printing String output is empty

Question

I'm new to programming , when I cout << str outside the loop , it prints nothing and when cout << str[0] << str[1] it prints the char in this position and also length of str outside of the loop is 0, what is the reason for that behavior.

Note: I tried to use concatenate str+= s[i]; and it's working in this case , I can cout << str;

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

int main()
{
    int n,k;
    cin >> n >> k;
    int z = k;
    string s = "abcdefghijklmnopqrstuvwxyz";
    string str;
    int repeated = n-k;
    for(int i{};i<n;++i){
        if(k > 0){      //distinct
        str[i] = s[i];
            k--;
            //cout << str[i];
        }else if(k==0 && repeated >0){    //repeated
            str[i] = str[i-z];
            repeated--;
            //cout << str[i];
        }
    }
    cout << str;
        //cout << str[0] << str[1] << str[2] << str[3] << str[4] ;
    return 0;
}

463035818_is_not_an_ai · Accepted Answer · 2021-02-17 12:49:07Z

2

string str; is an empty string with size 0 and nothing in your code changes that.

Accessing any element but str[0] invokes undefined behavior. There is no str[i] for i > 0.

To append to a string you can use + or std::string::append.

answered Feb 17, 2021 at 12:49

463035818_is_not_an_ai

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2 Comments

logan_92 Over a year ago

thanks for the answer , this means accessing any element in str doesn't change it's size and i have to append.

463035818_is_not_an_ai Over a year ago

@logan_92 exactly. std::map and std::unordered_maps operator[] is a weird exception, but other than that I don't know any standard container that implicity grows its size on element access, you either have to resize or push_back or append or +

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