why string in array just take 8 bytes?

Asked May 07 '19 at 02:50

Active May 07 '19 at 02:50

Viewed 51 times

one array which saves some strings. when I check everyone address of string in array, i found that one string just take 8 bytes, look like array just save point to string but not string object. I want to known the reason.

my code as shown below:

#include <string>
#include <iostream>

using namespace std;

int main()
{
    string nums[] = {"oneoneoneoneoneoneone",
                     "twotwotwotwotwotwotwotwotwo",
                     "threethreethreethreethreethreethreethreethreethreethree"};

    for (int i = 0; i < 3; ++i) {
        cout << &nums[i] << endl;
    }
}

the result of this program show:

0x7ffe432ba6f0 0x7ffe432ba6f8 0x7ffe432ba700

address of the second item is bigger than first item 8 bytes.

asked May 07 '19 at 02:50

Heng Zhang

The string object is a pointer to a string. – user253751 May 07 '19 at 02:55
3

My psychic powers suggest you really want `nums[i].size()` – selbie May 07 '19 at 02:59
My guess is that you're on a 32-bit system where size of `size_t` and pointers are 32-bits (four bytes) each. – Some programmer dude May 07 '19 at 03:11

why string in array just take 8 bytes?

0 Answers0