Counting the chars?

[Jyy]

Hello,

I tried to make a code that turns the letters of a word, in a compact way (so not with extra functions). Dere is my code:

#include <iostream>
#include <string>
#include <vector>

using namespace std;

main() {
char hoi[500];
int i; // i becomes the size of vector kip


cout << "Hello!" << endl << "type something: ";
cin.getline(hoi,500);

vector<char> kip (hoi, hoi + sizeof(hoi) / sizeof(int) ); // standard coding


  for (i=kip.size(); i >= 0; i--) // as long as i isn't less than 0, the vector is printed. 
    cout << kip[i];
cin.get();
return 0;
}

and what I get is this:

C:\CPPs\hoi>hoi
Hello!
type something: damn it
¶ ☻ å*@ ↕²t Aí@ ↕²╚Zz■░ ↕²< AíÇ @╪☼ ↕²╠ ↕²¿ → @;H z Z ¶N≡ A{¼ ↕■░ A¥( å1É ☻ ↕²á
@╖ⁿ ti nmad

So what happens is that empty chars are getting stored in the vector, and the vector prints them as strange chars.

Can someone help me?

Jyy.

[Gumbercules]

You can use strlen from <cstring> to give you the length up to (but not including) the null, which marks the end. andm you shouldn't be dividing by sizeof(int), hoi + strlen(hoi) will point to the last letter.

[Jyy]

Thanks! for the one's who want to know, this is my code now:

#include <iostream>
#include <string>
#include <vector>

using namespace std;

main() {
char hoi[500];
int i; // i becomes the size of vector kip


cout << "Hello!" << endl << "type something: ";
cin.getline(hoi,500);

vector<char> kip (hoi, hoi + strlen(hoi)); // standard coding


  for (i=kip.size()-1; i >= 0; i--) // as long as i isn't less than 0, the vector is printed. 
    cout << kip[i];
cin.get();
return 0;
}

[PanGalactic]

Why use "char hoi[500];" rather than "std::string hoi;"?