Why do i have an out of range exception working with c++ strings

Question

The code gets an exception of type out of range and i don't know why. It seems to work when i debug it, teh string is converted to what i want it to be. First time on stack overflow btw:)

#include <iostream>

using namespace std;

string s;
string alpha = "abcdefghijklmnopqrstuvwxyz";
string crypto(string& s);

int main()
{
    cin >> s;

    cout << crypto(s);

    return 0;
}

string crypto(string& s)
{

    size_t i = 0;
    while (i < s.length()) {
        for (size_t j = 0; j < alpha.length(); j++) {
            if (s.at(i) == alpha.at(j)) {
                s.at(i) = alpha.at(alpha.length() - 1 - j);
                ++i;
            }
        }
    }

    return s;
}

`++i` is not put at the good place. At next `j` iteration, you might have an out of bound access — Damien, Jan 28 '21 at 17:19
I don't understand. You are modifying the parameter *and* returning a copy of the modified parameter. Usually functions do one or the other, not both. — Thomas Matthews, Jan 28 '21 at 19:32

score 0 · Answer 1 · edited Jan 28 '21 at 17:28

0

Think about the case: if s.length() < alpha.length().

edited Jan 28 '21 at 17:28

D-RAJ

3,263
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answered Jan 28 '21 at 17:23

Nitai Chandra Banik

102
9

Blastfurnace · Answer 2 · 2021-01-28T17:38:50.800

The inner loop of the crypto function can increment i past the end of the string. There's no test to stop it.

You could avoid this by breaking out of the loop when a letter is changed (that's more correct and potentially faster). That then means you should increment i outside the inner loop which means the outer loop can be a for loop as well.

string crypto(string &s) {
  for (size_t i = 0; i < s.length(); ++i) {
    for (size_t j = 0; j < alpha.length(); ++j) {
      if (s.at(i) == alpha.at(j)) {
        s.at(i) = alpha.at(alpha.length() - 1 - j);
        break;
      }
    }
  }
  return s;
}

score 0 · Answer 3 · answered Jan 28 '21 at 17:33

The problem was the i++ at the wrong place. You should format your code better, then this is much easier to spot.

Also avoid non-constant global variables and using namespace std:

#include <iostream>
#include <string> //Include the headers you use

const std::string alpha="abcdefghijklmnopqrstuvwxyz";
void crypto(std::string &s);

int main(){

    std::string s;
    crypto(s);
    std::cout<<s;
    return 0;
}

void crypto(std::string &s) //If you take the string by reference, it is changed, so you do not have to return it
{   
    for(std::size_t i = 0; i < s.length(); ++i) { //The for loop avoids the mistake completely
      for(size_t j=0; j < alpha.length(); ++j){
        if(s.at(i)==alpha.at(j)){
          s.at(i)=alpha.at(alpha.length()-1-j);
        }   
      }
    }
}

To not solve your problem completely, there is still a bug in the code, that was also in yours. Try to find the error yourself.

score 0 · Answer 4 · answered Jan 28 '21 at 17:37

The i++ was not put at the right place.

Moreover, there is another issue in the code: once an element has been replaced, you must leave the inner loop immediately (break). If not, you can have a -> z -> a in the same loop.

Input

 abcyz

Output

zyxba
abcyz

#include <iostream>
#include <string>

std::string crypto(std::string& s) {
    const std::string alpha = "abcdefghijklmnopqrstuvwxyz";
    for (size_t i = 0; i < s.length(); ++i) {
        for (size_t j = 0; j < alpha.length(); j++) {
            if (s.at(i) == alpha.at(j)) {
                s.at(i) = alpha.at(alpha.length() - 1 - j);
                break;
            }
        }
    }
    return s;
}

int main() {
    std::string s;
    std::cin >> s;
    std::cout << crypto(s) << std::endl;
    std::cout << crypto(s) << std::endl;
    return 0;
}

Why do i have an out of range exception working with c++ strings

4 Answers4