CODEWITHSUNDEEP
c++string
herohuyongtao
herohuyongtao

Reputation: 50667

Different memory address for &string and &string[0]?

From Strings in Depth, I learn that the exact implementation of memory layout for the string class is not defined by the C++ Standard. But how come &string and &string[0] will have different memory addresses?

string variable("String Test");
cout << (void *) &variable << endl;         // 003EFE58
cout << (void *) &variable[0] << endl;      // 003EFE5C
cout << (void *) &variable[1] << endl;      // 003EFE5D
... ...

But they do share the same content, check out:

cout << ((string)*(&variable))[0] << endl;  // 'S'
cout << *(&variable[0]) << endl;            // 'S'

Can anyone explain what's going on here?

Upvotes: 1

Views: 980

Answers (3)

zackery.fix
zackery.fix

Reputation: 1816

This happens to confuse a lot of programmers, so I will do my best to explain it.

// here, you take the address (&) of the first item in the variable array...
//       you then convert that address to a string object
//       an implicit cast occure, which constructs a string instance and copies the contents of 
//       the referenced item in variable (which is a dereferenced char* char('S'))
//       this line is similar to:
//         char c = *(&variable[0]); // dereference
//         string S = string(c);     // copy
//         printf("%c\n", S);        // print
printf("%c\n", ((string)*(&variable))[0]);  // 'S'

// here, you do the same dereference of the first item in variable,
//       except you don't copy it into a string object
//       similar to
//         char c = *(&variable[0]);
//         printf("%c\n", c);
printf("%c\n", *(&variable[0]));            // 'S'

Upvotes: 0

Kirill Kobelev
Kirill Kobelev

Reputation: 10557

String is an object. It has some fields in its beginning. This is implementation specific. In your case there is just one 4 bytes field. The sting that it contains starts with an offset. In fact, the sting can be in a heap buffer. It may not be in the object itself.

Upvotes: 2

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272477

Because one is the address of the class instance, the other is the address of the first character, which is typically contained in dynamically-allocated memory managed by the class (or somewhere in the instance itself in the case of SSO).

Upvotes: 6

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