *p = "Hello World!";
cout << p;

it is that overloaded version of << that makes sure the "Hello World!" string itself is sent to output, not the pointer value.

是重载的<< < <确保“hello world!”字符串本身被发送到输出，而不是指针值。< p>

And this is exactly what makes your code to output the entire string as well. Since C++11 std::string objects are required to store their data as zero-terminated strings and &a[0] is nothing else than a pointer to the beginning of the string stored inside your a object.

这就是代码输出整个字符串的原因。由于C+ 11 std::string对象被要求将其数据存储为零终止字符串，而[0]只不过是存储在&a对象中的字符串开头的指针。

#2

&a[0] yields type char*. This is a type for which operator<<() is overloaded. This particular overload prints the characters starting at the address until it finds a null-character, '\0'. It won't print the address like you'd expect.

和[0]收益率char *类型。这是一个操作符<<()被重载的类型。这个特殊的重载打印从地址开始的字符，直到找到一个空字符'\0'。它不会像你预期的那样打印地址。

Since you need the address, there's std::addressof() in the standard library:

由于需要地址，标准库中有std::addressof():

std::cout << std::addressof(a[0]);

you can also cast to void* which is almost like the above variant:

你也可以施放到void*，这几乎和上面的变体一样:

std::cout << static_cast<void*>(&a[0]);

#3

When printing a pointer to a standard library output stream, if it's char* or const char*, the null-terminated string pointed to will be printed, rather than the address itself. If you want to have the address printed:

当打印指向标准库输出流的指针时，如果是char*或const char*，则将打印指向的以null结尾的字符串，而不是地址本身。如要印上地址:

cout << static_cast<const void*>(&a[0]);

(Trivia: if the pointer type isn't convertible to const void* either---because it is a function pointer or a pointer to member or it is volatile---then it is converted to bool.)

(琐事:如果指针类型也不能转换为const void*——因为它是一个函数指针或成员指针，或者它是易变的——那么它就被转换为bool。)

#1