Why there is so much gap in address of two vector elements?

Question

I'm learning about vectors in c++. On printing address of two adjacent vector<string> elements I'm getting result which is confusing me a little bit.
Program

#include<iostream>
#include<vector>
using namespace std;

int main()
 {  
   vector<string> word;

   //Enter any two words
     string ch;
     for(int i=1;i<3;i++)
      {
       getline(cin,ch);
       word.push_back(ch);
       }

   //Display the words
     for(auto i:word)
      {
       cout<<i<<" ";
      }  
     cout<<endl;

    //Display address of first two elements

      cout<<word.data()<<" "<<(word.data()+1)<<endl;

     return 0;
  }

INPUT Data: hello
world
OUTPUT : hello world
0x1b9f060 0x1b9f080

What I'm not getting is that there are only 10 character in total(5 in first input and another five in second)and each character take 1 byte space.
So why there is gap of 20 bytes between both vector elements?
I checked with vector<int> but the gap was only of four bytes.

Those are 32 bytes, the addresses are printed with base 16 (hexadecimal). — Ulrich Eckhardt, Oct 14 '18 at 12:03
what I'm trying to say is if it is 32 byte then difference should be more than 20 between both the addresses. &Ulrich — Yogesh Tripathi, Oct 14 '18 at 12:07
Compare `0x20` (the difference between the two addresses printed) with `sizeof(std::string)`. There will be a distinct relationship. — Peter, Oct 14 '18 at 12:37
Possible duplicate of [Why is sizeof(std::string) only eight bytes?](https://stackoverflow.com/questions/34560502/why-is-sizeofstdstring-only-eight-bytes) — Alan Birtles, Oct 14 '18 at 13:24

score 5 · Answer 1 · edited Oct 14 '18 at 13:05

5

First of all, the actual "gap" is 32 bytes since the addresses are hexadecimal values 0x1b9f080 - 0x1b9f060 = 0x20 = 32.

This is the size of the element of the vector which in this case is std::string. Most importantly std::string is a class and its size is roughly determined by the sizes of its members just like any other class.

What are the actual members of std::string and what size does it have as the result is dependent on the implementation of the standard library you are currently using. The most important thing, however, is that the size of std::string doesn't depend at all on the length of the actual string it's storing because it only holds a pointer to it and the string itself is located elsewhere (on the heap).

edited Oct 14 '18 at 13:05

Jesper Juhl

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answered Oct 14 '18 at 12:08

r3mus n0x

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Is this also a big difference between C++ `std::string and C string ? – Yogesh Tripathi Oct 14 '18 at 12:14
2

@user4156958, yes, but you can't really adequately compare those because they are different entities altogether. In C a string is just an array of `char`s, while `std::string` is a class that hides this array from you and manages it itself. – r3mus n0x Oct 14 '18 at 12:16
`std::string` is C++ specific. It is not part of C at all. – Peter Oct 14 '18 at 12:38

score 1 · Answer 2 · answered Oct 14 '18 at 12:01

1

A string is not an array of characters, it's a pointer to an allocated space (of given size) that contains what you think is the string. There is usually also additional information like the actual size of the string (because C++ strings can have a \0 character in them), the allocated space perhaps...

There are exceptions, like Small Array Optimizations, but let's say it's not your actual string for the sake of clarity.

The int in your vector<int> has a size of 4, that's what we expect indeed.

answered Oct 14 '18 at 12:01

Matthieu Brucher

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So string in C and c++ are totally different? @matthieu Brucher – Yogesh Tripathi Oct 14 '18 at 12:05
3

Yes. How would you store a variable length string otherwise in a vector? – Matthieu Brucher Oct 14 '18 at 12:21

Why there is so much gap in address of two vector elements?

2 Answers2