size of string

[John]

A text book says "a string's size is independent of the length of the
string it addresses". Why?
For example,
string st1( "foobar" );
string st2( "a mighty oak" );
cout << "st1: " << sizeof( st1 )
<< " st2: " << sizeof( st2 );

The output is:
st1: 12 st2: 12

Why? The length of any string is 12?

Thanks.

Johm

[Siemel Naran]

"John" <jo*********@yahoo.com> wrote in message

A text book says "a string's size is independent of the length of the
string it addresses". Why?
For example,
string st1( "foobar" );
string st2( "a mighty oak" );
cout << "st1: " << sizeof( st1 )
<< " st2: " << sizeof( st2 );

The output is:
st1: 12 st2: 12

Why? The length of any string is 12?

On your implementation, yes. The sizeof(any class) is the sum of the sizeof
of each of its data members. A non reference counted string normally has 3
data members: (1) a pointer to the data usually of type char* and
sizeof(char*) is 4 on your implementation, (2) an integer holding the length
of the string, (3) an integer holding the length of the data buffer.

[Matthias Käppler]

John wrote:

A text book says "a string's size is independent of the length of the
string it addresses". Why?
For example,
string st1( "foobar" );
string st2( "a mighty oak" );
cout << "st1: " << sizeof( st1 )
<< " st2: " << sizeof( st2 );

The output is:
st1: 12 st2: 12

Why? The length of any string is 12?

Thanks.

Johm

Since your string object only holds a pointer to the actual string data, and
since pointers are usually 4 bytes long on 32-Bit architectures, the size
of your two string objects is the same.

Think of an std::string to be a more convenient way to handle character
arrays, instead of an actual string. Maybe that helps.

- Matthias

[Rolf Magnus]

Siemel Naran wrote:

"John" <jo*********@yahoo.com> wrote in message

A text book says "a string's size is independent of the length of the
string it addresses". Why?
For example,
string st1( "foobar" );
string st2( "a mighty oak" );
cout << "st1: " << sizeof( st1 )
<< " st2: " << sizeof( st2 );

The output is:
st1: 12 st2: 12

Why? The length of any string is 12?

sizeof gives you the size of the class, which is always a compile-time
constant. This is not limited to strings, but rather is true for every data
type. What the book refers to as "size" is not the number of characters
stored (which you would get with the size() member function), but the value
of sizeof.
On your implementation, yes. The sizeof(any class) is the sum of the
sizeof of each of its data members.

Not really. There can be alignment, making the class bigger than its data
member sizes added together. Try it with something like:

class X
{
public:
char a;
int b;
};

On many compilers, sizeof(X) will be two times sizeof(int), even though a
char is usually smaller than an int.
Also, classes that contain virtual functions ususally get the size of one
pointer added.

[Ron Natalie]

Siemel Naran wrote:

On your implementation, yes. The sizeof(any class) is the sum of the sizeof
of each of its data members.

It might be bigger than that. However, sizeof (some_class) is always
a constant that most likely is the sizeof all the non-static data members
plus possibly a small bit of overhead.

[John]

Rolf Magnus <ra******@t-online.de> wrote in message news:<cn*************@news.t-online.com>...

Siemel Naran wrote:

"John" <jo*********@yahoo.com> wrote in message

A text book says "a string's size is independent of the length of the
string it addresses". Why?
For example,
string st1( "foobar" );
string st2( "a mighty oak" );
cout << "st1: " << sizeof( st1 )
<< " st2: " << sizeof( st2 );

The output is:
st1: 12 st2: 12

Why? The length of any string is 12?

sizeof gives you the size of the class, which is always a compile-time
constant. This is not limited to strings, but rather is true for every data
type. What the book refers to as "size" is not the number of characters
stored (which you would get with the size() member function), but the value
of sizeof.

On your implementation, yes. The sizeof(any class) is the sum of the
sizeof of each of its data members.

Not really. There can be alignment, making the class bigger than its data
member sizes added together. Try it with something like:

class X
{
public:
char a;
int b;
};

On many compilers, sizeof(X) will be two times sizeof(int), even though a
char is usually smaller than an int.
Also, classes that contain virtual functions ususally get the size of one
pointer added.

Thanks.

John