CODEWITHSUNDEEP
c++string
motiur
motiur

Reputation: 1680

Size of string data type in C++ array

This is a very simple code in C++. The address of the strings are separated by a constant gap of 28 bytes. What does these 28 bytes contains. I am trying to find an analogy with the gap of 4 bytes of an array containing integers. As far as I know the 4 bytes ensures the upper limit of the value of an integer that can be reached. What happens in case of the 28 bytes. Does it really contain 28*8 bits of character data - I do not believe that. I have tried giving in a large text of data, and it still prints without any issues.

    string str[3] = { "a", "b", "c" };
    for (int i = 0; i < 3; ++i) {
        cout << &str[i] << endl;
    }

Upvotes: 0

Views: 106

Answers (1)

eerorika
eerorika

Reputation: 238301

What does these 28 bytes contains.

It contains the object of type string. We don't know any more unless we know how you have defined that type.

If string is an alias of std::string, then it is a class defined by the standard library. The exact contents and thus the exact size depend on and vary between standard library implementations, and the target architecture.

If we consider what some implementation might do in practice:

Does it really contain 28*8 bits of character data - I do not believe that.

Believe it or not, (modern) string implementations really do contain ~ sizeof(string) (sans potential overhead) bytes of character data when those characters fit on that space.

They use advanced tricks to change the internal layout to support longer strings. For those, they use pointers. Typically, there would be a pointer to beginning, pointer to end of string (storing offset is another option) and pointer (or offset) to the end of dynamic storage. This representation is essentially identical to a vector.

If you read the standard library headers that you use, you'll find the exact definition of the class there.

Upvotes: 1

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